Yatzy - discussion of Game 1
The probability of a Yahtzee is 347897/7558272 = .04603, or about 1 in 22. Beating 827 requires
at the very least 6 Yahtzees (500 bonus points). The process of doing that can be made quicker
by only starting a game with a Yahtzee - which on the average will occur in around 2 minutes.
The game is then continued until no good possibility is seen to get the 6 Yahtzees (there is
an optimal point to quit, which would be a very complicated set of calculations), which will
usually happen fairly quickly. So I am estimating an average of 2 minutes to start a game and
another 36 seconds or so to finish it - 2.6 minutes per game.
The probability of getting 5 or more Yahtzees in 12 turns is:
Sum of n = 5-12 of b(n,12,.04603)
i.e., the probability of success for at least n of 12 binomial trials with probability .04603.
b(n,12,.04603) = [12!/((12-n)!n!)](.04603)^n(.95397)^(12-n)
Thus the probability is:
12!/5!7!(.04603)^5(.95397)^7 = 1.17671984586432401101347121065874e-4
+ 12!/6!6!(.04603)^6(.95397)^6 = 6.6240884852413919303655698255073e-6
+ 12!/7!5!(.04603)^7(.95397)^5 = 2.73959013064796530487784670933704e-7
+ 12!/8!4!(.04603)^8(.95397)^4 = 8.26174655084317660563805361650852e-9
+ 12!/9!3!(.04603)^9(.95397)^3 = 1.7717221190757753582165738171028e-10
+ 12!/10!2!(.04603)^10(.95397)^2 = 2.5646205585413987779083899745661e-12
+ 12!/11!1!(.04603)^11(.95397)^1 = 2.2499179850264127945156116121302e-14
+ 12!/12!0!(.04603)^12(.95397)^0 = 9.046731453012654217948293491615e-17
= 1.24578473590711546022534967668964e-4 = .000124578473590711546022534967668964
Because 1/.000124578... = 8027.07, that is about 1 of every 8027 games on the average.
(1 - .000124578...)^5564 = 0.499974..., so on the average 5564 games would be required for
this to happen the first time. Using my estimate of 2.6 minutes per game gives an estimate
of (5564)(2.6) = 14466 minutes to get a game with 6 Yahtzees or more - which is about 241
hours - could be done in about 2 months playing 4 hours a day. Anything much less than that
would require this extremely lucky game to occur much sooner than the laws of probability
indicate, and it could easily be more than twice as long. A group of 20 people may accomplish
this spending no more than 10-25 hours on it each - if they use good strategy (see below).
So now the question is, would a game with 6 Yahtzees or more beat 827? The games with more than
6 almost certainly would, but make a small portion of the total probability. For a game with 6,
straights and Full House may be a problem because according to the rules a bonus Yahtzee must
be used in the Left Section if that category is open. If it isn't, it is best to use them first
in Large Straight then perhaps even Small Straight (see below). If 6 Yahtzees in a game were
obatined and the game score was less than 827, that would set the person back another couple
months ;)
The records for Games 2 & 3 were much more beatable and require more skill - especially 2 - so
didn't require nearly so much time as Game 1 would. The record for Game 4 would be very tough
to beat, and that game greatly differs from normal Yahtzee. Game 1 is the best game IMO when
Yahtzees are not used as jokers for straights or Full House and can be placed anywhere anytime
with no bonus points. Records are made to be broken, and Game 2 is probably the least difficult
at this point (9/25/2006).
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A score of 333 (with no bonus Yahtzees) can be obtained with the minimum of 63 points in the
Left Section:
1's: 3 3ok: 30
2's: 6 4ok: 30
3's: 9 FH: 25
4's: 12 SS: 30
5's: 15 LS: 40
6's: 18 Yahtzee: 50
Total: 63 CH: 30
Bonus: 35 Right Total: 235
Left Total: 98 Grand Total: 333
Because Yahtzee bonus rolls must be used in the Left Section if the category is open, the Left
Section will never be a problem in a game with 6 Yahtzees. The main problem is getting the
straights, Full House, and a decent score in 4-of-a-kind.
A more realistic game with 6 Yahtzees (if there is such a thing) would look something like this:
1's: 1 3ok: 27
2's: 10 4ok: 14
3's: 6 FH: 25
4's: 12 SS: 30
5's: 25 LS: 40
6's: 30 Yahtzee: 50
Total: 84 CH: 23
Bonus: 35 Right Total: 209
Left Total: 119 Yahtzee Bonus: 500
Grand Total: 828
The box of FH, SS, LS, & Yahtzee total to 145, so in this case a total of 64 points would be
needed among the other 3 categories - an average of about 21-22 for each. As mentioned above,
any bonus Yahtzee with the Left Section category filled should be placed in LS first, then
probably FH (or 4ok) instead of SS. Their probabilities (according to my calculations) are:
Full House (FH): 5485535/15116544 = .36288
Small Straight (SS): 1508506033/2448880128 = .61600
I don't have the one for LS, but recall it is something in the 20-25% range.
The expected (average) score for Chance (CH) when rolling optimally for it (keeping 6's & 5's
after the first roll, then those & 4's after the second) is 23 1/3 (23.333...).
If someone is rolling optimally *for Yahtzee* - as would be done when attempting to get 6 or
more in one game - the probabilities for each category (according to my calculations) are:
1-of-a-kind: .00079
2-of-a-kind: .25601
3-of-a-kind: .45240
4-of-a-kind: .24476
Yahtzee: .04603
LS: .00026
SS: .06040
FH: .11064
4-of-a-kind is not easy to get - comparable to LS if rolling for it, though you'll see more of
them in the course of a game when rolling for Yahtzees - so perhaps a Yahtzee with 6's or 5's
should be used there if the Left Section category is already used.
©2006, Joseph Bartlo
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