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Odds question (for a card game)

I'm no math major, so this may be simple, or it may be difficult. What I know is that I don't have any clue what the answer is, so I'm hoping someone here might. :)

It relates to Double Pinochle, the rules for which you can find at my web site[1], though the rules are not necessary to solve this problem.

A "Double Pinochle" deck consists of 80 cards: the 10s, Jacks, Queens, Kings, and Aces from 4 decks of cards. Thus there are four of each card in each suit (i.e., there are four Ace of Spaces, four Ace of Clubs). Each player is dealt 20 cards, one at a time, in clockwise order, starting with the player to the left of the dealer.

Two questions:

What are the odds of one of the players being dealt "Triple Aces Around" (that is, their hand contains three aces from each of the four suits, and either other unimportant cards)?

Does the position of the player make an statistical difference?

Remember, I don't know what the answer is. ;) The reason I ask is that I put card game instructions on my web site, and someone had this sort of a hand come up after only a few dozen played hands, and they wondered what the odds were. They did some web research and couldn't find an answer. Needless to say, I've never seen such a great hand personally. ;)

Thanks!
Brad

P.S. Michael, sorry to hear about your dog. :(

[1] http://www.quality.nu/dotnetguy/games/fog0000000030.aspx

Brad Wilson
Friday, March 08, 2002

Oops, left out one detail in the deal description: there are always four players, so all 80 cards are always dealt (20 to each player).

Brad Wilson
Friday, March 08, 2002

Short answer: 4*4*4*4*(64 choose 8) / (80 choose 20).

Longer answer:

Pretend all the cards are marked so that you can
distinguish between the different aces of spaces,
the different tens of clubs, and so on.

How many possible hands are there? Well, there are
80 ways to choose the first card, 79 ways to choose the
second card, ..., 61 ways to choose the 20th. So,
80*79*...*61 ways. But wait! It doesn't matter which
order we list the cards in, so we've counted each
possible hand several times. How many? Well, pick
any possible-hand and list its cards in any order.
We want to know how many possible orders there
are to list its cards in. Answer: there are 20 ways to
choose the first card in order, then 19 ways to choose
the second, ..., 1 way to choose the last. So,
20*19*...*1 possibilities. Therefore, the number
of possible hands is (80*...*61) / (20*...*1). That's
3535316142212174320 possible hands.

Next: how many possible hands are there that are
"triple aces around"? Well, there are 4 ways to choose
three aces of spaces; 4 ways to choose three aces
of hearts; 4 ways to choose three aces of diamonds;
4 ways to choose three aces of clubs. When you've
done that, you need to choose another 8 cards out of
the 64 non-aces. How many ways are there to do
*that*? It's just the same calculation as we did above,
but with different numbers: (64*63*...*57) / (8*7*...*1).
So, in all, we have 4*4*4*4*(64*...*57) / (8*...*1).
That's 1133098334208 possible hands.

All possible hands are equally likely. So the probability
of getting "triple aces around" is
1133098334208 / 3535316142212174320.
That turns out to be about1 in 3,000,000.

So, the chances of one person having it happen
in their first 30 hands are about one in 100,000.

I've been assuming that "triple aces around" means
*exactly* 3 aces of each suit. If it means *at least*
3 aces of each suit, the probability is a bit higher.

Gareth McCaughan
Sunday, March 10, 2002

Well, "triple aces around" means at least 3 aces of each suit. You could have 4 as well, but not all four suits could have 4 aces, because then that would be "quad aces around". So "triple aces around" could mean 3/3/3/3, 3/3/3/4, 3/3/4/4, or 3/4/4/4, but not 4/4/4/4.

Does that sufficiently confuse the problem? ;)

Thanks for the analysis!

Brad Wilson
Tuesday, March 12, 2002

To cover the remaining cases, you just have to add in the hands where some suits have all four aces.  How many are there?  Well, there's four cases each of the 3/3/3/4 and 3/4/4/4 distributions, and six of 3/3/4/4.

Rather than go through all the computation for these, wouldn't it be nicer if we could work off of Gareth?  There are 4*4*4*4*(64 ch 8) 3/3/3/3 hands.  The four 4s come from the fact that for any given suit, there are 4 ways to pick three aces.  Well, there's only one way to pick all four, so that lops off a factor of 4.  There are four ways to get a 3/3/3/4, though, so the 4 returns.  However, 13 cards are chosen for 3/3/3/4, so the (64 ch 8) becomes (64 ch 7) - a factor of 57/8 goes away (this hand is roughly 1/7 as likely as a 3/3/3/3).

For a 3/3/4/4 hand, two 4s go away (two suits have one way to choose aces), but a factor of 6 comes in (6 ways to choose two suits), and (64 ch 8) becomes (64 ch 6), so 58*57/8*7 goes away.  3/3/4/4 hands are roughly 1/22 as likely as 3/3/3/3s.

For a 3/4/4/4 hand, three 4s go away, one comes back, and 5 extra cards can be chosen.  A total factor of 4*4*59*58*57/8*7*6 goes away, meaning 3/4/4/4 hands are very unlikely - 1/9288 as likely as a 3/3/3/3.

In summary, by the time you saw 9300 3/3/3/3 hands, you would also have seen about 130 3/3/3/4s, about 42 3/3/4/4s, and 1 3/4/4/4.  Not very likely at all!

Paul Brinkley
Wednesday, March 13, 2002

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