# Problem 120 Solution

Let Ex = expected number of more rolls to obtain a yahtzee given the player already has an x of a kind.

The problem is to find E1.

E5=0 since the player already has a yahtzee and needs no further rolls.

E4=1+(5/6)*E4+(1/6)*E5

6*E4=6+5*E4+E5

E4=6+E5

E4=6

Next solve for E3

E3 = 1 + (25/36)*E3 + (10/36)*E4 + (1/36)*E5

36*E3 = 36 + 25*E3 + 10*E4 + E5

11*E3 = 36 + 10*E4 + E5

11*E3 - 10*E4 + E5 = 36

Next solve for E2

At this point it starts to get less obvious what the probabilities are to advance closer to a yahtzee. Lets assume the two of a kind is of the number a. Below are the number of ways to advance closer and the number of corresponding combinations.

abc: 3!*combin(5,2)=60
abb: 3*5=15
bbb: 5

So there are 60+15+5=80 ways to progress to a three of a kind.

aab: 3*5=15 So there are 15 ways to progress to a four of a kind.

There is obviously only 1 way to progress to a five of a kind.

cde: 3!*combin(5,3)=60
bbc: 5*4*3=60
So there are 60+60=120 ways to stay at a pair.

There are 63=216 ways to roll 3 dice.

E2 = 1 + (120/216)*E2 + (80/216)*E3 + (15/216)*E4 + (1/216)*E5

216*E2 = 216 + 120*E2 + 80*E3 + 15*E4 + 1*E5

96*E2 = 216 + 80*E3 + 15*E4 + 1*E5

96*E2 = 216 + 80*E3 + 15*E4

96*E2 - 80*E3 - 15*E4 = 216

Next solve for E1. Lets assume that the one of a kind is already an a.

There is obviously 1 way to get to a yahtzee.

Going to a four of a kind:

aaab: 5*4=20
bbbb: 5

20+5=25

Going to a three of a kind:

aabc: combin(5,2)*4!/2! = 10*24/2 = 120
aabb: 5*combin(4,2) = 5*6 = 30
abbb: 5*4 = 20
bccc: 5*4*4 = 80
120+30+20+80 = 250

Going to a pair:

abcd: combin(5,3)*4! = 240
abbc: 5*4*4!/2! = 240
bbcc: combin(5,2)*combin(4,2) = 60
bbcd: 5*combin(4,2)*4!/2! = 360
240+240+60+360 = 900

Staying at one of a kind:

bcde: combin(5,4)*4! = 120

E1 = 1 + (120/1296)*E1 + (900/1296)*E2 + (250/1296)*E3 + (25/1296)*E4 + (1/1296)*E5

1296*E1 = 1296 + 120*E1 + 900*E2 + 250*E3 + 25*E4 + 1*E5

1176*E1 = 1296 + 900*E2 + 250*E3 + 25*E4 + 1*E5

1176*E1 = 1296 + 900*E2 + 250*E3 + 25*E4

1176*E1 - 900*E2 - 250*E3 - 25*E4 = 1296

So now we have four equations and four unknowns and can solve with matrix algebra. To summarize here is what we know thus far:

1176*E1 - 900*E2 - 250*E3 - 25*E4 = 1296

96*E2 - 80*E3 - 15*E4 = 216

11*E3 - 10*E4 + E5 = 36

E4=6

Here is the matrix that needs to be solved.

 1176 -900 -250 -25 1296 0 96 -80 -15 216 0 0 11 -10 36 0 0 0 1 6

I'll leave the solving to the matrix to you. Here are the final answers.

E1 = 151496136/13660416 =~ 11.09015538

E2 = 11046/1056 =~ 10.46022727

E3 = 96/11 =~ 8.727272727

E4 = 6

Thanks to Nick Hobson for suggesting a similar problem.

Michael Shackleford, ASA