Chance of winning simple dice game

Question

Tossing two fair dice, if the sum is 7 or 11, then I win; if the sum is 2, 3 or 12, then I lose; if the sum is one of rest of numbers then I toss the two dices again. What is probability of winning?

What I have tried: every tossing, p(win)=$\frac{2}{9}$, and p(lose)=$\frac{1}{9}$, p(tossing again)=$\frac{2}{3}$, so the probability of winning is $\frac{2}{9}+\frac{2}{3}\frac{2}{9}+(\frac{2}{3})^2\frac{2}{9}+(\frac{2}{3})^3\frac{2}{9}+.....=\frac{2}{9}\sum^{\infty}_{n=0}(\frac{2}{3})^n=\frac{2}{3}$

Does anyone could tell me why my idea is wrong? What is the correct solution? Thanks!

Answer 1

First: how many ways can a win occur? On a 7 or an 11 roll you win (which can occur as 1,6;2,5;3,4;4,3;5,2;6,1 or 5,6;6,5), occurring with a probability then of $\frac{8}{36} = \frac{2}{9}$.

Losing occurs with a 2, 3, or 12 (which can occur as 1,1; 6,6; 1,2; or 2,1) with a probability of $\frac{4}{36} = \frac{1}{9}$.

I will assume at the moment that the rules for the game are such that if you roll a different total sum you get to reroll. If you again roll a different different total sum, you still get to reroll, and reroll, potentially indefinitely.

You can describe this with an absorbing Markov chain with three states. There are several different methods to handle markov chains, one of which you correctly showed above in your post (in the special case that all states but one are absorbing).

The method I like to use personally is the following: Set up the transition matrix (where the $i$th row $j$th column entry corresponds to the chance to change from the $j$th state to the $i$th state) with the possible states being in this case Win-Lose-Redo:

$A = \begin{bmatrix} 1 & 0 & \frac{2}{9}\\ 0 & 1 & \frac{1}{9} \\ 0 & 0 & \frac{6}{9}\end{bmatrix}$

This matrix is in the form $A = \begin{bmatrix} I & S\\0& R\end{bmatrix}$ and $\lim\limits_{n\to\infty} A^n = \begin{bmatrix}I&S(I-R)^{-1}\\0&0\end{bmatrix}$

$I-R = [\frac{1}{3}]$ and $(I-R)^{-1} = [3]~~~\dagger$

As such, $\lim\limits_{n\to\infty} A^n = \begin{bmatrix}1&0&\frac{2}{3}\\0&1&\frac{1}{3}\\0&0&0\end{bmatrix}$

Multiplying by our initial state distribution of $\begin{bmatrix}0\\0\\1\end{bmatrix}$ we get a final distribution of $\begin{bmatrix}\frac{2}{3}\\\frac{1}{3}\\0\end{bmatrix}$.

Interpreting the answer, it says that 2/3 of the time you will win, 1/3 of the time you will lose, and according to $\dagger$ above, it takes an average of three throws of the dice for the game to end.

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Assuming that the rules of the game are that you either get to win on the first die roll, or on the second, but if you fail to win on the second die roll then you automatically lose (even if the sum rolled was not a 2,3, or 12).

In this case, it is easier to think in terms of tree diagrams. Your chances of winning on the first turn are $\frac{2}{9}$ (by having rolled a winning number) and you chances of winning on the second turn are $\frac{2}{3}\cdot\frac{2}{9} = \frac{4}{27}$ (since it requires first having rolled a redo-number and then a winning number). Adding these together you get $\frac{2}{9}+\frac{4}{27}= 0.\overline{370}$.

If you are allowed a maximum of three rolls, then it becomes $\approx 0.469$. If allowed a maximum of four rolls, it becomes $\approx 0.535$.

Answer 2

With your edited solution i think you are now correct with 2/3.

Since $$\frac{\frac{2}{9}}{\frac{2}{9}+\frac{1}{9}}=\frac{2}{3}$$

Answer 3

Your idea is correct, and so is your answer $\frac23$. However, there is a route that avoids calculating infinite sums and is in my view more elegant.

Let $W$ denote the event that you win and let $X$ denote the sum at the first toss. Then:

$$\mathbb{P}\left(W\right)=\mathbb{P}\left(W\mid X\in\left\{ 7,11\right\} \right)\mathbb{P}\left(X\in\left\{ 7,11\right\} \right)+\mathbb{P}\left(W\mid X\in\left\{ 2,3,12\right\} \right)\mathbb{P}\left(X\in\left\{ 2,3,12\right\} \right)+\mathbb{P}\left(W\mid X\notin\left\{ 2,3,7,11,12\right\} \right)\mathbb{P}\left(X\notin\left\{ 2,3,7,11,12\right\} \right)$$

Of course $\mathbb{P}\left(W\mid X\in\left\{ 7,11\right\} \right)=1$ and $\mathbb{P}\left(W\mid X\in\left\{ 2,3,12\right\} \right)=0$.

But also we have: $$\mathbb{P}\left(W\mid X\notin\left\{ 2,3,7,11,12\right\} \right)=\mathbb{P}\left(W\right)$$ This because you just start over under this condition and the past looses its influence.

This leads to: $\mathbb{P}\left(W\right)=\frac{2}{9}+\mathbb{P}\left(W\right)\frac{2}{3}$ and consequently $\mathbb{P}\left(W\right)=\frac{2}{3}$.