In the domino game Mexican
Train, holding a double can be a
problem, since when a double is
played, a second domino of the
same denomination must be played
immediately, or else a domino from
the bone pile must be drawn; and,
if it doesn’t match the double, you
lose control of your train.
The other
night, I was playing Mexican Train
with friends using a double 12 set
(a double 12 set includes every
two number combination from 0-0
through 12-12); and, in picking
a hand of 11 dominos, I got four
doubles.
What is the probability of
this?
91 in set = 13 choose 2 plus 13
This set of 91 dominoes has 13 doubles and 78 non-doubles.
The probability of getting 4 doubles out of 11 is:
COMB(11,4) * (13*12*11*10*78*77*76*75*74*73*72) / (91*90*89*88*87*86*85*84*83*82*81)
=330 * (13*12*11*10*78*77*76*75*74*73*72) / (91*90*89*88*87*86*85*84*83*82*81)
The requested probability is:
(I think reduced to lowest fraction)
16144957400 / 404490084579
approx 0.039914346520518396
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Posted by Larry
on 2025-01-02 10:16:16 |
Solution
