Rolling Two Dice, Who Is lucky?

2015/10/09

Question 4-33 in the book Probability, Random Variables and Stochastic Processes 4th by Athanasios Papoulis and S. Unnikrishna Pillai is an interesting one. The question can be rephrased as: “Players $X$ and $Y$ roll two dice alternately starting with $X$. The player that rolls eleven wins. What is the probability that $X$ wins?” There are multiple approaches to solve it. I have found two approaches. One of them is suggested by the hint given in the book. The thinking process behind both two approaches is worth noting

Let us see an example of how this game will play out, which can help us get better understanding of the game.

round	#1	#2	#3	#4
X	12	5	8	11
Y	10	2	4

In this realization of a game, $X$ wins. $X$ wins at $4^{th}$ round.

The approach I found was to model the game differently. Observe that the possible outcomes of event we are interesting in is binary; getting eleven and not getting eleven. Furthermore, if we think $X$ and $Y$ rolling two dice as a single sequence of trails, we can see that it is a geometric distribution. Let it be $Z$, then for example, we could transform the previous table,

N	1	2	3	4	5	6	7
Z	12	10	5	2	8	4	11

Note that the event that $X$ wins will be translated to N is an odd number.

Now let us go a little formal. $N$ be the geometric random variable, with parameter $\alpha$. The $\alpha$ will be $\mathbb{P}(outcome\ is\ eleven)$. Then,

\[\begin{align} \mathbb{P}(X\ wins) &= \mathbb{P}(N \in \{2n-1|n \in \mathbb{Z}\}) \nonumber \\ &= \lim_{n\to\infty} \sum_{i=1}^{2n-1} P(N=i) \nonumber\\ &= \lim_{n\to\infty} \sum_{i=0}^{n} (1-\alpha)^{2i}\alpha \nonumber\\ \end{align}\]

This series will converge and an analytic solution can be obtained. Or we could use computer to approximate it.

Another approach is suggested by the hint in the book. The method requires us to spot the recurrent relationship in the problem. To do that, let us define some events: $A =$ {$X$ wins}, $M =$ {eleven shows up at first try}, then

\[\begin{align} P(A) = P(A|M)P(M) + P(A|\overline{M})P(\overline{M}) \end{align}\]

$P(A|M)$, $P(M)$ and $P(\overline{M})$ are $1$, $\alpha$ and $1-\alpha$ respectively.Then we need $P(A|\overline{M})$ to find $P(A)$. let us define $\beta$ be $P(A)$, then carefully looking at the problem, we can obtain $P(A|\overline{M}) = 1-\beta$. The argument is that, if the first trial is not eleven, the game can be think of as if $Y$ start first, which mean probability of $Y$ winning will be $\beta$, so $P(A|\overline{M}) = 1-\beta$. To sum up, the we obtain:

\[\beta = \alpha+(1-\beta)(1-\alpha) \\ \Rightarrow \beta = \frac{1}{2-\alpha}\]