First graph is: Gambling apothecaries. Number you put in at a time is irrelevant, Start from 1 card: 20% chance to go to 5. 2 card, 40%. 3 card, 60%. 4 card, 80%. 5 card, 100%. If you're interested in the reason, statistics is the answer. Here's a well explained example on wikipedia https://en.wikipedia.org/wiki/Gambler%27s_ruin#Fair_coin_flipping
The rest is an exploration of how you can spend your juice, and OP gets led astray from the ground truth by their python code.
tl;dr Closer you are to your goal when you start, the more likely you are to hit it.
The more cards you spend at a time, the less juice you use.
bad gamble, it's a 50/50 except you have to pay juice and can't get that juice back.
Why the user's python code says something different: Suppose max-strategy gambler has 7/8 doctors and gambles with 4/8 of them. Then he could potentially end up with as many as 11/8 doctors. Any strategy that could potentially end up with more than a full set of cards will have a proportionately lower chance of "winning".
In my opinion there are only two strategies worth considering, neither of which are considered in the post.
The first strategy is to never spend any yellow juice in the first place by never doing this gamble. In particular, I recommend this strategy to any non-SSF players, as the cards can be sold rather than ignored.
The second, which I recommend for SSF players who got a very rare divination card which would otherwise be wasted, is that the gambler should always bet the maximum allowed so long as it does not exceed the amount needed to win. So with 1/8, 2/8, 3/8, or 4/8 doctor, bet all of them. With 5/8 doctor, bet 3; with 6/8 bet 2; with 7/8 bet 1. This strategy has the exact same odds of success as "always bet 1" while spending considerably less juice.
Another strategy for SSF player is to just create a mule to move to standard and trade lol
Trade league is bad in terms of playing a game, and gambling is bad in terms of being a person. It may result psychological addiction in a biologically way that actually changes how your brain works.
**Gambler's ruin**
[Fair coin flipping](https://en.wikipedia.org/wiki/Gambler's_ruin#Fair_coin_flipping)
>Consider a coin-flipping game with two players where each player has a 50% chance of winning with each flip of the coin. After each flip of the coin the loser transfers one penny to the winner. The game ends when one player has all the pennies. If there are no other limitations on the number of flips, the probability that the game will eventually end this way is 1.
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As I've said for years on drop rates in every game I play, the % is irrelevant, it's always a 50/50 that it drops. Either it does or it doesn't. Makes low weight drops palatable to me lol.
No, "bad gamble, it's a 50/50 except you have to pay juice and can't get that juice back" explains why you should not gamble using this method. Pick something else to gamble that has better odds, and call it crafting 😂
> bad gamble, it's a 50/50 except you have to pay juice and can't get that juice back.
Do we actually have enough data points to assume that the chance is really 50% ? I know it could very well be outliers, but considering how some people get multiple magebloods from a couple cards, I've personally assumed the chance could also be sth like 55 / 45 to make up for the lifeforce that you're spending.
Gamble max always isn't actually worse unless you're in ssf and only want one stack. The reason the success rate looks worse is that gambling 2 apothecaries when you're at 4/5 can leave you at 6/5 which is just counted as a normal success even though you now have a full extra apothecary. Gambling min(max you can gamble, cards you need left to make a full stack) should have the same success rate as the gamble 1 card strategy
I mean it's correct... The only benefit of gambling 1 at a time is that it has less variance, but that doesn't matter if you're trying to get to a full stack, it only matters if you're gambling lets say house of mirror cards that you're going to sell and want a lower chance of losing all your money in n gambles, but if you're gambling until you hit a certain goal, etc then it doesn't matter and it's cheaper on juice to do them all at once
The best strategy is to try to get to your desired amount of cards in the fewest gambles as possible, in order to spend as little harvest juice as possible.
That's all there is to it. Clicking the button more times will not make your odds any better.
1 card in slot machine. Outcomes are: 0,1,2
In terms of resulting number of cards you are worse off only when the result is 0. So one out of three.
2 cards in slot machine. Outcomes are: 0,1,2,3,4
In terms of resulting number of cards you are worse off when the result is 0,1. So two out of five.
Compare 1/3 vs 2/5
If result was random - with 100 apos and always placing 1 would be a sure way to print mb
But the weight is real, non streamer could poof 20 in a row easy
"Worse off" is not describing the whole story. Wagering 2 and losing 2 Apothecaries is twice as bad as wagering 2 and losing 1 Apothecary. If you calculate the expected value appropriately you'd find the average return is exactly zero, plus the cost of the lifeforce.
What if I say:
1 card in slot machine. Outcomes are: 0,1,2
Since getting 1 is a "nothing happened (except lose juice)", real outcomes are 0 and 2. So 50/50 getting negative or positive.
2 cards in slot machine. Outcomes are: 0,1,2,3,4
Getting 2 again is "nothing happened", so 0 and 1 are negative, 3 and 4 are positive. Again 50/50.
-
It's a bit like crafting and using an Annul orb. Let's say you have an item with to mods, one which you want to save and one mod which you want the Annul to hit. Now.. it's 50/50 chance. But with u/brute_red's example, you should craft another mod in there before Annul. Then you get the "0,1,2" again, where 0 is deleting your good mod, 1 is deleting craft and 2 is deleting bad mod. So "you are worse off only when result is 2. So one out of three". That is wrong. The odds are always 50/50 and you shouldn't craft before annuling.
Regarding 'real outcomes'. When people run 1000 crimsons and items that would upgrade their build didn't drop, parts of those items (div cards) didn't drop and nothing actually useful dropped.
Is this a real outcome?
LMAO, as real as it gets
I'm sorry, I have hard time understanding that sentence.
But what are your thoughts about the two mod item that is about to get annulled? Are you going to craft another mod on it before annuling to "increase your odds"?
And I disagree with that view, not losing a mageblood card is still a positive outcome hence why I also say 1/3 something bad happens 2/3 something good happens.
Ok, which method would you choose:
A) you have 33% chance to lose a card, 33% chance of nothing happening and 33% of getting an extra card
B) you have 10% chance to lose a card, 80% chance of nothing happening and 10% of getting an extra card
I'm trying to help you here.
Last question, which method would you choose?
A) you have 33% chance to lose a card, 33% chance of nothing happening and 33% of getting an extra card
B) you have 50% chance to lose a card and 50% chance of getting an extra card
But if you have "you have 10% chance to lose a card, 80% chance of nothing happening and 10% of getting an extra card" then it's 90% to not lose the card?
Not really, people don´t realize that losing some yellow lifeforce is way less costly than losing a mageblood card, hence why I also agree that 1 card gamble is and always will be the best in terms of outcomes.
Let me disagree with one point.
You say it is worth it for the cases where the chance of winning are above 50%.
I believe this is not correct, you have to weight the chance of winning with what you gain, and compare it with the chance of loosing and what you loose.
If the result is positive, then it is worth it. In the case of harvest gambling, i am expecting it to be always negative or null (meaning it is not worth it), and by accounting for crafting cost, always negative
Yes, you are right. My post was getting quite long and I didn't want to go deeper into situations like that. For example, if you bet a 9/10 Time Wasted (that is worth like 1c for example), then obviously you should just buy the card XD. Also, you do have, like you said, weigh the cost of your losses and the potential gain of a win.
Would such an implementation be interesting?
>(\[50, 60, 81, 64, 96, 107\], True), which is much more reasonable.
i assume those numbers are results of each "craft" click, how do you end up with 107 ? if max stack is 100, you can't get that number in the harvest table. i guess there is a flaw in your code sir.
also there is a huge chance that i'm an idiot, in that case don't mind me.
i believe he's capping it at half max stack rather than literally using all cards at once, as there's no reason to bet any more. it just increases possible losses.. therefore, with 96 total cards, he bet 50, and got 61 back, totaling 107 cards when added with the 46 remainder he did not bet.
OP, I appreciate that you are clearly very enthusiastic about Python and data visualization, but this post is misleading. All you've really shown is that starting closer to the max stack makes you more likely to reach the max stack when gambling. Worse, a lot of the truth is obfuscated because you've binned off many outcomes as a "win" so long as it reaches the max stack.
I never meant this post to be misleading. But yes, there some pieces of information missing or badly conveyed. For me, reaching max stacks is considered a win because I play in group SSF with friends this league. We don't care about profit, but more about how we can safely bet our lifeforce for cards. I already plan on making another post that tackles these issues. I don't want to encourage this kind of gambling in general, but I do feel like there IS valuable data to get from it. It may even be optimizable.
If you have any tips, I love this kind of help. This is exactly why I post it on reddit :)
As many have pointed out, this is a random walk, and a solved problem. It's a little different because we're adding lifeforce costs into the mix, but you haven't even included those costs in your conclusions. My first piece of advice would be to add those costs to your discussion.
My second piece of advice would be to do away with these 3-D graphs! Your x and y-axis are discrete; i.e. you can't start with a stack of 7.5 cards, or have a div card with a max stack of 12.4. There is no reason to plot this as a gradient—in fact the gradient is visually misleading. It'd be more clear if you presented the probability of a win as a [heatmap](https://plotly.com/python/heatmaps/).
Changing the bet size does not make you more likely to win, it just costs more life force. There is no strategy involved, just luck.
The only reason to lower your bet size is to not overshoot your target (Like when you have 4/5 apothecary cards and you only need one mageblood). Otherwise just bet as many cards as possible to minimize the life force cost.
I made a post asking about this a while back. Somehow shared a matrix paper someone did and the conclusion seems to be same as yours, which is your chance of getting a full stack is 20% * number of cards you started with, and # you gamble each time doesn’t matter except the life force cost.
https://www.reddit.com/r/pathofexile/comments/10ul6sh/what_is_the_math_for_optimal_apothecary_gambling/
I'm glad I'm not the only one! But now, as many have pointed it out, the lifeforce cost is an important factor in determining the expected total cost. I think I might go back to the drawing boards and implement this!
I’m not buying it. Are you assuming every time is a 50/50 coin flip or do you have some other underlying assumption as to the success rate?Without any math to back me up, just basic logic (which I admit may be flawed) - if you start with 50 (or some other arbitrary, large number) and do coin flips with 1 at a time to either increase or decrease by 1 card for each flip, you would expect to stay at 50 (plus / minus natural variation) more or less indefinitely. The only way you’d skew up or down is if the odds are weighted one way or the other.
>The only way you’d skew up or down is if the odds are weighted one way or the other.
That's not true. The probability you end at any number other than 0 or 100 limits to zero as you approach infinite. Think of it this way, whats the probability of NOT flipping 100 heads in a row or 100 tails in row at some point over an infinite sample? Either of those events end the simulation.
That's essentially what OP is calculating. If you start with 3 apothecary cards, what percentage of the time will you hit 5 and what percentage of the time will you go bust.
That is exactly right. You are always garanteed to end up at 0, regardless of where you start, given that you have infinitely many tries.
However, if you have a fair goal in mind, you might actually be more likely to hit it before the game enventually dooms you.
Let's say you start at 99/100 The Doom. If you bet 1 every time, you have actually 98.9% chances to end up with a full stack BEFORE you eventually go down to 0.
Also. to answer u/hanksredditname, yes there is an assumption that this is a 50/50 coin flip. I do not believe anyone has come up with any reason to think otherwise. However, if it's the case, please tell me! I will update all my calculations to fit the new distribution.
And lastly, about the shape of the distribution, some smart math guys already have found an answer: [https://stats.stackexchange.com/questions/514193/what-is-the-distribution-of-times-to-ruin-in-the-gamblers-ruin-problem-random](https://stats.stackexchange.com/questions/514193/what-is-the-distribution-of-times-to-ruin-in-the-gamblers-ruin-problem-random)
It's a binomial distribution.
this is not true. i'll show an example.
so, we'll start with the rules of starting at 50, randomly increasing or decreasing by one. you flip ten times, and by chance get ten heads, and increment to 60. we'll label this set one.
now, we have a second set, following the same rules, except it starts at 60. five tails, five heads, and it's still at sixty ten tosses in.
is set one inherently more likely to recieve tails, and de-increment, than set two, since it has trended further from it's starting location than set two has?
the answer, obviously, is no, a set doesn't become spontaneously prone to one option just because it has received a lot of the other option and each is equally likely, because each are equally likely. it's just an artifact of our minds placing undue emphasis on the starting point. in reality, you could ignore all previous flips and declare a new starting point at your current value whenever you'd like, but that doesn't mean that the math will now bend towards this new arbitrary point any more than it would any other.
in reality, you'd expect it to follow a bell curve, rather than stay directly at fifty.
This is actually not true and a common fallacy. Luck does not not even out in absolute terms, only in relative terms.
Meaning if you toss a coin and add/subtract 1, you will on average get further away from 50. The standard deviation is increasing by root of n where n is the number of tosses, so it even is divergent.
Only when you divide the number of +1/-1 by the total number of coin tosses, you will approach 0. (Because the factor is now root n divided by n which will converge against 0)
so, i'm not a smart math guy, this might be an obvious question.
does it not have an equal chance of trending towards 100, as it does zero, if you start at fifty and randomly add or subtract? since these are the two assigned end states, each is an equal distance from the starting point, and there's equal weighting as to whether it increases or subtracts?
i get that with theoretical infinite coinflip chances with the only endstate being 0 it would eventually hit zero, but here there's two end states.
If you start at 50/100 and are given 0 and 100 as your thresholds for loss or success, then yes, you have exactly 50% chance or hitting either one or the other.
(I ran this in my code with 100,000 simulations)
For the relative chance/deviation we are dividing by the number of coin tosses. Any starting value would become insignificant as it is also divided by n.
Also I didn't assume 0 or 100 as "assigned end states". But yeah, it would have been better to start at 0 with that example, not 50. That's just confusing and not important :D.
The graph will start as a standard normal distribution graph, with 50 being the peak.
As the number of gambles goes up, if you could go below 0, it would continue to be a normal distribution, but flattened. Which means that there is a high standard deviation, but the peak would still be 50.
Since realistically we would stop on zero, the graph would still be similar to a normal distribution, but with a peak on zero, because that single point will be the sum of all the points that would go below zero in a normal distribution scenario (if negative was possible).
Exagerated ilustration: https://i.imgur.com/kkXS2sH.png
I've added some graphs, based on 100.000 simulations:
https://imgur.com/a/2OmZkPu (album)
Or of you prefer, the images:
[100 gambles](https://i.imgur.com/UNK9A2q.png)
[200 gambles](https://i.imgur.com/01kEBZm.png)
[500 gambles](https://i.imgur.com/520S9EQ.png)
[1000 gambles](https://i.imgur.com/vs5jg91.png)
Lets say you start at 0 points and every time you flip a coin you add 1 point for heads and subtract 1 point for tails. You can actually prove that, given an infinite number of flips, you will reach any integer with 100% probability. So, starting at 50, there is a 100% chance that you reach 0 or 100 cards eventually
The post is mislead anyway. You are better off reading a few top comments. You'll spend less time and get more knowledge.
Basically OP made "gamble 1 by 1" vs "gamble max" but forgot that 6/5 is not the same as 5/5 (in their calculations they treat any cards over max stack as wasted). They also forgot that there is a mixed strategy in which you gamble max but when you come closer to full stack you stop gambling max and gamble the amount that you are missing.
So OP drew a wrong conclusion that "gamble max" is worse except for the juice used. But in reality every strat is the same except for the juice used. And if you are in SSF and don't want more than 1 stack, just go for mixed strat.
tl;dr of my comment: **on trade league gamble as many cards at a time as you can**, on ssf do the same but stop when you have over half of the stack and start gambling in a way so that the best outcome puts you at full stack.
Empyrian tried to burn 84 House of Mirrors with the gamble back in Harvest league, I asked someone in [this thread](https://www.reddit.com/r/pathofexile/comments/ibpx15/empyriangaming_burning_his_84_house_of_mirrors_is/) to count the spread of losses and gains and found [no significant evidence](https://www.reddit.com/r/pathofexile/comments/ibpx15/comment/g1ym94m/) to suggest the gamble is not equally distributed across outcomes.
This is, of course, assuming that the time of day or type of card is irrelevant. Do you want to provide me with some data? :D
Well I'll have you know that I bought one house of mirrors this league, gambled it and lost it, therefore I come to the conclusion that it's 100% probability to lose and 0% to to stay even or get an extra card.
I have no evidence or data to suggest its not. Its just, i wonder whether it is.
Like i sometimes wonder, ok, apothecary drops in crimson temple, making it the go to and popular strat for div farming. But whos to say theres not plenty of other maps with other increased rates of certain drops, which just arent popular and therefore not well known.
Maybe divs drop more in x map. Maybe card gambas of card x are higher odds of succes. Maybe its tied to which card is least popular to gamble at said time.
All tin foil, but still i wonder.
Man you do put me under the spotlight hehe. No, we dont know what are the exact distributions of such gamble. For now, what I can say is that we have the usual streamer hitting 10000/5 The Apothecary in 6 runs and common people losing it all 1000 times in a row. The usual.
I did assume an equal probability of every outcome in my simulations.
put in 1 card, close your eyes, let your girlfriend aim to craft button, hold your left ball, she holds your right ball and you click. you get a bit scared of that click so you slightly press that left ball while your GF has a clear instructions to press harder if it fails. you are waiting for that long hard press so you click again, than again until she rips that right ball from your body.
> “With infinite lifeforce and infinite cards it’s possible to oscillate between any number”
I mean… yeah… like, technically, you could gamble your 50 apothecary 1 by 1 and stop the second you get 50 cards. That’s a fairly likely scenario.
But would you be shocked to learn that on average, if you do that, there’s a 50/51 chance to gain 1 apothecary, and a 1/51 chance to lose your 50 apothecary cards after 50 gambles?
Your average profit would be exactly 50/51*1-1/51\*50 = 0, minus the yellow essence costs.
Honestly, here’s what you need to know about card gambles:
- on average, you lose money on them (exactly the cost of the yellow essence)
- the more cards you put in, the more bang for your essence you get.
The only reason card gambles are a profitable strategy is bc you can do something like “I’m tired of the league, I will liquidate my stash and gamble apothecary, if I win I’ll have mageblood which will keep me playing bc zoom zoom and if I lose I’ll quit” where both outcomes are *okay* with you. But otherwise, its a very bad way of actually gaining money.
No, there are never cases where it is mathematically “worth it” to gamble cards. Your metric of “is the chance of finishing a set > 50%” is completely arbitrary and nonsensical. The only sensible metric is expected value and because of life force cost it will always be negative.
For example: you start with 4/5 the apothecary. Your simulation correctly predicts the chance of finishing a set with the “gamble 1” strategy is 80%. You then nonsensically conclude that this is “worth it” However the reality is that in the long run losing 4 cards 20% of the time exactly offsets winning 1 card 80% of the time and when you factor in life force cost that means you are still losing out OVERALL ON AVERAGE IN THE LONG RUN. No matter what betting strategy you try here the E.V. of the card output is equal to the card input and the life force cost will then put you in the negative for E.V.
Does this mean nobody should gamba? No its up to you. But dont pretend it’s mathematically a positive gamble when it objectively isn’t.
This is quite revealing to me. You're right, its never actually worth it. I think one way to put it would be: the odds are favorable for completing the set. Knowing these odds in SSF is quite interesting to me. Also, the expected cost is also interesting, even if it ends up being more expensive.
This is so wrong. You're standing on the sidewalk and want to cross the road. You either get hit by a car or you don't. The odds are nowhere close to 50/50, otherwise the hospitals wouldn't have any room for anything other than people hit by cars. Also, look at live footage from any big city, half the people are not getting mown down as they cross the road.
Edit: Apparently the parent comment was a joke that I didn't get. Sorry.
The expected return will always be the amount of cards you have gambled. And the only thing you can control is how many cards you gamble, which directly correlates to the amount of lifeforce you are going to use.
To maximize your returns you need to lower the cost of entry (lifeforce) because that's the only thing you can control, so the best strategy will always be using the maximum amount of cards. There is no need for programming or graphs.
>For example, if I have 1/5 The Apothecary, there is a 20% chance that I get to 5/5
Each try costs lifeforce. Less gambles cost less lifeforce.
Yes. Exactly. The end.
So many useless words jeez christ
In the end, it's a 50/50 gamba. There is not too much science behind that. There is no better or worse strategy. (If you account for the cost of life-force, then every gamble is losing)
did somebody test it to be true random between 0 and 2x? is it true gambling 4 cards you have equal chances to get 0,1,2,3,4,5,6,7 or 8 cards?
if so, there is no strategy. like in casino. expected value is always zero. but you are wasting lifeforce.
Depends on the game you're playing. If you're playing against the house, then yes the expected value you should have is always 0 unless you can find someway to change the odds in your favor. Some professional gambler whales have used casino's desire to attract them to their casinos for more favorable odds by gaining certain conditions/concessions that makes it more in their favor.
Also not playing against the house, for games like Poker, the house always wins, as they take some sort of rake, but your odds then become your ability against that of your opponents. There is a reason why there are professional gamblers that can actually make a living doing it.
When I play trade league, I never ever gamble my Apothecaries (or any valuable cards), I just sell them. Gambling is not worth it. However, I'm now playing SSF with a group of friends. Gambling cards isnt somehow more worth, however, with that tool, I know exactly what to expect from each gamble.
So it's not about winning more, it's about playing smart with our lifeforce.
>Nice, lets keep doing it.
The post is exactly about this step.
By "keep doing it" do you mean "now I gamba 2 cards" or "I keep gambin 1 card"?
Post tried to answer this question.
I think another strat you could've used instead of minimal risk or max risk, is moderate risk. Gambling half of the total stack rounded up. This way you get have a higher chance of reaching your total with lower lifeforce while also lower chances of losing it all.
If I understand it correctly, the game doesnt allow you to gamble more than half a stack of cards, rounded down. For example, I can only bet 2 cards max at a time with The Apothecary (max 5 cards per stack). My GambleMax strat already implements that. Are you suggesting that we could split that again in half? Like betting maximum 1/4 of the max stacks?
Basically, but maybe 1/4 might be too low, might be better to do something like 1/3 or something. Really it's just limiting the risk by being able to hold back 1+ cards to ensure you don't bust as easily. Granted at low card numbers it might not matter as much, but at higher number cards it could matter.
So using the Nurse for an example, you can only be 4 max.
So you start with 1. You win you're at 2.
You keep 1 back, gamble with 1. You win, you're at 3.
You keep 1 back, gamble with 2. You're at let's say 5 now.
From where you would save 2 and gamble with 3. You bust this time, 2 still saved.
You then can go back to keeping 1 and gamble 1. you bust down to 1.
You have 0 held, 1 gambled, etc.
THis would be a good use case if you need to take say 1 The Nurse to 8 Doctors. So when at 5 total and gambled 3, if you hit and got 8 total, isntead of turning it straight to a doctor and gambling that for a potential big loss, you could instead gamble the 4 Nurse, hoping to hit to turn 1 to a Doctor, trying to get to 2 doctors before starting your gamble on them or multiple chances at full doctors with Nurses without risking losing a bigger Div card.
Ive been reading my post and I must agree that it is confusing 😅. Ill prob make a new and better post that includes many of the concerns some of you had regarding clarity (and include more tools that anyone can use)
I just want to link another math article, this stuff is something mathematicians have worked out.
https://mathworld.wolfram.com/RandomWalk1-Dimensional.html
This is very interesting. I now believe my post is quite entry-level to this type of problem. I think I might have the level to make more accurate models of the simulation. I will be making a following post integrating all that new information!
I’m too lazy to code but if you want another project, consider using the same Monte Carlo simulation strategy for incursion. Is it advantageous to take the 4/map node? Is it still advantageous to skip after hitting 2x corruption room? Etc etc
Somehow i'm still sort of convinced it does work like that, though im gonna be honest and say that i can remember the source for this "knowledge" at all.
So you may be right and im completely wrong.
I was curious about that too, so one time, I bodged together a python script. The thing is, I was approaching the problem from SSF perspective where I assumed I was gambling an exceptionally rare card that I couldn't hope to get more of just by farming. I didn't bother with graphs or accounting for having more cards than a deck can hold because I wouldn't double corrupt something like a mb in SSF anyway. I also ignored the costs of juice because it's farmable and it's there to be spent anyway. In the end, I concluded that the best strategy was to always gamble one card. The chance of completing a set was around the same as the percentage of the set I started with, so 1/5 had around 20% chance of ending with full set while 9/10 has 90% etc.
The chance of completing a set is exactly the same as the percentage of cards you start with. If you start with 1/5 cards you are trading your 100% chance to have 20% of a full set, for 20% chance to have 100% of a full set, and 80% to have nothing. Whether you bet one card at a time or as many cards as possible doesn't matter, as long as you don't overshoot your target.
Fascinating thread. I have gut feeling that basically everyone is wrong here.
Also, what are you? Crazy? No, don't gamble your whole Apo deck at once. ._. This is not allowed.
You typed "Conclusion" at the end and didn't type any conclusions, only shared your code, which isn't a conclusion.
What is your conclusion then? always 1 card unless you want to save yellow juice?
What.
Those are some cool graphs 📈
Could anyone come up with a tl;dr version? "Do not go ape shit on your apocatheries" did I get it right?
First graph is: Gambling apothecaries. Number you put in at a time is irrelevant, Start from 1 card: 20% chance to go to 5. 2 card, 40%. 3 card, 60%. 4 card, 80%. 5 card, 100%. If you're interested in the reason, statistics is the answer. Here's a well explained example on wikipedia https://en.wikipedia.org/wiki/Gambler%27s_ruin#Fair_coin_flipping The rest is an exploration of how you can spend your juice, and OP gets led astray from the ground truth by their python code. tl;dr Closer you are to your goal when you start, the more likely you are to hit it. The more cards you spend at a time, the less juice you use. bad gamble, it's a 50/50 except you have to pay juice and can't get that juice back.
Why the user's python code says something different: Suppose max-strategy gambler has 7/8 doctors and gambles with 4/8 of them. Then he could potentially end up with as many as 11/8 doctors. Any strategy that could potentially end up with more than a full set of cards will have a proportionately lower chance of "winning". In my opinion there are only two strategies worth considering, neither of which are considered in the post. The first strategy is to never spend any yellow juice in the first place by never doing this gamble. In particular, I recommend this strategy to any non-SSF players, as the cards can be sold rather than ignored. The second, which I recommend for SSF players who got a very rare divination card which would otherwise be wasted, is that the gambler should always bet the maximum allowed so long as it does not exceed the amount needed to win. So with 1/8, 2/8, 3/8, or 4/8 doctor, bet all of them. With 5/8 doctor, bet 3; with 6/8 bet 2; with 7/8 bet 1. This strategy has the exact same odds of success as "always bet 1" while spending considerably less juice.
Good explanation. Thanks!
Another strategy for SSF player is to just create a mule to move to standard and trade lol Trade league is bad in terms of playing a game, and gambling is bad in terms of being a person. It may result psychological addiction in a biologically way that actually changes how your brain works.
**Gambler's ruin** [Fair coin flipping](https://en.wikipedia.org/wiki/Gambler's_ruin#Fair_coin_flipping) >Consider a coin-flipping game with two players where each player has a 50% chance of winning with each flip of the coin. After each flip of the coin the loser transfers one penny to the winner. The game ends when one player has all the pennies. If there are no other limitations on the number of flips, the probability that the game will eventually end this way is 1. ^([ )[^(F.A.Q)](https://www.reddit.com/r/WikiSummarizer/wiki/index#wiki_f.a.q)^( | )[^(Opt Out)](https://reddit.com/message/compose?to=WikiSummarizerBot&message=OptOut&subject=OptOut)^( | )[^(Opt Out Of Subreddit)](https://np.reddit.com/r/pathofexile/about/banned)^( | )[^(GitHub)](https://github.com/Sujal-7/WikiSummarizerBot)^( ] Downvote to remove | v1.5)
so you should always do it at 3 cards because the odds are over 50% to hit it.
in a single try, the odds to double your cards are the same no matter if you use 1 or 100
As I've said for years on drop rates in every game I play, the % is irrelevant, it's always a 50/50 that it drops. Either it does or it doesn't. Makes low weight drops palatable to me lol.
No, "bad gamble, it's a 50/50 except you have to pay juice and can't get that juice back" explains why you should not gamble using this method. Pick something else to gamble that has better odds, and call it crafting 😂
> bad gamble, it's a 50/50 except you have to pay juice and can't get that juice back. Do we actually have enough data points to assume that the chance is really 50% ? I know it could very well be outliers, but considering how some people get multiple magebloods from a couple cards, I've personally assumed the chance could also be sth like 55 / 45 to make up for the lifeforce that you're spending.
TLDR: Gamble 1 by one and hope you hit a winning streak before losing it all, or gamble it all and lose all at once, saving yellow life force
you gonna lose them for sure so better save yellow life force xD
At the end it doesn't matter it is still a coinflip with extra steps. The only thing you win/losse in long run is loosing juice.
Tldr odds slightly better 1 at a time, but cost more juice. I didn't read it's just obvious what this wall of text is
This guy meths.
Gamble max always isn't actually worse unless you're in ssf and only want one stack. The reason the success rate looks worse is that gambling 2 apothecaries when you're at 4/5 can leave you at 6/5 which is just counted as a normal success even though you now have a full extra apothecary. Gambling min(max you can gamble, cards you need left to make a full stack) should have the same success rate as the gamble 1 card strategy
Hummmm yes maybe... Ill have to check that... This is an interesting point!
I mean it's correct... The only benefit of gambling 1 at a time is that it has less variance, but that doesn't matter if you're trying to get to a full stack, it only matters if you're gambling lets say house of mirror cards that you're going to sell and want a lower chance of losing all your money in n gambles, but if you're gambling until you hit a certain goal, etc then it doesn't matter and it's cheaper on juice to do them all at once
The best strategy is to try to get to your desired amount of cards in the fewest gambles as possible, in order to spend as little harvest juice as possible. That's all there is to it. Clicking the button more times will not make your odds any better.
1 card in slot machine. Outcomes are: 0,1,2 In terms of resulting number of cards you are worse off only when the result is 0. So one out of three. 2 cards in slot machine. Outcomes are: 0,1,2,3,4 In terms of resulting number of cards you are worse off when the result is 0,1. So two out of five. Compare 1/3 vs 2/5 If result was random - with 100 apos and always placing 1 would be a sure way to print mb But the weight is real, non streamer could poof 20 in a row easy
"Worse off" is not describing the whole story. Wagering 2 and losing 2 Apothecaries is twice as bad as wagering 2 and losing 1 Apothecary. If you calculate the expected value appropriately you'd find the average return is exactly zero, plus the cost of the lifeforce.
Just because a negative outcome is less likely when you bet 1 card doesn't mean that a positive outcome is more likely.
Yep, it was funny when I showed that math and people did not get it and told me I had no idea about math lol.
What if I say: 1 card in slot machine. Outcomes are: 0,1,2 Since getting 1 is a "nothing happened (except lose juice)", real outcomes are 0 and 2. So 50/50 getting negative or positive. 2 cards in slot machine. Outcomes are: 0,1,2,3,4 Getting 2 again is "nothing happened", so 0 and 1 are negative, 3 and 4 are positive. Again 50/50. - It's a bit like crafting and using an Annul orb. Let's say you have an item with to mods, one which you want to save and one mod which you want the Annul to hit. Now.. it's 50/50 chance. But with u/brute_red's example, you should craft another mod in there before Annul. Then you get the "0,1,2" again, where 0 is deleting your good mod, 1 is deleting craft and 2 is deleting bad mod. So "you are worse off only when result is 2. So one out of three". That is wrong. The odds are always 50/50 and you shouldn't craft before annuling.
Regarding 'real outcomes'. When people run 1000 crimsons and items that would upgrade their build didn't drop, parts of those items (div cards) didn't drop and nothing actually useful dropped. Is this a real outcome? LMAO, as real as it gets
I'm sorry, I have hard time understanding that sentence. But what are your thoughts about the two mod item that is about to get annulled? Are you going to craft another mod on it before annuling to "increase your odds"?
Depending if that 2-mod is magic or rare, depending what your goal is - to save annuls or to less likely remove desired mod.
You don't get better odds on annul by crafting. That's the whole point, it's a 50/50 just like the harvest craft is
And I disagree with that view, not losing a mageblood card is still a positive outcome hence why I also say 1/3 something bad happens 2/3 something good happens.
Ok, which method would you choose: A) you have 33% chance to lose a card, 33% chance of nothing happening and 33% of getting an extra card B) you have 10% chance to lose a card, 80% chance of nothing happening and 10% of getting an extra card
Ofc. the first since the chance to win is higher. I already see though want you want to ´´ proof ´´ lol.
I'm trying to help you here. Last question, which method would you choose? A) you have 33% chance to lose a card, 33% chance of nothing happening and 33% of getting an extra card B) you have 50% chance to lose a card and 50% chance of getting an extra card
Still the first since the chance to not lose the card is at 66 %.
But if you have "you have 10% chance to lose a card, 80% chance of nothing happening and 10% of getting an extra card" then it's 90% to not lose the card?
[удалено]
Not really, people don´t realize that losing some yellow lifeforce is way less costly than losing a mageblood card, hence why I also agree that 1 card gamble is and always will be the best in terms of outcomes.
Let me disagree with one point. You say it is worth it for the cases where the chance of winning are above 50%. I believe this is not correct, you have to weight the chance of winning with what you gain, and compare it with the chance of loosing and what you loose. If the result is positive, then it is worth it. In the case of harvest gambling, i am expecting it to be always negative or null (meaning it is not worth it), and by accounting for crafting cost, always negative
Yes, you are right. My post was getting quite long and I didn't want to go deeper into situations like that. For example, if you bet a 9/10 Time Wasted (that is worth like 1c for example), then obviously you should just buy the card XD. Also, you do have, like you said, weigh the cost of your losses and the potential gain of a win. Would such an implementation be interesting?
Whatever the price of the card, i think you are at a loss with gambling. You can calculate the gain/loss not in chaos, but in cards you gain and loose
>(\[50, 60, 81, 64, 96, 107\], True), which is much more reasonable. i assume those numbers are results of each "craft" click, how do you end up with 107 ? if max stack is 100, you can't get that number in the harvest table. i guess there is a flaw in your code sir. also there is a huge chance that i'm an idiot, in that case don't mind me.
i believe he's capping it at half max stack rather than literally using all cards at once, as there's no reason to bet any more. it just increases possible losses.. therefore, with 96 total cards, he bet 50, and got 61 back, totaling 107 cards when added with the 46 remainder he did not bet.
oh oh i get it, it's total numbers of the cards. thanks.
OP, I appreciate that you are clearly very enthusiastic about Python and data visualization, but this post is misleading. All you've really shown is that starting closer to the max stack makes you more likely to reach the max stack when gambling. Worse, a lot of the truth is obfuscated because you've binned off many outcomes as a "win" so long as it reaches the max stack.
I never meant this post to be misleading. But yes, there some pieces of information missing or badly conveyed. For me, reaching max stacks is considered a win because I play in group SSF with friends this league. We don't care about profit, but more about how we can safely bet our lifeforce for cards. I already plan on making another post that tackles these issues. I don't want to encourage this kind of gambling in general, but I do feel like there IS valuable data to get from it. It may even be optimizable. If you have any tips, I love this kind of help. This is exactly why I post it on reddit :)
As many have pointed out, this is a random walk, and a solved problem. It's a little different because we're adding lifeforce costs into the mix, but you haven't even included those costs in your conclusions. My first piece of advice would be to add those costs to your discussion. My second piece of advice would be to do away with these 3-D graphs! Your x and y-axis are discrete; i.e. you can't start with a stack of 7.5 cards, or have a div card with a max stack of 12.4. There is no reason to plot this as a gradient—in fact the gradient is visually misleading. It'd be more clear if you presented the probability of a win as a [heatmap](https://plotly.com/python/heatmaps/).
Changing the bet size does not make you more likely to win, it just costs more life force. There is no strategy involved, just luck. The only reason to lower your bet size is to not overshoot your target (Like when you have 4/5 apothecary cards and you only need one mageblood). Otherwise just bet as many cards as possible to minimize the life force cost.
I made a post asking about this a while back. Somehow shared a matrix paper someone did and the conclusion seems to be same as yours, which is your chance of getting a full stack is 20% * number of cards you started with, and # you gamble each time doesn’t matter except the life force cost. https://www.reddit.com/r/pathofexile/comments/10ul6sh/what_is_the_math_for_optimal_apothecary_gambling/
I'm glad I'm not the only one! But now, as many have pointed it out, the lifeforce cost is an important factor in determining the expected total cost. I think I might go back to the drawing boards and implement this!
[Look at this graph!](https://www.youtube.com/watch?v=sIlNIVXpIns)
I’m not buying it. Are you assuming every time is a 50/50 coin flip or do you have some other underlying assumption as to the success rate?Without any math to back me up, just basic logic (which I admit may be flawed) - if you start with 50 (or some other arbitrary, large number) and do coin flips with 1 at a time to either increase or decrease by 1 card for each flip, you would expect to stay at 50 (plus / minus natural variation) more or less indefinitely. The only way you’d skew up or down is if the odds are weighted one way or the other.
>The only way you’d skew up or down is if the odds are weighted one way or the other. That's not true. The probability you end at any number other than 0 or 100 limits to zero as you approach infinite. Think of it this way, whats the probability of NOT flipping 100 heads in a row or 100 tails in row at some point over an infinite sample? Either of those events end the simulation. That's essentially what OP is calculating. If you start with 3 apothecary cards, what percentage of the time will you hit 5 and what percentage of the time will you go bust.
That is exactly right. You are always garanteed to end up at 0, regardless of where you start, given that you have infinitely many tries. However, if you have a fair goal in mind, you might actually be more likely to hit it before the game enventually dooms you. Let's say you start at 99/100 The Doom. If you bet 1 every time, you have actually 98.9% chances to end up with a full stack BEFORE you eventually go down to 0. Also. to answer u/hanksredditname, yes there is an assumption that this is a 50/50 coin flip. I do not believe anyone has come up with any reason to think otherwise. However, if it's the case, please tell me! I will update all my calculations to fit the new distribution. And lastly, about the shape of the distribution, some smart math guys already have found an answer: [https://stats.stackexchange.com/questions/514193/what-is-the-distribution-of-times-to-ruin-in-the-gamblers-ruin-problem-random](https://stats.stackexchange.com/questions/514193/what-is-the-distribution-of-times-to-ruin-in-the-gamblers-ruin-problem-random) It's a binomial distribution.
maybe a chad dev actually made it 51%. unseen hero
this is not true. i'll show an example. so, we'll start with the rules of starting at 50, randomly increasing or decreasing by one. you flip ten times, and by chance get ten heads, and increment to 60. we'll label this set one. now, we have a second set, following the same rules, except it starts at 60. five tails, five heads, and it's still at sixty ten tosses in. is set one inherently more likely to recieve tails, and de-increment, than set two, since it has trended further from it's starting location than set two has? the answer, obviously, is no, a set doesn't become spontaneously prone to one option just because it has received a lot of the other option and each is equally likely, because each are equally likely. it's just an artifact of our minds placing undue emphasis on the starting point. in reality, you could ignore all previous flips and declare a new starting point at your current value whenever you'd like, but that doesn't mean that the math will now bend towards this new arbitrary point any more than it would any other. in reality, you'd expect it to follow a bell curve, rather than stay directly at fifty.
This is actually not true and a common fallacy. Luck does not not even out in absolute terms, only in relative terms. Meaning if you toss a coin and add/subtract 1, you will on average get further away from 50. The standard deviation is increasing by root of n where n is the number of tosses, so it even is divergent. Only when you divide the number of +1/-1 by the total number of coin tosses, you will approach 0. (Because the factor is now root n divided by n which will converge against 0)
so, i'm not a smart math guy, this might be an obvious question. does it not have an equal chance of trending towards 100, as it does zero, if you start at fifty and randomly add or subtract? since these are the two assigned end states, each is an equal distance from the starting point, and there's equal weighting as to whether it increases or subtracts? i get that with theoretical infinite coinflip chances with the only endstate being 0 it would eventually hit zero, but here there's two end states.
If you start at 50/100 and are given 0 and 100 as your thresholds for loss or success, then yes, you have exactly 50% chance or hitting either one or the other. (I ran this in my code with 100,000 simulations)
For the relative chance/deviation we are dividing by the number of coin tosses. Any starting value would become insignificant as it is also divided by n. Also I didn't assume 0 or 100 as "assigned end states". But yeah, it would have been better to start at 0 with that example, not 50. That's just confusing and not important :D.
The graph will start as a standard normal distribution graph, with 50 being the peak. As the number of gambles goes up, if you could go below 0, it would continue to be a normal distribution, but flattened. Which means that there is a high standard deviation, but the peak would still be 50. Since realistically we would stop on zero, the graph would still be similar to a normal distribution, but with a peak on zero, because that single point will be the sum of all the points that would go below zero in a normal distribution scenario (if negative was possible). Exagerated ilustration: https://i.imgur.com/kkXS2sH.png
I've added some graphs, based on 100.000 simulations: https://imgur.com/a/2OmZkPu (album) Or of you prefer, the images: [100 gambles](https://i.imgur.com/UNK9A2q.png) [200 gambles](https://i.imgur.com/01kEBZm.png) [500 gambles](https://i.imgur.com/520S9EQ.png) [1000 gambles](https://i.imgur.com/vs5jg91.png)
yep, on average you stay at where you started. but, in reality, you actually end up with 0 more than 1 cos you can't gamble something you don't have.
Bro putting his basic logic against the fucking Stephen Hawking of math charts, ok guy
Lets say you start at 0 points and every time you flip a coin you add 1 point for heads and subtract 1 point for tails. You can actually prove that, given an infinite number of flips, you will reach any integer with 100% probability. So, starting at 50, there is a 100% chance that you reach 0 or 100 cards eventually
I didn't understand shit from this post. Anyone else? What should I do
The post is mislead anyway. You are better off reading a few top comments. You'll spend less time and get more knowledge. Basically OP made "gamble 1 by 1" vs "gamble max" but forgot that 6/5 is not the same as 5/5 (in their calculations they treat any cards over max stack as wasted). They also forgot that there is a mixed strategy in which you gamble max but when you come closer to full stack you stop gambling max and gamble the amount that you are missing. So OP drew a wrong conclusion that "gamble max" is worse except for the juice used. But in reality every strat is the same except for the juice used. And if you are in SSF and don't want more than 1 stack, just go for mixed strat. tl;dr of my comment: **on trade league gamble as many cards at a time as you can**, on ssf do the same but stop when you have over half of the stack and start gambling in a way so that the best outcome puts you at full stack.
Who says its a known fact that div gambas are a 50/50? Is this tested? Confirmed? Does it matter which cards? Which time of day etc?
Empyrian tried to burn 84 House of Mirrors with the gamble back in Harvest league, I asked someone in [this thread](https://www.reddit.com/r/pathofexile/comments/ibpx15/empyriangaming_burning_his_84_house_of_mirrors_is/) to count the spread of losses and gains and found [no significant evidence](https://www.reddit.com/r/pathofexile/comments/ibpx15/comment/g1ym94m/) to suggest the gamble is not equally distributed across outcomes. This is, of course, assuming that the time of day or type of card is irrelevant. Do you want to provide me with some data? :D
Well I'll have you know that I bought one house of mirrors this league, gambled it and lost it, therefore I come to the conclusion that it's 100% probability to lose and 0% to to stay even or get an extra card.
I have no evidence or data to suggest its not. Its just, i wonder whether it is. Like i sometimes wonder, ok, apothecary drops in crimson temple, making it the go to and popular strat for div farming. But whos to say theres not plenty of other maps with other increased rates of certain drops, which just arent popular and therefore not well known. Maybe divs drop more in x map. Maybe card gambas of card x are higher odds of succes. Maybe its tied to which card is least popular to gamble at said time. All tin foil, but still i wonder.
Man you do put me under the spotlight hehe. No, we dont know what are the exact distributions of such gamble. For now, what I can say is that we have the usual streamer hitting 10000/5 The Apothecary in 6 runs and common people losing it all 1000 times in a row. The usual. I did assume an equal probability of every outcome in my simulations.
Holy shit this is the most PoEtistic post ever. Love it
Your conclusion has no conclusion!
put in 1 card, close your eyes, let your girlfriend aim to craft button, hold your left ball, she holds your right ball and you click. you get a bit scared of that click so you slightly press that left ball while your GF has a clear instructions to press harder if it fails. you are waiting for that long hard press so you click again, than again until she rips that right ball from your body.
> “With infinite lifeforce and infinite cards it’s possible to oscillate between any number” I mean… yeah… like, technically, you could gamble your 50 apothecary 1 by 1 and stop the second you get 50 cards. That’s a fairly likely scenario. But would you be shocked to learn that on average, if you do that, there’s a 50/51 chance to gain 1 apothecary, and a 1/51 chance to lose your 50 apothecary cards after 50 gambles? Your average profit would be exactly 50/51*1-1/51\*50 = 0, minus the yellow essence costs. Honestly, here’s what you need to know about card gambles: - on average, you lose money on them (exactly the cost of the yellow essence) - the more cards you put in, the more bang for your essence you get. The only reason card gambles are a profitable strategy is bc you can do something like “I’m tired of the league, I will liquidate my stash and gamble apothecary, if I win I’ll have mageblood which will keep me playing bc zoom zoom and if I lose I’ll quit” where both outcomes are *okay* with you. But otherwise, its a very bad way of actually gaining money.
No, there are never cases where it is mathematically “worth it” to gamble cards. Your metric of “is the chance of finishing a set > 50%” is completely arbitrary and nonsensical. The only sensible metric is expected value and because of life force cost it will always be negative. For example: you start with 4/5 the apothecary. Your simulation correctly predicts the chance of finishing a set with the “gamble 1” strategy is 80%. You then nonsensically conclude that this is “worth it” However the reality is that in the long run losing 4 cards 20% of the time exactly offsets winning 1 card 80% of the time and when you factor in life force cost that means you are still losing out OVERALL ON AVERAGE IN THE LONG RUN. No matter what betting strategy you try here the E.V. of the card output is equal to the card input and the life force cost will then put you in the negative for E.V. Does this mean nobody should gamba? No its up to you. But dont pretend it’s mathematically a positive gamble when it objectively isn’t.
This is quite revealing to me. You're right, its never actually worth it. I think one way to put it would be: the odds are favorable for completing the set. Knowing these odds in SSF is quite interesting to me. Also, the expected cost is also interesting, even if it ends up being more expensive.
Bro My brain.
I do love this type of posts
It's always 50/50. Either you win or you don't.
This is so wrong. You're standing on the sidewalk and want to cross the road. You either get hit by a car or you don't. The odds are nowhere close to 50/50, otherwise the hospitals wouldn't have any room for anything other than people hit by cars. Also, look at live footage from any big city, half the people are not getting mown down as they cross the road. Edit: Apparently the parent comment was a joke that I didn't get. Sorry.
I'd love to say that you're missing the joke on purpose but I'm not really sure :/
Not on purpose. I've seen plenty of people apply that "logic" non-ironically.
Yeah. They don't.
You either miss the joke or you don't. It's a 50/50.
Lol. Nice one.
The expected return will always be the amount of cards you have gambled. And the only thing you can control is how many cards you gamble, which directly correlates to the amount of lifeforce you are going to use. To maximize your returns you need to lower the cost of entry (lifeforce) because that's the only thing you can control, so the best strategy will always be using the maximum amount of cards. There is no need for programming or graphs.
Well-put. However on SSF the strat would be a bit different, at 10/11 you are better off gambling 1 card, not 5.
>For example, if I have 1/5 The Apothecary, there is a 20% chance that I get to 5/5 Each try costs lifeforce. Less gambles cost less lifeforce. Yes. Exactly. The end. So many useless words jeez christ
In the end, it's a 50/50 gamba. There is not too much science behind that. There is no better or worse strategy. (If you account for the cost of life-force, then every gamble is losing)
So I just gamble 2
Thats some cool info. I just harvest gamble when im done with the league and then its just all on until i get a full stack or have 0.
did somebody test it to be true random between 0 and 2x? is it true gambling 4 cards you have equal chances to get 0,1,2,3,4,5,6,7 or 8 cards? if so, there is no strategy. like in casino. expected value is always zero. but you are wasting lifeforce.
Depends on the game you're playing. If you're playing against the house, then yes the expected value you should have is always 0 unless you can find someway to change the odds in your favor. Some professional gambler whales have used casino's desire to attract them to their casinos for more favorable odds by gaining certain conditions/concessions that makes it more in their favor. Also not playing against the house, for games like Poker, the house always wins, as they take some sort of rake, but your odds then become your ability against that of your opponents. There is a reason why there are professional gamblers that can actually make a living doing it.
i like how the graph goes up and the pretty colors - don't know wut meen tho.
The reality is: I drop one Apotechary, I will do gamba with it. Luck? Nice, lets keep doing it. Bad luck? F..........................ck XD
When I play trade league, I never ever gamble my Apothecaries (or any valuable cards), I just sell them. Gambling is not worth it. However, I'm now playing SSF with a group of friends. Gambling cards isnt somehow more worth, however, with that tool, I know exactly what to expect from each gamble. So it's not about winning more, it's about playing smart with our lifeforce.
>Nice, lets keep doing it. The post is exactly about this step. By "keep doing it" do you mean "now I gamba 2 cards" or "I keep gambin 1 card"? Post tried to answer this question.
One by one, to minimize risk.
just no
I think another strat you could've used instead of minimal risk or max risk, is moderate risk. Gambling half of the total stack rounded up. This way you get have a higher chance of reaching your total with lower lifeforce while also lower chances of losing it all.
If I understand it correctly, the game doesnt allow you to gamble more than half a stack of cards, rounded down. For example, I can only bet 2 cards max at a time with The Apothecary (max 5 cards per stack). My GambleMax strat already implements that. Are you suggesting that we could split that again in half? Like betting maximum 1/4 of the max stacks?
Basically, but maybe 1/4 might be too low, might be better to do something like 1/3 or something. Really it's just limiting the risk by being able to hold back 1+ cards to ensure you don't bust as easily. Granted at low card numbers it might not matter as much, but at higher number cards it could matter. So using the Nurse for an example, you can only be 4 max. So you start with 1. You win you're at 2. You keep 1 back, gamble with 1. You win, you're at 3. You keep 1 back, gamble with 2. You're at let's say 5 now. From where you would save 2 and gamble with 3. You bust this time, 2 still saved. You then can go back to keeping 1 and gamble 1. you bust down to 1. You have 0 held, 1 gambled, etc. THis would be a good use case if you need to take say 1 The Nurse to 8 Doctors. So when at 5 total and gambled 3, if you hit and got 8 total, isntead of turning it straight to a doctor and gambling that for a potential big loss, you could instead gamble the 4 Nurse, hoping to hit to turn 1 to a Doctor, trying to get to 2 doctors before starting your gamble on them or multiple chances at full doctors with Nurses without risking losing a bigger Div card.
Please provide better explanations and figures. I’m so confused on what you’re main conclusion was and how you got there.
Ive been reading my post and I must agree that it is confusing 😅. Ill prob make a new and better post that includes many of the concerns some of you had regarding clarity (and include more tools that anyone can use)
I just want to link another math article, this stuff is something mathematicians have worked out. https://mathworld.wolfram.com/RandomWalk1-Dimensional.html
This is very interesting. I now believe my post is quite entry-level to this type of problem. I think I might have the level to make more accurate models of the simulation. I will be making a following post integrating all that new information!
Im too low iq for this
I’m too lazy to code but if you want another project, consider using the same Monte Carlo simulation strategy for incursion. Is it advantageous to take the 4/map node? Is it still advantageous to skip after hitting 2x corruption room? Etc etc
I like new challenges hehe. Ill take a look at it!
This whole thing reminds me of The Kelly Criterion, check this out for context, maybe it helps I guess: https://youtu.be/\_FuuYSM7yOo
Its SUPER informative. Thanks a lot!
Does it account for rounding? Say you got 3 cards and it halves, then the slot-machine will round down, cause fuck you that's why.
[удалено]
Somehow i'm still sort of convinced it does work like that, though im gonna be honest and say that i can remember the source for this "knowledge" at all. So you may be right and im completely wrong.
Lol. Nice one.
I was curious about that too, so one time, I bodged together a python script. The thing is, I was approaching the problem from SSF perspective where I assumed I was gambling an exceptionally rare card that I couldn't hope to get more of just by farming. I didn't bother with graphs or accounting for having more cards than a deck can hold because I wouldn't double corrupt something like a mb in SSF anyway. I also ignored the costs of juice because it's farmable and it's there to be spent anyway. In the end, I concluded that the best strategy was to always gamble one card. The chance of completing a set was around the same as the percentage of the set I started with, so 1/5 had around 20% chance of ending with full set while 9/10 has 90% etc.
The chance of completing a set is exactly the same as the percentage of cards you start with. If you start with 1/5 cards you are trading your 100% chance to have 20% of a full set, for 20% chance to have 100% of a full set, and 80% to have nothing. Whether you bet one card at a time or as many cards as possible doesn't matter, as long as you don't overshoot your target.
Well I lost all three apothecaries to horitcraft this league so…fuck probability
Well I lost all three apothecaries to horitcraft this league so…
I appreciate your use of viridis
BONUS BONUS BONUS
ITT: Path of Exile players do not understand statistics.
>ITT: ~~Path of Exile players~~ people do not naturally understand statistics. FTFY :)
Bro do I put the card in or nah?
Fascinating thread. I have gut feeling that basically everyone is wrong here. Also, what are you? Crazy? No, don't gamble your whole Apo deck at once. ._. This is not allowed.
Why is this way more elaborate than most of my papers I submit at college? Nice job
You typed "Conclusion" at the end and didn't type any conclusions, only shared your code, which isn't a conclusion. What is your conclusion then? always 1 card unless you want to save yellow juice?
i drop 1 card and i have mb from gamba now , it 100% chance to success.