Problem 120 Solution

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.

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.