The problem is partly that 8C3 you're throwing in - that shouldn't be
in there at all - but mainly that you have forgotten the other five
cards you are drawing. ;-)
What is the probability of getting, say, 7-8-9 hearts? Well, you have
to choose those three cards, and the other five cards can be any of
the remaining 29 cards. So there are 29C5 ways to get this specific
combination, and the probability is 29C5 / 32C8 = 1/4960.
Now you might be tempted to think the answer is therefore 24/4960 =
3/620. Unfortunately it isn't. 8-) The problem is that some of those
hands have more than one three-card combination, and are therefore
being counted twice. Even worse is that some of them overlap (e.g.
7-8-9-10 of hearts counts as 7-8-9 and 8-9-10), so sorting them all
out is actually very tricky.
I'll look at the second part of this problem before I get into that,
because it will illustrate the issue, but it's much simpler.
How many ways can we get a specific 4 of a kind, e.g. four Jacks?
Well, we have to have the four Js, and the other four cards can be
anything, so there are 28C4 = 20475 ways of doing it.
Now, how many ways can we get any 4 of a kind?
We have 8 possible combinations of cards (for 4 of a kind there is
only one possible combination of suits, and 8 possible denominations
in this deck).
So we get 8 * 20475 = 163800 ways. But it's possible that an 8-card
hand can contain two sets of 4-of-a-kind, and we've counted those
hands twice. So we have to subtract the number of these hands from the
total.
How many hands have two sets of 4 of a kind? There are 8C2 = 28 ways
to choose the denominations, and once we have done that we have
completely determined the hand drawn. So there are 28 such hands.
Hence the total number of hands with (at least) one 4-of-a-kind is
163772, and the probability is 163772/32C8 = 1.56% (to 3 sf).
Now, back to the main problem. There are two main sources of
complexity here: long runs of cards, as mentioned earlier, and the
fact that we are drawing enough cards to have lots of different ways
of getting card combinations in two different suits in the same hand.
(For the second part, this was only just barely possible, and we
didn't have to worry about getting 5 or more of a kind.)
I always find it helpful to list the various possibilities, like this:
- One run: length 3, 4, 5, 6, 7, 8
- Two runs in different suits: 3+3, 3+4, 3+5, 4+4
- Two runs in the same suit, with a gap in between: 3+3, 3+4 (don't
have to worry about 3+5 or 4+4 since they would make a single run of
8, already considered above)
We need to work out the probability of each of these, and also
consider which ones are included in which other ones. We work from the
most restrictive cases back to the most general ones. I'll explain the
calculations in a bit of detail at the start, but I'll speed it up
once we're past the first few.
Combinations involving 8 cards
------------------------------
1 run of 8: 4 possibilities, 1 from each suit.
Two runs in different suits:
(3+5) 4P2 = 12 ways to choose which suit has the 3 and which the 5
(order is important); 6 possible 3-card runs in the first suit; 4
possible 5-card runs in the second suit: 12*6*4 = 288 combinations.
(4+4) 4C2 = 6 ways to choose the two suits (order is not important
since they have the same number of cards); 5 possible 4-card runs in
each of the two suits: 6*5*5 = 150 combinations.
Combinations involving 7 cards
------------------------------
1 run of at least 7: 4 (suits) * 2 (possible runs) * 25C1 (select
other cards) = 200 combinations. Note that this includes the runs of 8
as well, and double-counts them. Subtracting these gives 196 runs of
at least 7, and 192 combinations with runs of exactly 7.
Two runs in different suits (must be 3+4): 4P2 * 6 * 5 * 25C1 = 7500
combinations. Double-counts 3+5 combinations (the 5-card run
corresponds to two different 4-card runs) and quadruple-counts 4+4
combinations. True total for 3+4 or better is therefore 7500-288-3*150
= 6762, and the total for exactly 3+4 is 7500-2*288-4*150 = 6324.
Two runs in the same suit: must be 7-8-9 / J-Q-K-A or 7-8-9-10 /
Q-K-A. SO there are 4*2*25C1 = 200 combinations, but that includes all
8-card runs (twice). So the total for 3+4 (same suit) or better is
196, or 192 for exactly this combination. Note this is the same as for
a single 7-card run, and it's easy to see why: both represent an
8-card run modified by replacing one of two specific cards with some
other card. For the 7-card run, we replace one of the two end cards;
for this combination, one of the two cards in the middle. We can also
calculate it by noting that for the last card, there aren't really 25
possibilities but 24, since we want to exclude the remaining card in
the given suit. This gives us 4*2*24C1 = 192 directly.
Combinations involving 6 cards
------------------------------
Now it starts to get tricky!
1 run of 6 cards: 4*3*26C2 = 3900 combinations, all 7-card runs are
counted twice, all 8-card runs are counted three times. Total for a
run of exactly 6 cards is 3900-2*192-3*4 = 3504.
3 + 3, different suits: 4C2*6*6*26C2 = 70200, includes the following
combinations: 3+4 (twice), 3+5 (three times), 4+4 (four times) for a
total of 70200-2*6324-3*288-4*150 = 56088
3 + 3, same suit: possibilities are 7-8-9/J-Q-K; 7-8-9/Q-K-A;
8-9-10/Q-K-A. We cannot include any other card in the same suit
without making a longer run somewhere (though we could if we were
using 9 cards in each suit instead of 8, and this calculation would be
much harder), so we get 4*3*24C2 = 3312. The other two cards cannot be
part of any 3-card run since they are the only two not in the chosen
suit, so that's it.
All combinations for lower numbers of cards involve a single run only.
Single run of 5 cards
---------------------
4*4*27C3 = 46800. Includes 5+3 (different suits), once; 6-card runs,
twice; 7-card runs, three times; 8-card runs, four times. Total is
46800-288-2*3504-3*192-4*4 = 38912.
Single run of 4 cards
---------------------
4*5*28C4 = 409500. Includes the following combinations:
5+3 (different suits) * 2
4+4 (different suits) * 2
3+4 (different suits) * 1
3+4 (same suit) * 1
5-card run * 2
6-card run * 3
7-card run * 4
8-card run * 5
Total is therefore 409500 - 288*2 - 150*2 - 6324 - 192 - 38912*2 -
3504*3 - 192*4 - 4*5 = 312984.
Single run of 3 cards
---------------------
4*6*29C5 = 2850120. Includes:
5+3 (different suits) * 4 (5-run counted 3 times, 3-run counted once)
4+4 (different suits) * 4
3+4 (different suits) * 3
3+4 (same suit) * 3
3+3 (different suits) * 2
3+3 (same suit) * 2
4-card run * 2
5-card run * 3
6-card run * 4
7-card run * 5
8-card run * 6
Total is 2850120 - 288*4 - 150*4 - 6324*3 - 192*3 - 56088*2 - 3312*2 -
312984*2 - 38912*3 - 3504*4 - 192*5 - 8*6 = 1952292.
Now we can add up all the individual types to get the total:
4 + 288 + 150 + 192 + 6324 + 192 + 3504 + 56088 + 3312 + 38912 +
312984 + 1952292 = 2374242 combinations out of a total of 32C8 =
10518300. The probability is thus 2374242/10518300 = 22.57% (to 4
s.f.) of getting a 3-card run or better. |