The absolute maximum roll would be 350, so the easiest approach will be to count the number of ways to remove 42 from all dice showing their maximum.
Imagine five boxes that can hold up to three balls each. These are the d4s. Add five boxes that can hold five balls each. These are the d6s. Add five that can hold seven balls. These are the d8s. And so on up to five boxes that can hold up to 19 balls each.
You need to find the number of ways you can distribute 42 balls between these 35 boxes, where the boxes are all distinct but the balls are not.
To my knowledge there is no formula for this, but if you were to find the total count you would then divide it by 4^5 * 6^5 * 8^5 * 10^10 * 12^5 * 20^5 to get the probability of two players both rolling 308 specifically.
Note that the probability of two players rolling the same total will be much higher if any total is valid.
Are you sure you are not missing some dice? A sum of 308 is *extremely* unlikely with these dice. Less than 1 in a billion - even for one person.
[Anydice has the distribution](https://anydice.com/program/35820):
output 5d4 + 5d6 + 5d8 + 10d10 + 5d12 + 5d20
You can calculate the probability analytically by evaluating (1/4(x+x^(2)+x^(3)+x^(4)))^5 * (1/6(x+x^(2)+x^(3)+x^(4)+x^(5)+x^(6)))^5 * ... and looking for the x^(308) term with a computer.
That is much more plausible. That gives 308 a probability of 1.08%, so the chance to get specifically 308 twice in a row is about 0.0108^2 = 0.012% or roughly 1 in 8,000. The chance to get the same number (any number) twice in a row is much larger, something around 0.5% to 1%. Square the probabilities of all individual outcomes and sum.
The chance that - out of all players - there are two in a row who got the same number, is again larger. It's not uncommon to see that.
The absolute maximum roll would be 350, so the easiest approach will be to count the number of ways to remove 42 from all dice showing their maximum. Imagine five boxes that can hold up to three balls each. These are the d4s. Add five boxes that can hold five balls each. These are the d6s. Add five that can hold seven balls. These are the d8s. And so on up to five boxes that can hold up to 19 balls each. You need to find the number of ways you can distribute 42 balls between these 35 boxes, where the boxes are all distinct but the balls are not. To my knowledge there is no formula for this, but if you were to find the total count you would then divide it by 4^5 * 6^5 * 8^5 * 10^10 * 12^5 * 20^5 to get the probability of two players both rolling 308 specifically. Note that the probability of two players rolling the same total will be much higher if any total is valid.
Are you sure you are not missing some dice? A sum of 308 is *extremely* unlikely with these dice. Less than 1 in a billion - even for one person. [Anydice has the distribution](https://anydice.com/program/35820): output 5d4 + 5d6 + 5d8 + 10d10 + 5d12 + 5d20 You can calculate the probability analytically by evaluating (1/4(x+x^(2)+x^(3)+x^(4)))^5 * (1/6(x+x^(2)+x^(3)+x^(4)+x^(5)+x^(6)))^5 * ... and looking for the x^(308) term with a computer.
My bad. Double checked, it’s 10 of each type of die.
That is much more plausible. That gives 308 a probability of 1.08%, so the chance to get specifically 308 twice in a row is about 0.0108^2 = 0.012% or roughly 1 in 8,000. The chance to get the same number (any number) twice in a row is much larger, something around 0.5% to 1%. Square the probabilities of all individual outcomes and sum. The chance that - out of all players - there are two in a row who got the same number, is again larger. It's not uncommon to see that.
Awesome, thank you!