Suppose we want a Yahtzee with any dice. The best way to attempt this is to keep whichever 
dice occur most after each roll and roll the rest each of the turns (as many as 3).

When rolling 5 6-sided dice, each of the 5 dice can occur any of 6 ways. For 1 dice, there 
are 6^1 possibilities, 6^2 possibilities for 2 dice, etc. Thus for 5 dice there are 6^5 = 
7776 possibilities.

After 1 roll, you can be left with anything between none of the same dice & 5 of-a-kind, 
i.e.:

X X X X X, D D X X X, D D E E X, D D D X X, D D D D X, or D D D D D

for which D & E represent the dice which are the same, and X any other dice.

The number of times each occurs is:

X X X X X : 6*5*4*3*2 = 720
D D X X X : (6)(10)(5*4*3) = 3600
D D E E X : (6)(5)(4)(15) = 1800
D D D X X : (6)(10)(5^2) = 1500
D D D D X : (6)(5)(5) = 150
D D D D D : 6

Explanation: For X X X X X, no dice are the same. The first dice can be any of 6 numbers, 
the next dice any of 5 numbers, etc. So there are 6*5*4*3*2 possibilities. For a pair D D 
with no other pair, the D D can be any of 6 dice, which can be in any of 10 positions: 
D D X X X, D X D X X, D X X D X, D X X X D, X D D X X, X D X D X, X D X X D, X X D D X, 
X X D X D, X X X D D. For the remaining 3 dice, the first can be any of 5 dice, the next 
any of 4, the next any of 3 - thus 5*4*3 ways. Thus (6)(10)(5*4*3) possibilities. For the 
2 pair D D E E, the DDEE can occur any of 6 ways: DDEE DEDE DEED EDDE EDED EEDD, and for 
each, the X can be in any of 5 positions - e.g. for DDEE: XDDEE DXDEE DDXEE DDEXE DDEEX, 
and X can be any of 4 dice (neither D nor E). There can be 15 sets of pairs for D & E: 
65 64 63 62 61 54 53 52 51 43 42 41 32 31 21. Thus there are (6)(5)(4)(15) possible ways 
of rolling 2 pairs. For 3 of-a-kind, D can be any of 6 dice and D D D can occur in any 
of 10 positions (see above, replacing D with X). The other 2 dice can be any of 5 dice, 
thus 5^2 ways. So D D D occurs (6)(10)(5^2) ways. For 4 of-a-kind, D can be any of 6 dice, 
and the X can be any of 5 dice in any of 5 positions. Thus there are (6)(5)(5) possibilities 
for D D D D. There are 6 possibilities of rolling D D D D D (Yahtzee) - one for each dice.

These add to 7776. For subsequent calculations, it does not matter which pair is chosen 
among D D E E X, so they can be combined to say that a pair occurs 3600 + 1800 = 5400 ways. 
Thus the table for rolling 5 dice can be written:

5 dice rolled, 7776 possibilities:

None of the same dice : 720
Pair : 5400
3 of-a-kind : 1500
4 of-a-kind : 150
Yahtzee : 6

If none of the same dice are rolled, it does not matter whether any one dice is chosen 
or all are rerolled - the probability is the same (as if the first dice rolled is the 
one chosen). Thus if at least a pair is not obtained, may as well roll all the dice again.


So during subsequent rolls, either none of the same dice, a pair, 3 of-a-kind, 4 of-a-kind, 
or Yahtzee will be kept. Thus similar tables must be made for those (except for Yahtzee).


Pair kept (D D kept) - 3 dice rolled, 6^3 = 216 possibilities:

X X X : 5^3 - 5 = 120
E E E : 5
D X X : (3)(5^2) = 75
D D X : (3)(5) = 15
D D D : 1

Explanation: Here X X X represents anything without a D and not 3 of the same dice (E E E). 
For no D, X can be any of 5 dice - thus 5^3 ways - except for the 5 ways of rolling 3 of 
the same other dice, which is E E E. Thus X X X occurs 5^3 - 5 ways. This must be noted, 
because if E E E is rolled, then those 3 dice are kept rather then the D D which was started 
with. Thus E E E is the same as D X X, both of which give 3 of-a-kind. D X X can occur any 
of 75 ways (D can be in 3 positions, and the X's can be any of 5 dice each). For D D X, the 
X can be in any of 3 positions and be any of 5 dice - thus (3)(5) ways. There is only one 
way of rolling D D D with 3 dice.

Thus the 3 dice table can be rewritten:

Pair kept - 3 dice rolled, 6^3 = 216 possibilities:

Pair : 120
3 of-a-kind : 80
4 of-a-kind : 15
Yahtzee : 1


Similarly for the others:

3 of-a-kind kept (D D D) - 2 dice rolled, 6^2 = 36 possibilities:

X X : 5^2 = 25
D X : (2)(5) = 10
D D : 1

Thus:

3 of-a-kind kept - 2 dice rolled, 6^2 = 36 possibilities: 

3 of-a-kind : 25
4 of-a-kind : 10
Yahtzee : 1


4 of-a-kind kept (D D D D), 6 possibilities:

X : 5
D : 1

Thus:

4 of-a-kind kept, 6 possibilities:

4 of-a-kind : 5
Yahtzee : 1


So to summarize, here are the tables:


5 dice rolled, 7776 possibilities:

None of the same dice : 720
Pair : 5400
3 of-a-kind : 1500
4 of-a-kind : 150
Yahtzee : 6


Pair kept - 3 dice rolled, 6^3 = 216 possibilities:

Pair : 120
3 of-a-kind : 80
4 of-a-kind : 15
Yahtzee : 1


3 of-a-kind kept - 2 dice rolled, 6^2 = 36 possibilities: 

3 of-a-kind : 25
4 of-a-kind : 10
Yahtzee : 1


4 of-a-kind kept, 6 possibilities:

4 of-a-kind : 5
Yahtzee : 1


Consider the table for a pair kept (and 3 dice rolled). The probability of being left 
with a pair again is 120/216, the probability of 3 of-a-kind is 80/216, the probability 
of 4 of-a-kind is 15/216, and the probability of a Yahtzee is 1/216. The same is 
applicable to all tables. E.g., the probability of rolling 3 of-a-kind with 5 dice 
is 1500/7776.

These probabilities can be used to calculate the probability of obtaining a Yahtzee 
using as many as 3 rolls. All possibilities must be considered. E.g., rolling none 
of the same dice, then none of the same dice, then a Yahtzee. The probability of that 
occurring is the product of each individual probability. I.e., (720/7776)(720/7776)(6/7776). 
Similarly for none of the same dice, then a pair, then a Yahtzee; none of the same dice, 
3 of-a-kind, then a Yahtzee, etc.

So the calculation goes as follows. Suppose none of the same dice are rolled first. 
The probability of obtaining Yahtzee this way is:

(720/7776)((720/7776)(6/7776) + (5400/7776)(1/216) + (1500/7776)(1/36) +
           (150/7776)(1/6) + (6/7776))

I.e., the probability of rolling none of the same dice times the sum of rolling none 
of the same dice then a Yahtzee, a pair then a Yahtzee (with 3 dice - 1/216), 3 of-a-kind 
then a Yahtzee, 4 of-a-kind then a Yahtzee, and a Yahtzee.

Similarly for a pair rolled first:

(5400/7776)((120/216)(1/216) + (80/216)(1/36) + (15/216)(1/6) + (1/216))

Note that rolling a pair takes us to the 3 dice table. E.g., the 2nd term of the sum 
represents the probability of being left with 3 of-a-kind after rolling the 3 dice 
remaining (80/216) then a Yahtzee after rolling the 2 dice remaining (1/36) - each 
such term of course multiplied by the probability of a pair rolled on the first roll 
(5400/7776).

Similarly for 3 of-a-kind rolled first:

(1500/7776)((25/36)(1/36) + (10/36)(1/6) + 1/36)

Similarly for 4 of-a-kind rolled first:

(150/7776)((5/6)(1/6) + 1/6)

Similarly for 5 of-a-kind rolled first:

6/7776



So above the probabilities of all possible ways of rolling a Yahtzee are listed. Now it 
is as simple as adding the terms:

(720/7776)((720/7776)(6/7776) + (5400/7776)(1/216) + (1500/7776)(1/36) + (150/7776)(1/6) +
           (6/7776)) +

(5400/7776)((120/216)(1/216) + (80/216)(1/36) + (15/216)(1/6) + (1/216)) +

(1500/7776)((25/36)(1/36) + (10/36)(1/6) + 1/36) +

(150/7776)((5/6)(1/6) + 1/6) +

6/7776

which leaves a huge denominator:

(720)(720 + (5400)(6) + (1500)(36) + (150)(216) + 7776)/(7776^2*1296) +

(5400)(6^3)(120 + (80)(6) + (15)(36) + 216)/(7776^2*1296) +

(1500)(6^5)(25 + (10)(6) + 36)/(7776^2*1296) +

(150)(6^7)((5 + 6)/(7776^2*1296) +

(6^10)/(7776^2*1296)

Here I rearranged the terms and multiplied them by the appropriate factors to get one 
denominator for all. Now they can be added:

(720)(720 + 32400 + 54000 + 32400 + 7776)/(7776^2*1296) +
(5400)(216)(120 + 480 + 540 + 216)/(7776^2*1296) +
(1500)(7776)(25 + 60 + 36)/(7776^2*1296) +
(150)(279936)(11)/(7776^2*1296) +
60466176/(7776^2*1296)

which equals:

(91653120 + 1581638400 + 1411344000 + 461894400 + 60466176)/78364164096

which equals:

3606996096/78364164096 = 347897/7558272 = .046028642530

A common factor for the numerator and denominator must be a combination of 2's & 3's, 
the denominator being 6^14. It so happens that this is 2^7 * 3^4 = 10368.

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