Probability of a specific Yahtzee
Suppose we have Yatzy & 3's left and want at least 62 points. What is the probability?
Consider the first turn of at least 3 rolls. Clearly, if an equal or greater number of
3's are rolled as any other dice, it is best to keep the 3's because either 12 in 3's
or Yatzy with them is acceptable. The situations which must be examined are those for
which more of another dice are rolled - in which case rolling them for the Yatzy may
increase the probability of success.
So suppose we roll a 3 and a pair of anything else. Which is best to keep? Keeping the
3 and rolling again, we can get any of 1296 possibilities:
3 X X X X : (5)(4)(3)(2) = 120
3 3 X X X : (4)(5)(4)(3) = 240
3 3 C C X : (12)(5)(4) = 240
3 3 3 X X : (6)(5^2) = 150
3 3 3 3 X : (4)(5) = 20
3 3 3 3 3 : 1
3 C C D D : (10)(6) = 60
3 C C X X : (5)(6)(4)(3) = 360
3 C C C X : (5)(4)(5) = 100
3 C C C C : 5
for which the C & D are the same dice but not a 3, and X any others but a 3 which would
give as many or more 3's than C's. Because it does not matter which of the C or D are
chosen and because 3's are chosen if the same number of each occur, the table can be
simplified:
3 X X X X : 120
3 3 X X X : 480
3 3 3 X X : 150
3 3 3 3 X : 20
3 3 3 3 3 : 1
3 C C X X : 420
3 C C C X : 100
3 C C C C : 5
If we kept the pair and rolled the 3 remaining dice, there are 216 possibilities:
C C X X X : (5)(4)(3) = 60
C C C X X : (3)(5^2) = 75
C C C C X : (3)(5) = 15
C C C C C : 1
C C D D X : (4)(3)(4) = 48
C C D D D : 4
C C 3 3 X : (3)(4) = 12
C C 3 3 3 : 1
With the same rules applied, this can be simplified as:
C C X X X : 108
C C C X X : 79
C C C C X : 15
C C C C C : 1
3 3 X X X : 12
3 3 3 X X : 1
Considering the case of keeping 2 3's rather than 3 of any other kind, after one roll
there are 216 possibilities:
3 3 X X X : (5^3) = 125
3 3 3 X X : (3)(5^2) = 75
3 3 3 3 X : (3)(5) = 15
3 3 3 3 3 : 1
If 3 of another kind are kept, there are 36 possibilities:
C C C X X : 24
C C C 3 3 : 1
C C C C X : 10
C C C C C : 1
With the above tables and other trivial ones such as rolling one dice, I can now compute
the probability of success with the various choices.
Consider a 3 being kept instead of a pair of something else after the first roll.
To get 4 3's after 2 more rolls:
(120/1296)(20/1296) + (480/1296)(15/216) + (150/1296)(10/36) + (20/1296)(5/6) +
(420/1296)(20/1296) =
((120)(20) + (480)(15)(6) + (150)(10)(36) + (20)(5)(216) + (420)(20))/6^8 =
(240 + 43200 + 54000 + 21600 + 8400)/6^8 = 127440/1679616 = 295/3888
To get a Yatzy after 2 more rolls or less:
(120/1296)(1/1296) + (480/1296)(1/216) + (150/1296)(1/36) + (20/1296)(1/6) + (1/1296) +
(420/1296)(1/1296) + (100/1296)(1/36) + (5/1296)(1/6) =
(120 + 2880 + 5400 + 4320 + 1296 + 420 + 3600 + 1080)/6^8 = 19116/1679616 = 59/5184
If 4 3's are obtained, a Yahtzee is needed - which has the probability 347897/7558272.
If a Yatzy is obtained, at least 4 3's are needed - which has the probability
12274918019/117546246144. Thus the probability of success is:
(295/3888)(347897/7558272) + (59/5184)(12274918019/117546246144) =
.0034924 + .0011885 = .0046809
Now we consider keeping the pair and rolling at least twice more. If in the process we
get at least as many 3's as those, we must switch to the 3's:
To get a Yatzy:
(108/216)(1/216) + (79/216)(1/36) + (15/216)(1/6) + 1/216 + (12/216)(1/216) +
(1/216)(1/36) =
(108 + 474 + 540 + 216 + 12 + 6)/6^6 = 1356/46656 = 113/3888
To get 4 3's:
(12/216)(20/216) + (1/216)(10/36) = (240 + 60)/6^6 = 300/6^6 = 25/3888
Then the probability of success is:
(113/3888)(12274918019/117546246144) + (25/3888)(347897/7558272) =
.0030350 + .0002960 = .0033310
Clearly it is better to keep the 3 rather than a pair. Strictly, the same must be proven
after 2 rolls with those, though it seems obvious - as does the idea of keeping 3 of
another kind instead of a pair of 3's after 2 rolls will make little overall difference.
Perhaps I will prove it a bit later.
So now what if we are left with 2 3's and 3 of any other dice after the first roll? Which is
best to keep? To test that, the probabilities of each are computed. Though it may seem obvious
that keeping 3 of another kind is better than keeping a 3, it should at some point be proven
also.
2 3's kept rather than 3 of any other kind after one roll. To get 4 3's:
(125/216)(15/216) + (75/216)(10/36) + (15/216)(5/6) = (1875 + 4500 + 2700)/6^6 =
9075/46656 = 3025/15552
To get a Yatzy (with 3's):
(125/216)(1/216) + (75/216)(1/36) + (15/216)(1/6) + 1/216 =
(125 + 450 + 540 + 216)/6^6 = 1331/46656
Then the probability of success is:
(3025/15552)(347897/7558272) + (1331/46656)(12274918019/117546246144) =
.0089530 + .0029791 = .0119320 (rounding)
If we kept the 3 of any other kind:
To get a Yatzy:
(25/36)(1/36) + (10/36)(1/6) + 1/36 = (25 + 60 + 36)/6^4 = 121/1296
Then the probability of success is:
(121/1296)(12274918019/117546246144) = .0097497
It is shown again that it is better to keep the 3's than one more of the others.
Note also that keeping 3 of any other pair is much better than keeping only one 3.
Now we know what choices to make, and can proceed with the calculation. We must consider
all possibilities, all the way along keeping an equal or one less number of 3's than any
other combination if they are available. The tables required are (along with the trivial
ones of 2 dice left or less):
Rolling 5 dice, there are 7776 possibilities:
X X X X X : 5! = 120
3 X X X X : (5)(5)(4)(3)(2) = 600
3 3 X X X : (10)(5)(4)(3) = 600
3 3 3 X X : (10)(5)(4) = 200
3 3 3 3 X : (5)(5) = 25
3 3 3 3 3 : 1
3 C C X X : (3)(10)(5)(4)(3) = 1800
3 C C D D : (6)(10)(5) = 300
3 C C C X : (10)(5)(2)(4) = 400
3 C C C C : (5)(5) = 25
3 3 C C X : (6)(5)(5)(4) = 600
3 3 C C C : (10)(5) = 50
3 3 3 C C : (10)(5) = 50
C C X X X : (10)(5)(4)(3)(2) = 1200
C C D D X : (6)(10)(5)(3) = 900
C C C X X : (10)(5)(4)(3) = 600
C C C D D : (10)(5)(4) = 200
C C C C X : (5)(5)(4) = 100
C C C C C : 5
for which C & D are dice other than a 3, and X is anything else with no pair. Because of
the choosing rules, the table can be simplified as follows:
Rolling 5 dice, there are 7776 possibilities:
X X X X X : 5! = 120
3 X X X X : (5)(5)(4)(3)(2) = 2700
3 3 X X X : (10)(5)(4)(3) = 1250
3 3 3 X X : (10)(5)(4) = 250
3 3 3 3 X : (5)(5) = 25
3 3 3 3 3 : 1
C C X X X : (10)(5)(4)(3)(2) = 2100
C C C X X : (10)(5)(4)(3) = 1200
C C C C X : (5)(5)(4) = 125
C C C C C : 5
Rolling 4 dice with a 3 kept, there are 1296 possibilities:
3 X X X X : 540
3 3 X X X : 480
3 3 3 X X : 150
3 3 3 3 X : 20
3 3 3 3 3 : 1
3 C C C X : 100
3 C C C C : 5
Rolling 3 dice with 2 3's kept, there are 216 possibilities:
3 3 X X X : (5^3) = 125
3 3 3 X X : (3)(5^2) = 75
3 3 3 3 X : (3)(5) = 15
3 3 3 3 3 : 1
Rolling 3 dice with a non-3 (CC) pair kept, there are 216 possibilities:
C C X X X : 108
C C C X X : 79
C C C C X : 15
C C C C C : 1
3 3 X X X : 12
3 3 3 X X : 1
If 3 of another kind are kept (CCC), there are 36 possibilities:
C C C X X : 24
C C C 3 3 : 1
C C C C X : 10
C C C C C : 1
Now the calculation is done for a Yatzy and at least 4 3's in 2 turns. First we consider
getting 4 3's:
(120/7776)((120/7776)(25/7776) + (2700/7776)(20/1296) + (1250/7776)(15/216) +
(250/7776)(10/36) + (25/7776)(5/6))
(2700/7776)((540/1296)(20/1296) + (480/1296)(15/216) + (150/1296)(10/36) + (20/1296)(5/6)) +
(1250/7776)((125/216)(15/216) + (75/216)(10/36) + (15/216)(5/6)) +
(250/7776)((25/36)(10/36) + (10/36)(5/6)) +
(25/7776)(5/6)(5/6) =
(120)((120)(25) + (2700)(20)(6) + (1250)(15)(36) + (250)(10)(216) + (25)(5)(1296))/6^15 +
(2700)(6^2)((540)(20) + (480)(15)(6) + (150)(10)(36) + (20)(5)(216))/6^15 +
(1250)(6^4)((125)(15) + (75)(10)(6) + (15)(5)(36))/6^15 +
(250)(6^6)((25)(10) + (10)(5)(6))/6^15 +
(25)(6^8)(5)(5)/6^15 =
(120)(3000 + 324000 + 675000 + 540000 + 162000)/6^15 +
(2700)(36)(10800 + 43200 + 54000 + 21600)/6^15 +
(1250)(1296)(1875 + 4500 + 2700)/6^15 +
(250)(46656)(250 + 300)/6^15 +
(25)(1679616)(25)/6^15 =
(204480000 + 12597120000 + 14701500000 + 6415200000 + 1049760000)/6^15 =
34968060000/470184984576 = 121416875/1632586752
Now we consider a Yatzy:
(120/7776)((120/7776)(6/7776) + (2700/7776)(1/1296) + (1250/7776)(1/216) +
(250/7776)(1/36) + (25/7776)(1/6) + 1/7776 + (2100/7776)(1/216) + (1200/7776)(1/36) +
(125/7776)(1/6) + 5/7776)/6^15 +
(2700/7776)((540/1296)(1/1296) + (480/1296)(1/216) + (150/1296)(1/36) + (20/1296)(1/6) +
1/1296 + (100/1296)(1/36) + (5/1296)(1/6))/6^15 +
(1250/7776)((125/216)(1/216) + (75/216)(1/36) + (15/216)(1/6) + 1/216))/6^15 +
(250/7776)((25/36)(1/36) + (10/36)(1/6) + 1/36))/6^15 +
(25/7776)((5/6)(1/6) + 1/6))/6^15 +
1/7776 +
(2100/7776)((108/216)(1/216) + (79/216)(1/36) + (15/216)(1/6) + 1/216 + (12/216)(1/216) +
(1/216)(1/36))/6^15 +
(1200/7776)((24/36)(1/36) + (1/36)(1/216) + (10/36)(1/6) + 1/36)/6^15 +
(125/7776)((5/6)(1/6) + 1/6))/6^15 +
5/7776 =
(120)((120)(6) + (2700)(6) + (1250)(36) + (250)(216) + (25)(1296) + 7776 + (2100)(36) +
(1200)(216) + (125)(1296) + (5)(7776))/6^15 +
(2700)(6^2)(540 + (480)(6) + (150)(36) + (20)(216) + 1296 + (100)(36) + (5)(216))/6^15 +
(1250)(6^4)(125 + (75)(6) + (15)(36) + 216))/6^15 +
(250)(6^6)(25 + (10)(6) + 36))/6^15 +
(25)(6^8)(5 + 6)/6^15 + (6^10)/6^15 +
(2100)(6^4)(108 + (79)(6) + (15)(36) + 216 + 12 + 6)/6^15 +
(1200)(6^5)((24)(6) + 1 + (10)(36) + 216)/6^15 +
(125)(6^8)(5 + 6)/6^15 + (5)(6^10)/6^15 =
(120)(720 + 16200 + 45000 + 54000 + 32400 + 7776 + 75600 + 259200 + 162000 + 38880)/6^15 +
(2700)(36)(540 + 2880 + 5400 + 4320 + 1296 + 3600 + 1080)/6^15 +
(1250)(1296)(125 + 450 + 540 + 216)/6^15 +
(250)(46656)(25 + 60 + 36)/6^15 +
(25)(1679616)(11)/6^15 + 60466176/6^15 +
(2100)(1296)(108 + 474 + 540 + 216 + 12 + 6)/6^15 +
(1200)(7776)(144 + 1 + 360 + 216)/6^15 +
(125)(1679616)(11)/6^15 + (5)(60466176)/6^15 =
(83013120 + 1858075200 + 2156220000 + 1411344000 + 461894400 + 60466176 + 3690489600 +
6727795200 + 2309472000 + 302330880)/470184984576 = 19061100576/470184984576 =
22061459/544195584
Then the probability of success is:
(121416875/1632586752)(347897/7558272) + (22061459/544195584)(12274918019/117546246144) =
.0034232 + .0042334 = .0076566
Thus the chances of getting both the Yatzy and at least 4 3's in 2 turns is about 1 in 131.
In the game this situation presented itself, I rolled a 3 and a pair of 5's after the
first turn, and did keep the 3. I rolled no more of them though, and was left rolling
for a Yatzy and potentially a 348 game (which I didn't get).
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