# Thread: probability of winning dice rolling game

1. ## probability of winning dice rolling game

Struggling with a difficult (for me, at least) probability problem. Suppose we play a game where we repeatedly roll a dice. First me, then you, then me, etc.

When I roll three fours in a row, I win. When you roll four threes in a row, you win.

What's the probability of me winning the game?

2. ## Re: probability of winning dice rolling game

This is not an easy problem for me either. But this is how I would approach it:

1) Observe that you must roll three fours in a row at some point, and I must roll four threes in a row at some point. The game ends after you roll your third four in a row or I roll my fourth three.

2) Pretend that the game continues forever. Your rolls will be a list of randomly generated numbers in order, and so will mine. Let m represent the number of the roll in which you roll your third four in a row for the first time, and let n represent the number of the roll in which I roll my fourth three in a row for the first time. Then you win if m is less than or equal to n. So you must calculate the probability that m is less than or equal to n.

3. ## Re: probability of winning dice rolling game

Hmm yeah, the game could indeed go on forever. That will probably result in an infinite sum for the probability I need, and it's probably not gonna be a pretty one

I've been thinking some more about it, maybe this will get me anywhere:

I define P(k,n) to be the probability of me winning the game, eventually, if I currently have k fours in a row, and my opponent has n threes in a row, and it's my turn.
Likewise, Q(k,n) is the probability of me eventually winning the game when it's his turn.

Obviously, $0\leq k\leq 3$ and $0\leq n\leq 4$ and $P(3,n)=1\ \forall n$ and $Q(k,4)=0\ \forall k$.

What I'm looking for is P(0,0).

Now I can define:

$P(k,n) = \frac{1}{6}Q(k+1,n) + \frac{5}{6}Q(0,n)$
and
$Q(k,n) = \frac{1}{6}P(k,n+1) + \frac{5}{6}P(k,0)$

Since this is a limited situation (the boundaries being P(3,n)=1 and Q(k,4)=0) I'd say there should be an exact, explicit solution. But I'm not sure how to resolve this recurrence relation...?

Am I on the right track here or does anyone have a different idea?

PS: Maybe the boundaries should be P(k,n)=0 if $n\geq 4 \wedge k<3$, and Q(k,n)=1 if $k\geq 3 \wedge n<4$, i.e. I just threw my third four or he just threw his fourth three, and the game is now over. A situation where $k\geq 3 \wedge n\geq 4$ can never occur.

4. ## Re: probability of winning dice rolling game

K
Originally Posted by Maxim
Struggling with a difficult (for me, at least) probability problem. Suppose we play a game where we repeatedly roll a dice. First me, then you, then me, etc.

When I roll three fours in a row, I win. When you roll four threes in a row, you win.

What's the probability of me winning the game?
If you are to win 4 must be thrown on 1st,3rd,5th,……throws.Similarly if your opponent, say B is to win 3 must be thrown on 2nd,4th,6th,8th,……throws
Is it that you mean to say?
Explanation is required

5. ## Re: probability of winning dice rolling game

Originally Posted by Vinod
If you are to win 4 must be thrown on 1st,3rd,5th,……throws.Similarly if your opponent, say B is to win 3 must be thrown on 2nd,4th,6th,8th,……throws
Correct, the odd dice throws are 'my turns' (in which I can win, if I throw my 3th four at that time). The even dice throws are my opponent's turns (in which he can win, if he throws his 4th three at that time).

So the earliest throw at which I can win is 5th, and the earliest throw at which my opponent can win is 8th.