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### Yahtzee Probability

```Date: 07/20/2009 at 10:43:04
From: Bob
Subject: I rolled Yahtzee's on 3 consecutive turns 2 times

I rolled Yahtzee's on 3 consecutive turns (in other words within
those 9 rolls on 3 turns) and I actually have done this 2 times.
What is the odds of doing it once and what are the odds you could
ever do it again or for a second time please?

```

```
Date: 07/20/2009 at 15:51:17
From: Doctor Tom
Subject: Re: I rolled Yahtzee's on 3 consecutive turns 2 times

Hi Bob,

It's sort of a hard question.  In a real game of Yahtzee, if you
rolled a full-house, like {1,1,1,2,2}, you'd probably stop there,
depending on whether you had one already.  But I'll just assume that
on each series of rolls you're determined to get Yahtzee: with the
full house above, you'd re-roll the 2's.

In that case, you can set up the game as a Markov chain that can be
represented by a 5x5 matrix.  The situation at any stage, including
the start, is that you're in one of 5 states: no matches, a pair, a
triple, a quad, and a Yahtzee.  Before the game starts, you're in that
initial state, and may remain in that state if on your roll, all the
dice are different.

So the probabilities of being in any state are given by a vector like
this:

(a, b, c, d, e)

where a represents the probability of having no matches, b the
probability of having one match, and so on.  When you begin, that
vector looks like this:

(1, 0, 0, 0, 0)

Now you need to multiply that by the matrix that represents the
probabilities of going from one state to the next.   Here's that
matrix in Mathematica form:

M = {{720/7776, 0, 0, 0, 0},
{5200/7776, 120/216, 0, 0, 0},
{1500/7776, 85/216, 25/36, 0, 0},
{150/7776, 15/216, 10/36, 5/6, 0},
{6/7776, 1/216, 1/36, 1/6, 1}}

(This took a bunch of work for me to calculate: if you know something
about combinatorics, it's a good exercise to re-obtain it.  Remember
that if you start with a pair, there's a chance of getting a triple by
rolling three dice and getting three of a number that's not the
original pair).

The third power of the matrix above is:

{{125/157464, 0, 0, 0, 0},
{349375/1417176, 125/729, 0, 0, 0},
{10439375/22674816, 129625/279936, 15625/46656, 0, 0},
{460625/1889568, 5125/15552, 11375/23328, 125/216, 0},
{1036241/22674816, 22141/279936, 8281/46656, 91/216, 1}}

and the number you want is the probability of being in state 5,
starting from state 1, which is:

1036241/22674816 = 0.04570008418.

That's the chance of doing it once on any given set of up to three
rolls.  The chance of doing it three times in a row is just the third
power of this, or:

0.00009544452

which is a little less than one time in 10,000 attempts.  This will
never change:  Even if you just did it, and started again, it will
still happen about one time in ten thousand on the next set of three
Yahtzee series.

- Doctor Tom, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Probability

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