Re: Probabilities and the dominoes game
From: Carlos Moreno (moreno_at_mochima_dot_com_at_xx.xxx)
Date: 02/11/05
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Date: Fri, 11 Feb 2005 11:50:46 -0500
David M Einstein wrote:
> First there are C(28,7)=1184040 possible hands. Now if zero is missing
> there are only 21 dominos that can occur, and so there are
> C(21,7)=116280 hands that miss 0. It would be nice if the number of
> hands missing one number was 7 times this or 813960, unfortunately we
> have overcounted the hands that are missing two or more numbers. If we
> choose two specific numbers, say 0 and 1, then there are 15 dominos to
> choose from and C(15,7)=6435 hands. We can subtract out the 21
> possible combinations of two number, but then we would undercount the
> hands missing three or more numbers. Luckily, here we are done, as
> there are no hands missing four or more numbers, so there are
> 35*120=4200 hands missing exactly 3 numbers
>
> So we have
> Hands missing exactly 3 numbers = 4200
> Hands missing exactly 2 numbers = 122535
> Hands missing exactly 1 number=556290
> Hands missing no number=501015
>
> So the odds are slightly against getting a hand with no numbers missing.
Let me see if I understand the explanation.11
I see that the key point in the above argument is that for the case
of three misses, it is true that the number of hands with three
misses *is* C(10,7) = 120 times the number of combinations of three
numbers, since the number of hands with *at least* 3 misses is equal
to the number of hands with *exactly* 3 misses.
But then, I'm not sure where did the 122535 for the two misses come
from. My guess is that that should be the number of hands with at
least two misses minus the number of hands with at least three
misses. But the number of hands with at least two misses should
be C(15,7) * C(7,2) = 6435 * 21 = 135135. That includes the 4200
hands that have three misses, so the number hands with exactly two
misses should be 135135 - 4200 = 130935. Was it a mistake when you
did the calculations, or is there some detail that I'm missing? (or
am I making a mistake in the calculations? I don't see it)
I suspect that there is something I am missing, since your numbers
match *very closely* the statistics from the actual shufflings we
have -- in a set of 250000 hands drawn in the past few days, I get
the following percentages:
0 missess: 42.4758%
1 miss: 46.8377%
2 misses: 10.3315%
3 misses: 0.354991%
Your figure, 122535 gives 10.349%, whereas the number I'm getting
gives me 11.06%, which is significantly far from the statistics
obtained in practice)
Thanks for the beautiful solution! (I think I understood enough
to notice it is a neat solution! :-))
Carlos
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