The problem is to find E_{1}.

E_{5}=0 since the player already has a yahtzee and needs no further rolls.

E_{4}=1+(5/6)*E_{4}+(1/6)*E_{5}

6*E_{4}=6+5*E_{4}+E_{5}

E_{4}=6+E_{5}

E_{4}=6

Next solve for E_{3}

E_{3} = 1 + (25/36)*E_{3} + (10/36)*E_{4} + (1/36)*E_{5}

36*E_{3} = 36 + 25*E_{3} + 10*E_{4} + E_{5}

11*E_{3} = 36 + 10*E_{4} + E_{5}

11*E_{3} - 10*E_{4} + E_{5} = 36

Next solve for E_{2}

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 6^{3}=216 ways to roll 3 dice.

E_{2} = 1 + (120/216)*E_{2} + (80/216)*E_{3} + (15/216)*E_{4} + (1/216)*E_{5}

216*E_{2} = 216 + 120*E_{2} + 80*E_{3} + 15*E_{4} + 1*E_{5}

96*E_{2} = 216 + 80*E_{3} + 15*E_{4} + 1*E_{5}

96*E_{2} = 216 + 80*E_{3} + 15*E_{4}

96*E_{2} - 80*E_{3} - 15*E_{4} = 216

Next solve for E_{1}. 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

E_{1} = 1 + (120/1296)*E_{1} + (900/1296)*E_{2} + (250/1296)*E_{3} + (25/1296)*E_{4} + (1/1296)*E_{5}

1296*E_{1} = 1296 + 120*E_{1} + 900*E_{2} + 250*E_{3} + 25*E_{4} + 1*E_{5}

1176*E_{1} = 1296 + 900*E_{2} + 250*E_{3} + 25*E_{4} + 1*E_{5}

1176*E_{1} = 1296 + 900*E_{2} + 250*E_{3} + 25*E_{4}

1176*E_{1} - 900*E_{2} - 250*E_{3} - 25*E_{4} = 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*E_{1} - 900*E_{2} - 250*E_{3} - 25*E_{4} = 1296

96*E_{2} - 80*E_{3} - 15*E_{4} = 216

11*E_{3} - 10*E_{4} + E_{5} = 36

E_{4}=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.

E_{1} = 151496136/13660416 =~ 11.09015538

E_{2} = 11046/1056 =~ 10.46022727

E_{3} = 96/11 =~ 8.727272727

E_{4} = 6

Thanks to Nick Hobson for suggesting a similar problem.

Michael Shackleford, ASA