Four of a Kind in Half a Deck of Cards

Date: 11/26/1999 at 18:31:58
From: Kevin
Subject: Probability

If a full deck of cards is cut in half, what are the chances that at 
least one type of a card (i.e. kings, aces) will have all four suits 
present?

I considered two imaginary piles and proposed that a card could enter 
any pile, and had no bias toward a pile (1 in 2 chance of entering 
either pile.) Clearly the chances of getting all 4 are less than 
getting 2 or 3. To get 4 of a kind in a pile, I assume the chances are 
1 in 16. But we have 13 suits, so the chances are 13 times higher. 
Therefore I believe the chances are 13 in 16. Am I even close?

Kevin

Date: 11/28/1999 at 07:18:33
From: Doctor Mitteldorf
Subject: Re: Probability

Dear Kevin:

You've got some very intelligent, even clever elements in your 
approach. Unfortunately, probability problems often require a great 
deal of "careful" along with some "clever." The probability of the ace 
of spades being in the left half of the deck is 1/2. Same for the ace 
of diamonds, and the ace of clubs, and the ace of hearts. So you 
multiplied 1/2 * 1/2 * 1/2 * 1/2 to get 1/16. But there are two piles 
- aren't you counting "success" if all four aces are either in the 
left half or the right half? So maybe you should be using 1/8.

Here's one fly in the ointment: suppose you had a deck with 64 cards, 
so there were 16 suits of 4. Your calculation would come out to 16/16 
for the probability. How can that be? Worse, suppose there were 17 
suits of 4. Would you say the probability is 17/16? Of course, I'm 
telling you that you should be using 1/8 instead of 1/16 for each 
probability, so you're already up to 13/8.

Here's a hint about where that fly is: Suppose the probability of one 
event happening is A and another event happens with probability B. If 
you're looking for the probability of EITHER A or B, your intuition 
tells you that it should be A+B. But your intuition isn't exactly 
right. If A and B are both very unlikely, then A and B are both very 
small compared to 1, and the answer will come out pretty close to A+B, 
which is also a small number. But the catch is that you haven't 
thought about the probability that BOTH A and B are happening. This 
has to be subtracted from A+B, because you really counted it twice 
when you added A+B.

If A and B are "independent," then the probability of both happening 
is AB. Independent means that the probabilities are entirely separate, 
and the occurrence of A has no effect on the occurrence of B. So 
A+B-AB is a good answer in this case for the probability of "either A 
or B or both."

One problem with this kind of thinking is that when you add a third 
event C, you get A+B+C, and you need to subtract AB and AC and BC, but 
then what about the times ABC when ALL THREE occur? Clearly things get 
more and more complicated the more different events you're taking into 
account. What would be the formula for 13 different events?

A clever way around this is to think of the complementary probability 
- the probability of A NOT happening. The probability of "either A or 
B" is the complement of "not A" AND "not B." The probability that A 
won't happen is (1-A). The probability that B won't happen is (1-B). 
And the probability that neither A nor B will happen is (1-A)(1-B). 
Subtract this from 1 to get the probability of "either A 
or B or both.. Now it's much easier to see what to do with 13 suits.

Does this solve our problem about coming out larger than 1? Think 
about a lot of different probabilities A, B, C... that are all less 
than 1. Can it ever be that 1 - (1-A)(1-B)(1-C)...(1-Z) > 1?

One more fly in the ointment I want to alert you to: the part about 
"independent." If you know that the Ace of Spades is in the left half, 
the probability of the Ace of Diamonds also being in the left half is 
no longer exactly 1/2, but slightly less. Can you see why? These 
"slightlys" actually add up to something substantial when you take 
them all into account.

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/