Roll a dice, probability that game goes beyond one round

Question

There are two people $A$ and $B$. Person $A$ has $8$ chips, and person $B$ has $6$ chips. Each round, both people roll a die then take that number of chips from the other person. The game ends when one person has more chips than the other.

Question 1) What is the probability that the game goes beyond one round?

Question 2) What is a general formula for the probability that the game ends in round $n$?

My answer for (1) is $5/36$. There are five ways to continue the game (a tie): $$A1B2 \qquad A2B3 \qquad A3B4 \qquad A4B5 \qquad A5B6$$

I didn't figure out (2). Can someone help me out? Thanks!

Answer 1

The answer for 2nd part should be: $$ \frac{31}{36} \times \left( \frac{1}{6} \right)^{n-2} \times \frac{5}{6}$$ As after first round, both of them should get same no. until (n-1) round

Answer 2

We have players $A$ and $B$ with some fair dice $D_A=\{1,...,a\}$ and $D_B=\{1,...,b\}$ (i.e. uniform probability for any atomic event $\Pr(D=x)=\frac{1}{|D|}\ $) with number of sides $|D_A|=a$ and $|D_B|=b$ respectively.

Each player have some score that I will name $sA$ and $sB$ where the important thing is the difference between them i.e. $S=sA-sB$.

And we have that for any situation that compare the rolls

$$\Pr(D_A\ \mathbf R\ D_B)=\sum_{D_A=1}^{a}\frac{1}{a}\sum_{D_B=1}^{b}\frac{1}{b}[D_A\ \mathbf R\ D_B]=\frac{1}{ab}\sum_{D_A=1}^{a}\sum_{D_B=1}^{b}[D_A\ \mathbf R\ D_B]$$

for any kind of relation of comparison $\mathbf R$ where the brackets are Iverson brackets.

For any turn exists three different outcomes: $S>0$ where player A wins and the game ends, $S=0$ where is a tie and game continue, or $S<0$ where player B wins and game ends (I understand that these conditions are for all game but $S_0$, see below).

We start with $S_0=2$, the value of $S$ changes depending on the difference between the rolls of both players i.e. $S_{n+1}=S_n+2(D_A-D_B)_n$.

If the dice have equal number of sides $|D|=m$ we can simplify the above expression as

$$\Pr(D_A\ \mathbf R\ D_B)=\frac{1}{m^2}\sum_{1\le D_A,D_B\le m}[D_A\ \mathbf R\ D_B]$$

Because the game have 3 possible outcomes we have 3 different possible conditions:

$$A\ \text{wins} \iff D_A - S/2 > D_B\\ B\ \text{wins} \iff D_A - S/2 < D_B\\ \text{Tie} \iff D_A - S/2= D_B $$

and in any case $S_{n\ne0}/2=0$ and $S_0/2=1$, in other cases the game ends and we dont need take care about any value of $S/2$ anymore :)

Go to probabilities. For the first game we have:

$$\Pr_{n=1}(D_A -1< D_B)=\frac{m+1}{2m}\\ \Pr_{n=1}(D_A -1> D_B)=\frac{(m-2)^\overline{2}}{2m^2}\\ \Pr_{n=1}(D_A -1= D_B)=\frac{m-1}{m^2}$$

where $c^\overline{k}$ denotes the rising factorial. If the game continues then:

$$\Pr_{n>1}(D_A < D_B)=\Pr_{n>1}(D_A > D_B)=\frac{m-1}{2m}\\ \Pr_{n>1}(D_A = D_B)=\frac{1}{m}$$

Notice that in any turn the probabilities $\sum\Pr(X=x)=1$

The probabilities that game ends in the turn n is the probability that it doesnt on $n-1$ turns and it does just in the $n$th turn. Fixing the case for n=1 will have that the probability that game ends exactly in the turn n will be

$$\Pr(N=n)=\frac{(m-1)^2+m[n=1]}{m^{n+1}}$$

If we are using a standard dice with $m=6$ then

$$\Pr(N=n)=\frac{25+6[n=1]}{6^{n+1}}$$

Answer 3

The answer to the first part is ${5 \over 36}$ not ${31 \over 36}$. You're counting the number of ties divided by the number of outcomes.

For the second part, you need the following things to all happen:

1. Round $1$ is a tie.
2. Rounds $2$ to $n-1$ are ties.
3. Round $n$ is not a tie.

The probability of 1 is ${5 \over 36}$.
For 2, the outcomes in each round must be one of $A1B1$, $A2B2$ to $A6B6$. The probability of that happening in a single round is ${6 \over 36} = {1 \over 6}$, and for $n-2$ rounds ${1 \over 6^{n-2}}$.
The probability of round $n$ being the last one is $1 - $ the probability of a tie. That's ${5 \over 6}$.

Of course, each of these events is independent so you multiply to get $${5 \times 6 \over 36} \cdot{1 \over 6^{n-2}}$$