# Probability : Roll of Die

$\textsf{A}$ and $\textsf{B}$ are playing a game with $2$ standard dice.

• Both the dice are rolled together and the total is counted.
• $\textsf{A}$ says that a total of $2$ will be rolled first.
• $\textsf{B}$, whereas, says that two Consecutive totals of $7$′s will be rolled first.
• They keep rolling the dice till one of them wins !.

What is the probability that $\textsf{A}$ wins the game ?.

For a total of $2$, $\{(1,1)\}$ and for a total of $7$, $\{(1,6),(6,1),(2,4),(4,2),(3,4),(4,3)\}$ are the required scenarios. I don't understand how we need to incorporate the probabilities of $\textsf{A}$ winning, i.e., $1/36$ and $\textsf{B}$ winning, i.e. $6/36$ into a game of infinite rounds, i.e. until $\textsf{A}$ wins. –

• What scenario gives a total of 2? What scenarios give a total of 7? – Colm Bhandal Mar 16 '18 at 17:46
• For a total of 2, {(1,1)} and for a total of 7, {(1,6),(6,1),(2,4),(4,2),(3,4),(4,3)} are the required scenarios. I don't understand how we need to incorporate the probabilities of A winning, i.e., 1/36 and B winning, i.e. 6/36 into a game of infinite rounds, i.e. until A wins. – S.Rana Mar 16 '18 at 18:07

## 2 Answers

To me this sounds like a good application of an absorbing Markov chain. There are four states:

1. “Neutral” where neither A nor B is ahead.
2. “B up” where B can win on the next roll.
3. “A wins”
4. “B wins”

The first two states are transitional and the last two terminal. The transition matrix between these four states is $$P = \begin{bmatrix} 29/36 & 1/6 & 1/36 & 0 \\ 29/36 & 0 & 1/36 & 1/6 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} Q & R \\ 0 & I \end{bmatrix}$$ Here $p_{ij}$ is the probability of moving from state $i$ to state $j$. The $2\times 2$ matrices $Q$ and $R$ are the top left and top right blocks of $P$. The fundamental matrix of this Markov chain is $$N = (I-Q)^{-1} = \begin{bmatrix} 216/13 & 36/13 \\ 174/13 & 42/13 \end{bmatrix}$$ The probability of absorption matrix is $$NR = \begin{bmatrix} 7/13 & 6/13 \\ 6/13 & 7/13 \end{bmatrix}$$ This means that from neutral (state 1), A has a $7/13$ chance of winning (state 3). However, once a 7 is rolled (state 2), B has a $7/13$ chance of winning (state 4).

• So is your conclusion that, since the game starts in a neutral state, A's probability of winning is $7/13$? – Adam Bailey Mar 18 '18 at 14:38
• @AdamBailey: Yep, same as you. In fact, this Markov method just dresses up the same calculations you made. – Matthew Leingang Mar 18 '18 at 17:04

Let $T_n2$ be the event consisting of throwing a total of $2$ on the $n$th throw of a pair of dice, $T_n7$ be throwing a total of $7$, and $T_n2,7$ be throwing a total of either $2$ or $7$.

By counting cases it is readily shown that: $P(T_n2)=1/36$, $P(T_n7)=1/6$, $P(T_n2,7)=7/36$.

If a pair of throws gives any total other than $2$ or $7$, the status of the game at the next pair of throws will be exactly as at the start of the game. So we can focus on conditional probabilities, conditional on $T_n2,7$. We have:

$P(T_n2 | T_n2,7) = (1/36) / (7/36) = 1/7$

$P(T_n7 | T_n2,7) = (1/6) / (7/36) = 6/7$

In the first case A has won. In the second we need to consider what happens at throw $n+1$. We have:

$P(T_n7 \ \&\ T_{n+1}2 | T_n2,7) = (6/7)(1/36) = 1/42$

$P(T_n7 \ \&\ T_{n+1}7 | T_n2,7) = (6/7)(1/6) = 1/7$

If throw $n+1$ yields any total other than $2$ or $7$ the game reverts to its initial status, so it suffices to consider the relative probabilities of A and B winning during throws $n$ and $n+1$. We have (conditional on $T_n2,7$, and noting that B cannot win at $n$):

P(A wins) = P(A wins at $t$) + P(A wins at $t+1$) = 1/7 + 1/42 = 1/6

P(B wins) = P(B wins at $t+1$) = 1/7

If we now remove the conditionality, the relative probabilities will not change. Therefore:

P(A wins) $= \frac{1/6}{1/6 + 1/7} = \frac{7}{7+6}=\frac{7}{13}$