# probability of winning dice rolling game

• Jan 10th 2012, 05:00 AM
Maxim
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?
• Jan 11th 2012, 11:33 AM
icemanfan
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.
• Jan 13th 2012, 02:46 AM
Maxim
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, $\displaystyle 0\leq k\leq 3$ and $\displaystyle 0\leq n\leq 4$ and $\displaystyle P(3,n)=1\ \forall n$ and $\displaystyle Q(k,4)=0\ \forall k$.

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

Now I can define:

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

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 $\displaystyle n\geq 4 \wedge k<3$, and Q(k,n)=1 if $\displaystyle k\geq 3 \wedge n<4$, i.e. I just threw my third four or he just threw his fourth three, and the game is now over. A situation where $\displaystyle k\geq 3 \wedge n\geq 4$ can never occur.
• Jan 13th 2012, 07:11 AM
Vinod
Re: probability of winning dice rolling game
K
Quote:

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
• Jan 16th 2012, 02:10 AM
Maxim
Re: probability of winning dice rolling game
Quote:

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.
• Jul 17th 2013, 04:17 AM
Maxim
Re: probability of winning dice rolling game
Sorry to bump this 1.5 year old thread, but I never managed to solve it, and recently ran into this problem again. Would anyone have any idea?
• Jul 17th 2013, 06:07 AM
HallsofIvy
Re: probability of winning dice rolling game
There are three possible outcomes- you win, your opponent wins, or the game continues without either person getting their goal until you decide to give it up. Because the probability of rolling a four is the same as the probability of rolling a three, the probability of you winning is exactly the same as the probability of your opponent winning. The only question, then, is the probability of giving up the game with either winning- and that depends upon how long you are willing to go. Letting "P" be the probabability you win (and so also the probabiity your opponent wins) and Q the probability you give up the game with either winning, P+ P+ Q= 1 so 2P= 1- Q so P= (1- Q)/2. Everything depends on Q which depends on how long you are willing to continue the game.
• Jul 17th 2013, 06:34 AM
Maxim
Re: probability of winning dice rolling game
Quote:

Originally Posted by HallsofIvy
Because the probability of rolling a four is the same as the probability of rolling a three, the probability of you winning is exactly the same as the probability of your opponent winning.

Watch out:
Quote:

Originally Posted by Maxim
When I roll three fours in a row, I win. When you roll four threes in a row, you win.

But I see your point about the chance of giving up. For this particular situation we can assume this to be zero, and yes I agree that if we both needed 3 times the same number in a row to win, it'd be ˝ - ˝.

But in this case (sequence of 4 times vs sequence of 3 times to win) it's more complicated.