TITLE: Probability of matching faces on 2 separate pools of d10s
QUESTION [1 upvotes]: Thank you in advance for taking the time to read this.
I am trying to figure out the probability for one separate d10 to have a matching face with a pool of d10 comprised by up to 5 other d10s.
So just to make myself clear, if I roll up to 5 d10s and a separate d10 of a different color, what are the chances that said die matches with any of the other d10s?
I would like to know this probability for all 5 combinations (1d10,2d10,3d10,4d10 and 5d10 + the separate d10).
Also, after calculating the probability, if the "colored" die gets rolled twice, does the probability of at least one face matching simply double?
Cheers

REPLY [0 votes]: If you have one colored die, and you're comparing it to a pool of $N$ other dice, then the probability that any match your colored die is 1 minus the probability that they all mismatch.
The d10 has 10 sides, so the probability of a single mismatch is $(10-1)/10=0.9$.
And so the probability of $N$ mismatches is $0.9^N$.
The general formula for one d10 matching at least one die in a pool of $N$ other d10s is
$$P(N) = 1 - (0.9)^N$$
For the examples you gave,
$$P(1)=0.1$$
$$P(2)=0.19$$
$$P(3)=0.271$$
$$P(4)=0.3439$$
$$P(5)=0.40951$$
And the general formula for comparing an $X$-sided die versus a $N$-sized pool of $X$-sided dice would be
$$P(X,N) = 1 - ((X-1)/X)^N$$
Doubling the dice in the colored pool does not simply double the probability. I don't know the general formula, so I wrote up a matlab program to simulate all possible rolls, and got these values:
$$P_2(1)=0.19$$
$$P_2(2)=0.343$$
$$P_2(3)=0.4663$$
$$P_2(4)=0.5657$$
$$P_2(5)=0.646$$
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