Probability of rolling a 6 immediately after a 1 is rolled

Question

Question

Ann and Bob take turns to roll a fair six-sided die. The winner is the first person to roll a six immediately after the other person has rolled a one. Ann will go first. Find the probability that Ann will win.

Answer

$\mathbb{P} (\mathrm {Ann\ wins}) = \frac {36} {73}$

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I have thought long and hard about this question but I am unable to even start. I have tried considering cases, but got stuck along the way. For example, it is trivial to calculate the probability that Ann wins if there are only three rolls (which is the minimum number of rolls needed for Ann to win). However, the problem easily becomes very complicated when we consider five rolls and more.

The suggested solution by my professor uses first-step decomposition, but it is a new concept to me and I am struggling to understand it.

If anyone can provide a detailed and intuitive explanation as to how this problem should be solved, that will be greatly appreciated!

Answer 1

This works well with states.

There are only $4$ active states:

Let $(A,0)$ denote the state in which it is $A's$ turn and the prior toss was not a $1$ (this includes the START state).

Let $(A,1)$ denote the state in which it is $A's$ turn and the prior toss was a $1$

Let $(B,0)$ denote the state in which it is $B's$ turn and the prior toss was not a $1$

Let $(B,1)$ denote the state in which it is $B's$ turn and the prior toss was a $1$

For any state $S$ let $P(S)$ denote the probability that $A$ will eventually win, given that we are now in state $S$. We see that $$P(A,0)=\frac 16P(B,1)+\frac 56P(B,0)$$ $$P(A,1)=\frac 16\times 1+\frac 16P(B,1)+\frac 46P(B,0)$$ $$P(B,0)=\frac 16P(A,1)+\frac 56P(A,0)$$ $$P(B,1)=\frac 16\times 0 + \frac 16P(A,1)+\frac 46P(A,0)$$

This system is easily solved and confirms the official result, here

Answer 2

You should solve it using Markov Chains by constructing the set of states varying from the absorbing state $s_{-2}$ (Bob wins) and the absorbing state $s_3$ (Alice wins). I denote the starting state $s_{\emptyset}$. Unless I made a silly mistake somewhere, you should get this set of recurrent equations: $$ p_{\emptyset, 3} = \frac{5}{6}p_{1,3} + \frac{1}{6}p_{-1,3}\\ p_{1,3} = \frac{1}{6}p_{2,3} + \frac{5}{6}p_{\emptyset,3} \\ p_{2,3} = \frac{1}{6}p_{3,3} + \frac{4}{6}p_{\emptyset,3} + \frac{1}{6}p_{-1,3}\\ p_{-1,3} = \frac{1}{6}p_{-2,3} + \frac{1}{6}p_{2,3} + \frac{4}{6}p_{\emptyset,3} $$
The boundary values are $p_{3,3}=1, p_{-2,3} = 0$. I'm pretty sure of this setup, but perhaps I made a typo somewhere during calculations, as I keep getting $p_{\emptyset,3}=\frac{18}{73}$.

Answer 3

I am only guessing on the "First Step Decomposition" technique you referred to in your question, because although I know a method suitable for this question, I do not know the name for the method.

Suppose the answer is $p$. Notice that the first roll has three types of outcomes:

• Ann rolls 2, 3, 4, 5 or 6. The probability of Ann winning after that is $(1-p)$, since the game proceed with exactly the same rules, except Ann and Bob exchanges position.
• Ann rolls a 1. In this case, let's suppose Bob's winning probability becomes $q$. Either Bob rolls a 6 (probability $1/6$) and wins the game, or
  • Bob rolls 2, 3, 4, or 5. The probability of Ann winning after this is $p$ again.
  • Bob rolls 1 too. The probability of Ann winning is also $q$.

And let's summarize. The total winning probabilities add up to $p$: $$p = \frac56 (1-p) + \frac16 (1-q).$$ Also there is a relation between $p$ and $q$: $$q = \frac16 + \frac46 (1-p) + \frac16 (1-q).$$

Can you see why these relations hold?

Now solve the equations. You get $p = \frac{36}{73}$.

Answer 4

Let the probability of A winning from start be $s$, and from the brink of winning be $t$, then by symmetry, the corresponding state probabilities for $B$ will be $(1-s)\;$ and$\; (1-t)$, and a first step analysis gives

$\displaylines{s = \frac56(1-s) +\frac16(1-t)\\(1-s)=\frac56s +\frac16t\\(1-t) = \frac16\cdot0 +\frac46s +\frac16t}$

[Note: The zero multiplier in the last equation is because we want to find $\Bbb P$($A$ wins), so $B$ can't be allowed to win]

Wolfram gives the answer as $s= \frac{36}{73}, t = \frac{42}{73}$

Answer 5

This is my attempt at using Markov chains to solve the problem.

Following one of the answers posted, we have $4$ active states and $2$ absorbing states.

Let the $4$ active states be $(A, 0)$, $(A, 1)$, $(B, 0)$ and $(B, 1)$, where

$(A,0)$ is the state in which it is $A$'s turn and the prior flip was not a $1$,

$(A,1)$ is the state in which it is $A$'s turn and the prior flip was a $1$,

$(B,0)$ is the state in which it is $B$'s turn and the prior flip was not a $1$ and

$(B,1)$ is the state in which it is $B$'s turn and the prior flip was a $1$.

Let the $2$ absorbing states be $(A)$ and $(B)$, where

$(A)$ is the state in which $A$ wins and

$(B)$ is the state in which $B$ wins.

Note also that $(A, 0)$ shall be the state in which the game starts.

In order of $(A, 0)$, $(A, 1)$, $(B, 0)$, $(B, 1)$, $(A)$ and $(B)$, the corresponding transition matrix shall be

$$T = \begin{pmatrix} 0 & 0 & \frac 5 6 & \frac 1 6 & 0 & 0 \\ 0 & 0 & \frac 4 6 & \frac 1 6 & \frac 1 6 & 0\\ \frac 5 6 & \frac 1 6 & 0 & 0 & 0 & 0 \\ \frac 4 6 & \frac 1 6 & 0 & 0 & 0 & \frac 1 6 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix}.$$

Then, we are simply interested in $T^n_{1, 5}$ as $n \rightarrow \infty$, which can be easily computed to give approximately $0.493$, as required.

P.S. This is my very first time working with Markov chains and all of my learning regarding this concept is independent, so do feel free to provide any comments on how I can improve my working! I do believe I will be covering Markov chains soon, so this could be a nice head start!