>There are 47 unseen cards, and you draw 3, so there are 47 choose 3 or
>16,215 total possible redraws.
>
>Of these, there are 6 cards (4 deuces, 2 tens) that you're looking for, and
>you need two of them. That means there are 6 choose 2 or 15 ways to get the
>cards you need. Then the third card can be any of the other 41 cards. So
>there are 15*41=615 ways to make 4-of-a-kind. That makes the chance to draw
>exactly 4-of-a-kind 615/16,215 ~= 3.8%
>
>Ed
I've done a lot of such calculations, but I never trusted my knowledge
of probability to do it this way. I assumed my understanding of "47
choose 3" and "6 choose 2" was subject to error and I took a longer
route which made me feel more confident in my conclusion. It didn't
take any knowledge of probability, but only the abilities to do simple
calculations and think logically. On a question like this, I would
categorize the remaining 47 cards into the 6 cards that can help make
the 4 of a kind and the 41 that can't. To double check my work, which
I always strove to do, I would not just figure out how often 4 of a
kind occurs, but all other possibilities, even including 2 pair, in
spite of it not paying anything, and make sure it added up to what it
should. I would take it one card at a time and just be logical. This
was before I had a computer, so I used pen and paper. On the first
card, I would assume that the 6 cards that can help the 4 of a kind
along were, say, another ten and write "T - 6." Since I'll have to
calculate how often a full house occurs, I'll have to categorize the
other 41 cards into the 9 that had one of their denomination discarded
(assuming no pair was discarded so that I can assume that, say, a 9,
an 8, and a 7 were discarded) and I'll assume that one of the 9 cards
that had one of their denomination discarded is, say, a 9 and write "9
- 9" and assume that one of the 32 that didn't didn't have one of
their denomination discarded is, say, a 6 and write "6 - 32." Then,
I'd go on to the 2nd drawn card and proceed similarly. I'd take the
"T - 6" from the 1st card, which is now 3 tens, and, beside it, write
"T - 5," "9 - 9," and "6 - 32," I'd take the "9 - 9" from the first
drawn card and write "T - 6," "9 - 2" for how often a card that
matches the first drawn card will occur, "8 - 6" for a different card
of the original 9 that had one of their denomination discarded, and "6
- 32." I'd take the "6 - 32" and write "T - 6," "6 - 3," "5 - 28,"
and "9 - 9." Then, I'd add up all possibilities that I have from the
first 2 drawn cards. They would be:
TTTT 30 (6 x 5)
TTT9 108 (6 x 9 + 9 x 6)
TTT6 384 (6 x 32 + 32 x 6)
TT99 18 (9 x 2)
TT98 54 (9 x 6)
TT96 576 (9 x 32 + 32 x 9)
TT66 96 (32 x 3)
TT65 896 (32 x 28)
Since these add to 2162, or 47 x 46, I assume they're all correct,
which can only be wrong if I've made 2 or more offsetting mistakes.
Going on to the last card, proceeding similarly, I'd end up with:
TTTTT 120
TTTT9 810
TTTT6 2880
TTT99 324
TTT98 972
TTT96 10,368
TTT66 1728
TTT65 16,128
TT999 18
TT998 324
TT996 1728
TT987 162
TT986 5184
TT966 2592
TT965 24,192
TT666 192
TT665 8064
TT654 21,504
Since these add up to the desired 97,290, I assume they're all
correct, particularly since the total of the 4 of a kinds is 3690,
which matches Ed's 615 out of 16,215.
It might be hard to imagine with computer programs so easily
available, but going about this so laboriously put me ahead of most of
my competitors and, when computers made work like this obsolete, I had
the added advantage of knowing, directly, that the computer programs
were correct, so that I could have more confidence in them than those
who didn't do this kind of work.
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