TITLE: How do I calculate the distribution of an amount of dice throws?
QUESTION [1 upvotes]: Two players are throwing a (fair) dice in turns. Player 1 wins if he gets a 1 and Player 2 wins if he gets a 5 or a 6. If one of the players wins, the other one continues until he gets the required number.
If $X$ and $Y$ are the amount of throws of Player 1 and 2, how can I calculate the distributions of $X$ and $Y$?
I assume the distribution of getting any number is completely uniform (i.e $\frac{1}{6}$,$\frac{1}{6}$,$\frac{1}{6}$,$\frac{1}{6}$,$\frac{1}{6}$,$\frac{1}{6}$), but how do I formulate this in regards to the number of throws?
It seems to me that both distributions would be the same.

REPLY [3 votes]: According to your description, $X$ and $Y$ are statistically independent of one another. If player 1 at some point gets the number 1, only player 2 needs to continue. Likewise, if player 2 at some points gets the number 5 or 6, only player 1 needs to continue. This corresponds to a scenario in which both players keep throwing the dice until they hit (one of) the right number(s).
For $X$ to equal $n$, player 1 must fail on the first $n-1$ turns and succeed on the $n^{th}$ turn. This corresponds to a geometric distribution:
$$P[X=n] = \bigg(\frac{5}{6}\bigg)^{n-1}\frac{1}{6}$$
Similarly, we find:
$$P[Y=n] = \bigg(\frac{4}{6}\bigg)^{n-1}\frac{2}{6}$$
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