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Q: Card combinations & probability problem ( No Answer,   6 Comments )
Subject: Card combinations & probability problem
Category: Science > Math
Asked by: dedavai-ga
List Price: $10.00
Posted: 08 Nov 2005 02:32 PST
Expires: 08 Dec 2005 02:32 PST
Question ID: 590449
This is a combinations problem. Here's the setup. There is a standard
deck of 32 cards: 7, 8, 9, 10, J, Q, K, A. We are drawing 8 cards from
the deck. The problem is to find the probability of getting 3 suited
consecutive card combinations in our 8 card hand. By 3 suited
consecutive I mean 7-8-9 hearts or J-Q-K spades. The combination can't
wrap (i.e. K-A-7 doesn't count) and order doesn't matter (Q-J-10 clubs
is the same as J-Q-10 clubs).

I'm having trouble solving the problem because my probability comes
out too low. Here's my logic. There are 24 possible combinations that
fit the description (6 of each suit: 7.8.9, 8.9.10, 9.10.J, 10.J.Q,
J.Q.K, Q.K.A). There are 32C8 (32 choose 8) possible hands. Each and
any of those hands can be arranged in 8C3 3-card combinations. So
there are (32C8 x  8C3) = 589,024,800 possible 3-card combinations in
the deck. So the probability is 24/589,024,800, which is nearly 0,
although I'm sure it's much higher than that. Could you help and point
out the mistake in the logic here?

The 2nd part of the problem is to find the probability of catching 4
of a kind in the same setup (drawing 8 cards from a 32 card deck).

Thank you

Clarification of Question by dedavai-ga on 10 Nov 2005 12:21 PST

Thank you for a great answer! How can I make your comment the answer
for the question?

Clarification of Question by dedavai-ga on 18 Nov 2005 18:00 PST
manuka-ga, I have a quick question. When you are calculationg number
of ways to get one run of exactly 6 cards, you do 3900-2*192-3*4 =
3504 because all 7-card runs are counted twice, all 8-card runs are
counted three times. What if you want to find the number of ways to
get one run of AT LEAST 6 cards. Is it 3900-192-2*4 or 3900-196-2*4?
There is no answer at this time.

Subject: Re: Card combinations & probability problem
From: jack_of_few_trades-ga on 08 Nov 2005 09:24 PST
I'm no expert at these problems, but it seems to me that you should be
dividing 32C8 / 8C3 rather than multiplying like you did.  Because the
more possible combinations of 3 cards in your hand of 8 cards, the
32C8 / 8C3 = 187,827
24 / 187827 still seems low to me... but much more realistic than 24 / 589,024,800

Anyone with more insight?
Subject: Re: Card combinations & probability problem
From: manuka-ga on 09 Nov 2005 01:29 PST
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
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.
Subject: Re: Card combinations & probability problem
From: manuka-ga on 09 Nov 2005 15:42 PST
One slight correction to the first part of my answer - 29C5 / 32C8 is
not 1/4960 as I stated, but 7/620. The naive assumption would then be
for an overall answer of 24*7/620 = 42/155 = 27.10% (4 s.f.), but as
we saw this is almost 5% too high due to the combinations that are
being counted multiple times.
Subject: Re: Card combinations & probability problem
From: manuka-ga on 15 Nov 2005 20:15 PST
Thanks for the kind words. Unfortunately, that's all you can offer me
- as you can see, my name is shown in ordinary text, rather than as an
underlined blue hyperlink; this indicates that I am not an official
Google Answers Researcher, and can only contribute free comments. So
you've saved yourself $10 there. 8-)
Subject: Re: Card combinations & probability problem
From: ithandir-ga on 15 Nov 2005 21:33 PST
im not great at probability but i have learned combinatorics in the
open u. lets see the total amount of options: look at the problem as
choosing 3 cards of 32. the amount of 3 suited consequtive cards is
correct 24. the amount of options of picking 3 cards of 32 with no
matter of order is (binom of newton) 3 of 32. which is 32! / (3!*29!)
= 4960 options in total. which leaves ur final answer at picking one
option of 24 out of 4960 totals - each having an equal chance of
happening (the person takes the cards completely randomly) which makes
it a chance of 24/4960.
Subject: Re: Card combinations & probability problem
From: ithandir-ga on 15 Nov 2005 21:40 PST
ok correction i didnt notice u pull out 8 cards instead of 3. the
total amount of options would then be picking 8 out of 32 which is
32!/(8!*24!) = 10,518,300. if u want there to be 3 suited consequtive
cards then u pick one of 24 optinos for that and then u just have 5
cards to choose of the 29 remaining which is 29!/(5!*24!) = 118,755.
then the final answer is 118,755/10,518,300 ~= 1/88.571

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