Suppose we want a Yahtzee with 1's. The best way to attempt this is obviously to save all 
1's and roll the rest of the dice 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 0 & 5 1's, i.e.:

X X X X X, 1 X X X X, 1 1 X X X, 1 1 1 X X, 1 1 1 1 X, or 1 1 1 1 1

The number of times each occurs is:

X X X X X : 5^5 = 3125
1 X X X X : (5)(5^4) = 3125
1 1 X X X : (10)(5^3) = 1250
1 1 1 X X : (10)(5^2) = 250
1 1 1 1 X : (5)(5) = 25
1 1 1 1 1 : 1

which of course add to 7776.

Explanation: For X X X X X (X being anything other than 1), each dice can be any of 5 
possible numbers, so the number of ways this can occur is 5*5*5*5*5 = 5^5. For 1 X X X X, 
the 1 can be in any of 5 places, and among the remaining 4 dice, they can occur in any 
of 5^4 ways. Thus (5)(5^4) ways. For 1 1 X X X, the 1 1 can be in any of 10 places: 
1 1 X X X, 1 X 1 X X, 1 X X 1 X, 1 X X X 1, X 1 1 X X, X 1 X 1 X, X 1 X X 1, X X 1 1 X, 
X X 1 X 1, X X X 1 1; and among the remaining dice, they can occur in any of 5^3 ways. 
Thus (10)(5^3) = 1250 ways. Similarly for the others.


During subsequent rolls, anything between 5 & 1 dice will be rolled. Thus a similar 
table must be made for anything between 4 & 1 dice.


4 dice, 6^4 = 1296 possibilities:

X X X X, 1 X X X, 1 1 X X, 1 1 1 X, 1 1 1 1

The number of times each occurs is:

X X X X : 5^4 = 625
1 X X X : (4)(5^3) = 500
1 1 X X : (6)(5^2) = 150
1 1 1 X : (4)(5) = 20
1 1 1 1 : 1


3 dice, 6^3 = 216 possibilities:

X X X : 5^3 = 125
1 X X : (3)(5^2) = 75
1 1 X : (3)(5) = 15
1 1 1 : 1


2 dice, 6^2 = 36 possibilities:

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


1 dice, 6 possibilities:

X : 5
1 : 1


Consider the table for 3 dice. The probability of rolling 0 1's with 3 dice is 125/216, 
the probability of one 1 is 75/216, the probability of 2 1's is 15/216, and the probability 
of 3 1's is 1/216. The same is applicable to all tables. E.g., the probability of rolling 
2 1's with 5 dice is 1250/7776.

These probabilities can be used to calculate the probability of obtaining a Yahtzee with 
1's using as many as 3 rolls. All possibilities must be considered. E.g., rolling 0 1's, 
then 0 1's, then 5 1's. The probability of that occurring is the product of each individual 
probability. I.e., (3125/7776)(3125/7776)(1/7776). Similarly for 0 1's then one 1 then 
4 1's, 0 1's then 2 1's then 3 1's, etc.

So the calculation goes as follows. Suppose 0 1's are rolled first. Then the probability 
of obtaining Yahtzee this way is:

(3125/7776)((3125/7776)(1/7776) + (3125/7776)(1/1296) + (1250/7776)(1/216) +
            (250/7776)(1/36) + (25/7776)(1/6) + 1/7776)

I.e., the probability of rolling 0 1's times the sum of rolling 0 1's then & 5 1's, one 1 
then 4 1's (with 4 dice - 1/1296), 2 1's then 3 1's, 3 1's then 2 1's, 4 1's then one 1, 
and 5 1's.

Similarly for one 1 rolled first:

(3125/7776)((625/1296)(1/1296) + (500/1296)(1/216) + (150/1296)(1/36) +
            (20/1296)(1/6) + 1/1296)

Note that rolling one 1 takes us to the 4 dice table. E.g., the 3rd term of the sum 
represents the probability of rolling 2 1's with 4 dice (150/1296) then 2 1's with the 
remaining 2 dice (1/36) - each such term of course multiplied by the probability of 
one 1 rolled on the first roll (3125/7776).

Similarly for 2 1's rolled first:

(1250/7776)((125/216)(1/216) + (75/216)(1/36) + (15/216)(1/6) + 1/216)

Similarly for 3 1's rolled first:

(250/7776)((25/36)(1/36) + (10/36)(1/6) + 1/36)

Similarly for 4 1's rolled first:

(25/7776)((5/6)(1/6) + 1/6)

Similarly for 5 1's rolled first:

1/7776



So above the probabilities of all possible ways of rolling a Yahtzee with 1's are listed. 
Now it is as simple as adding the terms:

(3125/7776)((3125/7776)(1/7776) + (3125/7776)(1/1296) + (1250/7776)(1/216) +
            (250/7776)(1/36) + (25/7776)(1/6) + 1/7776) +

(3125/7776)((625/1296)(1/1296) + (500/1296)(1/216) + (150/1296)(1/36) +
            (20/1296)(1/6) + 1/1296) +

(1250/7776)((125/216)(1/216) + (75/216)(1/36) + (15/216)(1/6) + 1/216) +

(250/7776)((25/36)(1/36) + (10/36)(1/6) + 1/36) +

(25/7776)((5/6)(1/6) + 1/6) +

1/7776

which is not so simple actually - we are left with a huge denominator, but the calculation 
can be done:

(3125)(3125 + (3125)(6) + (1250)(36) + (250)(216) + (25)(1296) + 7776)/7776^3 +

(3125)(6^2)(625 + (500)(6) + (150)(36) + (20)(216) + 1296)/7776^3 +

(1250)(6^4)(125 + (75)(6) + (15)(36) + 216)/7776^3 +

(250)(6^6)(25 + (10)(6) + 36)/7776^3 +

(25)(6^8)(5 + 6)/7776^3 +

(1)(6^10)/7776^3

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

(3125)(3125 + 18750 + 45000 + 54000 + 32400 + 7776)/7776^3 +
(3125)(36)(625 + 3000 + 5400 + 4320 + 1296)/7776^3 +
(1250)(1296)(125 + 450 + 540 + 216)/7776^3 +
(250)(46656)(25 + 60 + 36)/7776^3 +
(25)(1679616)(11)/7776^3 +
60466176/7776^3

which equals:

(503284375 + 1647112500 + 2156220000 + 1411344000 + 461894400 + 60466176)/470184984576

which equals:

6240321451/470184984576 = 0.01327205601

The fraction above cannot be reduced because the denominator 6^15 is a product of 2's 
& 3's, and the numerator is divisible by neither.

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