## Tuesday, January 19, 2010

### A Neat Problem on Card Arrangements

Here's a neat problem on card arrangements. I learned about it on reddit. Here's the easiest way to state it:

Given a deck of 52 cards shuffled randomly -- a deal -- what is the probability that, somewhere in the deck, there is an Ace adjacent to a King?

Try to figure it out before looking at the analysis below.

At first glance, it doesn't seem so easy. For example, if we fix the positions of the Kings, then the Aces could appear on either side of a King, giving (it seems) 8 possibilities for an Ace to appear. This suggests that the number of deals where an Ace fails to be adjacent to a King is

(52 choose 4) : choose positions where the Kings are
* (4!) : choose the relative order of the 4 Kings
* (40 choose 4) : choose the positions for the Aces: not where the Kings are, or on either side of them
* (4!) : choose the relative order of the Aces
* (44!) : choose the positions of the remaining 44 cards.

This gives a probability of no Ace adjacent to a King of 9139/19458, or about 0.4697.

Unfortunately, this naive analysis is wrong. Why? If we're trying to compute the number of deals where an Ace fails to be adjacent to a King, we didn't consider the fact that in many deals, there will be two or more Kings adjacent to each other, thus increasing the number of places where Aces could go. Similarly, Kings could appear at the beginning or end of the deck. So the real probability of no Ace adjacent to a King should be somewhat higher than we computed.

Of course, we could go through all the possibilities of how many adjacent spots there are. But that is an awful calculation. I asked Ian Goulden, a professor in the Combinatorics & Optimization department at Waterloo, and he came up with this lovely argument - which I've modified slightly:

We start by choosing all the cards that are neither Kings nor Aces. There are 44 other cards, and so there are 44! ways of arranging them. Now we're going to insert the Aces. But to do so, we'll first figure out how the Aces are arranged by breaking them up into contiguous nonempty blocks. For example, you might have three Aces in a row, and then a 4th Ace somewhere later on. There can be 1, 2, 3, or 4 contiguous blocks; let the number of blocks be k (and we'll have to sum on k from 1 to 4).

Once we have the number of blocks, we need to assign the number of Aces in each block. It's not hard to see that this can be done in (3 choose k-1) ways. For example, for k = 1, there is only one way: put all 4 Aces in a single block. For k = 2 we can have the first block with 1 Ace, and the second with 3; or the first with 2 and the second with 2; or the first with 3 and the second with 1, etc. Finally, we can put the Aces themselves into these blocks in 4! ways. Now that we have the Aces in the blocks, we need to insert them into the 44 cards. There are 45 gaps between these 44 cards where we can insert the k blocks (including a gap to the left of each card and one at the right end), so that gives (45 choose k) ways of doing this.

We now have 48 cards laid out, and it remains to insert the 4 Kings. We'll do this in order: the King of Hearts first, the King of Spades next, then Diamonds, then Clubs. When we insert the first King, we can't put it immediately to the left of each block (there are k places); nor can we put it to the right of each Ace. So there are 49 positions, of which (4+k) are ruled out. So there are only 45-k places to put the King of Hearts. Then there are 46-k places to put the King of Spades, 47-k places for the King of Diamonds, and 48-k places for the King of Clubs.

Thus, the total number of deals where no King is adjacent to an Ace is
(44!) * Σk=14 (3 choose k-1)(4!)(45 choose k)(45-k)(46-k)(47-k)(48-k) =
41435794890014187172891516774574832972049361890932487618560000000000. Dividing this by 52! we get 300684703/585307450, or about 0.5137, as the probability that no King is adjacent to an Ace. So the probability that in a random deal there will be a King adjacent to an Ace is 284622747/585307450, or about .4863.

As a check, this agrees with a paper of David Singmaster entitled "The probability of finding an adjacent pair in a deck", Mathematical Gazette 75 (1991), 293-299. Goulden's formula is much simpler and more elegant than Singmaster's.

Generalizing the ideas above, Ian Goulden considered the more general situation where you have N cards, C cards in the first group, D cards in a second disjoint set, with C ≤ D. Then the probability that no card in the first group is adjacent to a card in the second in a random deal of N cards is given by

Σk=1C (C-1 choose k-1)*(N-C-k choose D)*(N-C-D+1 choose k)/( (N-D choose C)*(N choose D)).

Thanks to Ian Goulden for letting me post his beautiful solution.

D. Swart said...

I was about to demonstrate how a Monte Carlo method would achieve the same result especially as you had the final value approximated to 4 decimals (for some reason I often see 6 for such values).

However, while obtaining a decent amount of certainty for 3 digits is almost instantaneous, getting 4 digits takes on the order of 10 minutes. Which is no fun.

The Monte Carlo method gives no deeper level of understanding (as with the generalized equation you presented), it has helped me in the past win over "Monty-Hall-deniers"

And it has confirm your 0.4863

R0b said...

You know you're at a great school when it has a whole department dedicated to combinatorics and optimization. Combinatorics has a never-ending supply of problems, like the one above, that are guaranteed to keep you up at night and make you neglect your family.

student said...

i'm a bit curious, if you had instead asked a faculty from our from our actsci & statistics department, would he/she been able to answer correctly?

Jeffrey Shallit said...

Student:

Here it's not so much a matter of answering "correctly" - even I had an outline of an approach that would produce the correct answer, but the details would have been messy. It's more a matter of finding the combinatorially-satisfying solution. You also have to consider how much time you are willing to spend on it.

Ian Goulden, since he's an expert, could come up with the combinatorially-satisfying solution quickly. My guess is that many mathematicians could do it, but it might take a lot longer for them.

Filipe Calvario (from Brazil) said...

I have always liked problems about probability. However, after facing, a short time ago, "The Monty Hall problem", and failing miserably in it (after, obviously, have refused to understand and accept its right solution for a while), I'm in a sort of a temporary (I hope) lack-of-confidence crisis.

jason said...

After several days of thinking about it, I think I have a way of doing it . It involves using an extra card that is not an ace or a king (a joker, for instance), and then using inclusion-exclusion. Now I have to actually do the grunge to see if it works out...

Goulden's solution is a thing of beauty!

jason said...

It worked! .4863215... But now I won't be able to do arithmetic for a week without feeling sick.

My idea was this: the number of shuffles of a 52 card deck in which either AK or KA appears is the same as the number of shuffles of a 53 card deck (52 + joker) in which the first card is a joker, and at least one of the following sequences appears:

XAK
XKA
XAAK
XKKA
XAAAK
XKKKA
XAAAAK
XKKKKA

Here 'X' means any of the 44 + 1 = 45 cards that are neither Ace nor King. But that number is just 1/45 times the number of shuffles of 53 in which one of the above sequence appears, and the first card is an X. So you just subtract out the 53 card shuffles (containing at least one of the above sequences) that start with an Ace or a King. Having the 'X' as a separator allows you to make easy use of the inclusion-exclusion principle.

Takis Konstantopoulos said...

Here is another interesting probability problem.