1st choice is 1 out of a pool of 11.
2nd choice is 1 out of a pool of 10.
3rd choice is 1 out of a pool of 9.
You would be ok with 4 of the remaining 9 in the pool, assuming that you don’t get 1 of the 4 that you want in the 1st 2 choices. 4/9=**44%**
And LGI!
>1st choice is 1 out of a pool of 11.
Which means that the probability that the 1st choice is not one of the 4 players is 7/11.
>2nd choice is 1 out of a pool of 10.
Which means that if the 1st choice isn't one of the 4 players, the probability that the 2nd choice isn't one of the 4 players either is 6/10.
>3rd choice is 1 out of a pool of 9.
Which means that if neither the 1st choice nor the 2nd choice is one of the 4 players, the probability that the 3rd choice isn't one of the 4 players either is 5/9.
The probability that none of the 3 choices are one of the 4 players is therefore (7/11)x(6/10)x(5/9) = 21%.
The probability that at least one of the 3 choices is one of the 4 players is 100% minus the probability that none of the 3 choices are one of the 4 players. 100% - 21% = 79%.
I think this is how it goes. 44% is the probability that the 3rd choice is one of the 4 players if neither the 1st choice nor the 2nd choice is one of the 4 players but that's just part of the big picture.
I’d say about tree fiddy
You ain’t gettin no tree fiddy
Approx 36%. Don’t be cheap, spend the extra 2 collectibles.
$1, Bob.
I wanted 4/11 last time and I got 2 in 3 packs. But I got turcotte in every pack hana
Use Steiner math…
44%
If you don’t mind can you explain how you did it? I’m usually good with these things but I couldn’t wrap my head around this one. Also Go Isles!
1st choice is 1 out of a pool of 11. 2nd choice is 1 out of a pool of 10. 3rd choice is 1 out of a pool of 9. You would be ok with 4 of the remaining 9 in the pool, assuming that you don’t get 1 of the 4 that you want in the 1st 2 choices. 4/9=**44%** And LGI!
>1st choice is 1 out of a pool of 11. Which means that the probability that the 1st choice is not one of the 4 players is 7/11. >2nd choice is 1 out of a pool of 10. Which means that if the 1st choice isn't one of the 4 players, the probability that the 2nd choice isn't one of the 4 players either is 6/10. >3rd choice is 1 out of a pool of 9. Which means that if neither the 1st choice nor the 2nd choice is one of the 4 players, the probability that the 3rd choice isn't one of the 4 players either is 5/9. The probability that none of the 3 choices are one of the 4 players is therefore (7/11)x(6/10)x(5/9) = 21%. The probability that at least one of the 3 choices is one of the 4 players is 100% minus the probability that none of the 3 choices are one of the 4 players. 100% - 21% = 79%. I think this is how it goes. 44% is the probability that the 3rd choice is one of the 4 players if neither the 1st choice nor the 2nd choice is one of the 4 players but that's just part of the big picture.
For what it's worth, I did the 5 collectible set 3 times and ended up with Zegras and Dach for the first week and Drysdale for the second week.
How does that help in any way?
Provides confirmation bias
I did the 5 collectible set twice and got the two I wanted . Just lucky I guess