Question

You are three spaces from a win in Parcheesi.
On each turn, you will roll two dice. To win, you must roll
a total of 3 or roll a 3 on one of the dice. How many turns
might you expect this to take?

Answer 1

To win, the sum of two dice must equal 3 or one of the dice must equal 3.

Dice can sum to 3 in the following ways
1,2 or 2,1

There are 6*6 pairs of numbers with 2 dice and only 2 pairs of which will result in a win for this case, so probability that sum will be 3 is 2/36.

For the second case, you can get a 3 with 2 dice in the following ways
3,1
3,2
3,3
3,4
3,5
3,6
1,3
2,3,
4,3
5,3
6,3
or 11 ways out of 36, so the probability that either dice will be a 3 is 11/36

Each case is mutually exclusive, meaning one is not dependent on the other; so the total probability is 2/36 + 11/36 = 13/36.

So the number of turns required to win \(on~average\) will be the inverse of this or 36/13 or about 2.77 turns.

To see how the inverse makes sense for the number of turns, if your probability instead was 12/36=1/3 of winning, then 1/3 of the time you would expect to win or once in every 3 turns. So you win less frequently compared to when the probability was 13/36, where you win on average on 2.77 turns. So higher probabilities of winning lead to winning more frequently or in smaller number of turns, which makes sense intuitively.