Yahtzee Probabilities and a State Matrix

Date: 12/30/98 at 21:26:47
From: Bruce A Ryan
Subject: Rolling dice

In playing the game Yahtzee, what are the odds of rolling 5 of a kind 
within three rolls of the dice?

Date: 01/01/99 at 04:54:08
From: Doctor Pat
Subject: Re: Rolling dice

I'll start by making some assumptions. First, the goal is only to get 
five of a kind. If we get two or three or four of a kind we pick up the 
remaining dice and roll to improve our position, and for two pair or 
full house we pick up (randomly) the best of two or the best hand, and 
so on.

Given that I think the probabilities on the first roll are :

   P(5 of a kind) = 1/6^4
   P(4 of a kind) = 25/1296
   P(3 of a kind) = 125/648
   P(2 of a kind) = 75/108
   P(no two alike) = 5/54 (here we pick up all five and roll again)

This gives a transition matrix from roll to roll equal to the matrix 
below:

  from   to    1          2          3          4           5 

   1          5/54      75/108    125/648     25/1296      1/6^4
   
   2           0         5/9*      10/27**    15/6^3       1/6^3

   3           0          0        25/36       5/18        1/6^2
 
   4           0          0          0         5/6         1/6
 
   5           0          0          0          0           1

        * If all three of the other dice come up the the same value,
          that's a transition from state 2 to state 3:
          p = 125/216 - 5/216 = 120/216 = 5/9.

       ** If all three of the other dice come up with the same value,
          that's a transition from state to to state 3... but with a
          different value:  p = 75/216 + 5/216 = 80/216 = 10/27
             
Taking the third power of this matrix gives us the probability of 
ending in any of the five states. I got the following probabilities 
for the result after three rolls:

   P(no two alike) = .00079
   P(two alike)    = .256
   P(three alike)  = .4524
   P(four alike)   = .2447
   P(five alike)   = .046

If those don't add up to exactly one, the reason is roundoff error. In 
all, there is only about a 5% probability of getting a "Yahtzee."

For more information on transition matrices, please see:

   Probability Transition Matrices
   http://mathforum.org/library/drmath/view/54265.html   

- Doctor Pat, The Math Forum
  http://mathforum.org/dr.math/   

Date: 01/01/99 at 17:45:47
From: Bruce Ryan
Subject: Re: Rolling dice

Dear Dr. Pat,

I received your answer to my question. I had asked: in playing the game 
Yahtzee, what are the odds of rolling five of a kind within three rolls 
of the dice? However, I think I misled you. My question actually 
pertains more to just rolling 5 dice for 5 of a kind. 

I probably should have written that in a game we have at our social 
quarters, we are allowed to roll 5 dice just 3 times for a quarter. If 
you roll 5 of a kind you win the built up jackpot. You are only allowed 
to roll the dice using all 5 dice each time you roll. In other words, 
if you roll 3 of a kind on your first roll, that doesn't count. You 
have to pick up all 5 dice and roll again. What are the odds of rolling 
5 of a kind out of these 3 rolls? 

Thank you very much for your time.

Bruce A. Ryan

Date: 01/02/99 at 02:22:48
From: Doctor Pat
Subject: Re: Rolling dice

Bruce,

Boy, that is a lot easier. I'm glad you didn't ask that first or I 
would not have had the fun of doing all the other stuff. But now about 
the new situation:

The probability, as I said in my first response, is 1/6^4 or about 
.00077 that you will succeed on any roll. This means that in three 
tries the probability is 770 out of a million that you will win on the 
first try. The probability that you will win on the second and third 
try is almost exactly the same (actually minutely less.) This means 
the overall probability is about two chances out of a thousand roughly 
(3*.00077). If it happens to someone more than once, be very 
suspicious. 

Good luck,

- Doctor Pat, The Math Forum
  http://mathforum.org/dr.math/   

Date: 01/12/99 at 10:16:14
From: Florian
Subject: Dice probability - Yahtzee

Hi, 

I'm sorry but I cannot follow your thoughts completely. How did you 
get the transition matrix? How does it work? How can you get the 
probability from this matrix?

Sincerely,
Florian

Date: 01/12/99 at 16:25:04
From: Doctor Pat
Subject: Re: Dice probability - Yahtzee

Florian,

The matrix is called a state matrix, and it gives the probability that 
you will move from any "state" to the same or another state. In your 
problem the "states" would be the maximum number of dice showing the 
same number after each roll. If you draw five circles in sort of a 
pentagon and put the numbers one to five in them, you can draw a graph 
of what the matrix shows. If you think of the one as the position of no 
two alike, this is where you are before you roll. On the first roll, 
five things may happen:

1. You may roll five different numbers. This would be indicated on your 
graph by a line from (1) looping back to (1). The probability of this 
event is given in row 1 (from 1) column 1 (to 1) of the matrix. 

2-5. You may have two alike of one or more numbers. This would be 
indicated on your graph by a line from (1) to (2). The probability of 
this event is given in row 1, column 2. The same thing happens for 
three of a kind (from (1) to (3)) - row 1 column 3, etc.
  
After the first roll you may be at any one of the positions. Of course 
if you are at five, you quit and win. This means the circle (5) only 
has a loop back to itself, and the probability in row 5 is all zero 
except for column 5 which has probability of one.  

On any subsequent roll, the probability that you will move from, say, 
circle (3) to circle (4) is given by the matrix row 3, column 4. The 
matrix first row, then, gives you the probability of being in any state 
after one roll. A great Russian mathematician named Markov showed that 
if you multiply this matrix times itself, you get the probabilities for 
positions after two rolls, and if you raise it to the third power, you 
get the probabilities after three rolls.  

To get the actual values in each cell, I drew out a tree diagram of 
every possible position after three rolls and computed individual 
probabilities, but once I had the transition matrix, all else was easy.  
You can look this up using the search engine at the Math Forum. Try 
the search words:

   Markov matrix

Good luck,

- Doctor Pat, The Math Forum
  http://mathforum.org/dr.math/   

Date: 05/01/2002 at 21:39:17
From: Phil Lindsey
Subject: Yahtzee Odds

Would you please check the answer given by Dr. Pat above?  
Specifically, I think that line 2 in the Transition Matrix should be 
as follows:

To                  1         2         3         4        5

From 2              0      125/216    75/216    15/216    1/216


I will be able to sleep better if I know I am not going crazy!
Thank you.   
Phil Lindsey

Date: 05/02/2002 at 14:38:40
From: Doctor Ian
Subject: Re: Yahtzee Odds

Hi Phil,

The first difference is that you get 125/216 for the probability of 
going from 2 of a kind back to 2 of a kind, while Dr. Pat gets 5/9. 

I think your reasoning is that we already have two of some value, so 
the next three dice all have to be different from this value: 
(5/6)^3 = 125/216.  

However, this doesn't take into account that if all three of the dice 
come up with the same value, you're no longer going from state 2 to 
state 2, but from state 2 to state 3. So you have to exclude the 
possibility that the three dice come up with the same value.

That can happen in 5 ways, so the probability is

  125/216 - 5/216 = 120/216 = 5/9

The second difference is that you get 75/216 for the probability of 
going from 2 of a kind to 3 of a kind.  

I think your reasoning is that we already have two of some value, so 
one of the next three dice has to be the same, and the other two have 
to be different. That can happen in three ways, so: 

    (1/6 * 5/6 * 5/6) + (5/6 * 1/6 *5/6) + (5/6 * 5/6 * 1/6)

  = 3(1/6 * 5/6 * 5/6)

  = 75/216

However, suppose that all three dice turn up the same? Then we've also 
moved into state 3... but with a different value!  

The probability of rolling 3 of a kind with three dice is 1/216, and 
we have five possible values to choose from, so the transition 
probability is

  75/216 + 5/216 = 80/216 = 10/27

Does this help? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/