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Card Colm
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June 2006
We'll combine a simple probabilistic observation with a brand new card effect from Martin Gardner---the prince of recreational mathematics for the past fifty years---and also with a different card sorting principle from the best man at his wedding, the late Bill Simon, which was explored in the last Card Colm. The upshot is two different ways to control who gets the better poker hand when ten cards are randomly selected from a standard deck. We only consider winning hands that are at least as good as a matching pair. (Winning with nothing more impressive than some unrelated higher valued cards can also be controlled by the methods discussed below, but we leave that to the interested reader to pursue.)
This problem can be tackled using a common approach to the classic Birthday Problem, which concerns the number of people required to ensure a greater than 50% chance of having at least one birthday match. The surprisingly small answer there is 23 people (although, in reality a smaller number works, as pointed out by Diaconis and Holmes).
The key to estimating such probabilities is to turn things around, and focus on
the chances of there being absolutely no match, noting that Prob(at least
one match) = 1 - Prob(no match). In our case, if k cards are picked
at random, then since there are four cards of each value, it is not hard to show
that for 2< k < 14, we have:
Prob(at least one match) = 1 - (52/52)(48/51)(44/50)...(52-4k+4)/(52-k+1).
Tabulating most of these probabilities yields:
| k | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Prob | 0.0588 | 0.1718 | 0.3239 | 0.4929 | 0.6548 | 0.7898 | 0.8879 | 0.9491 | 0.9810 |
Hence, we need to pick at least 6 cards to be at least 50% sure of a birthday card match. Given 8 or 9 cards, there is a high probability of a match, and with 10 cards, it's almost certain to occur. The following tricks exploit this last fact, which we'll refer to as the Birthday Card Match Principle.
The gist of this is that provided you get to go first, and you start by taking the first card in the row, you can ensure that you end up with all the cards in the odd positions, whereas if you start by taking the last card in the row, you can ensure that you get the cards in the even positions. Note that after your initial selection, you invariably take the card next to the one your opponent just took; this observation saves you trying to keep track of odd/even positions as the row shrinks. (However, don't make it too obvious: feel free to take your time and appear to give the cards at each end due consideration, before deciding which one to remove!)
You may wish to point out that on three occasions, your opponent gets to decide between the card on the left and the card on the right. Hence, you could claim: "As a result, you essentially determined one of 23 = 8 possible outcomes for your cards. Yet I predicted the final result."
Hand the deck out for shuffling and ask for any ten cards. Pick them up and glance at their faces briefly. Drop them face down on the table, two or three at a time, commenting that you will give a spectator lots of free choices for the selection of five of these. The idea is to work with what you are given, assuming there is at least one matching pair, and drop clumps of cards so that there are two interwoven poker hands, one occupying the odd positions, and the other the even positions. Let's assume that the odds contain the matching pair, or whatever good cards you get, hence providing a better poker hand than the evens.
For instance, if there are two 5s, and nothing else of interest to poker players, just arrange it so that those two cards are in any of the first, third, fifth, seventh or ninth positions. If you get three of a kind, do likewise. If you get two pairs, you can get the higher one in odd positions, and the lower one in even positions, or you can be greedy and arrange all four cards in odd positions. No matter what ten cards you are handed, just massage them into interwoven hands as indicated, with the winning hand in the odd positions.
Pick up the packet, turn it face down, and shuffle some more. We suggest several types of shuffle which will preserve the essential interweaving property here, while maintaining the winning cards in the odd positions. (Try these with a face-up packet of ten cards alternating red/black and you'll soon be convinced.) For instance, fix a number k between 5 and 10, and do the "quad run" false shuffle implicit in Low Down Triple Dealing: running off k cards from the top of the packet into the other hand, and then dropping the rest on top, restores the packet to its original order if done four times total. Alternatively, do the following an even number of times: run off any odd number of cards from the top of the packet and drop the rest on top. Or you can simply cut any even number of cards from top to bottom. You may also combine these ideas to increase the illusion of fair shuffling.
Now, lay out the cards face-down, being careful to silently note which end of the row includes a card from the winning hand. As long as you start by selecting that card, and always subsequently pick the card next to the one which your opponent chooses, you are certain to end up with the victorious hand. Don't have any cards turned over until the end.
Hand the deck out for shuffling and ask for any ten cards to be handed to you. Glance at their faces briefly. Drop them, face down on the table, two or three at a time, commenting that you will give a spectator virtually free rein for the selection of five of these. Once again, the idea is to work with what you are given, assuming that there is at least one matching pair. The goal this time is to have the good cards---namely, the matching pair or better---somewhere in the group of four cards starting with the third from the top of the face-down packet. These will later contribute to the spectator's winning hand. The last four cards will end up in your hand, and the top two cards should be such that no matter which one you get and which one the spectator gets, you are still sure to have the losing hand.
Pick up the packet and shuffle it some more while maintaining the two/four/four groupings just described, with the good cards in the middle group. For instance, the "quad run" false shuffle mentioned earlier is appropriate here, shuffling off the same number of cards (at least five) four times. (There are other, more complex shuffling possibilities.)
"First, we each need a hole card. You decide which one I get and which one you get." Invite the spectator to take the top two cards, look at them, and put one in front of each of you, face down, as hole cards. "Next, you'll decide on two cards for your pile, then two for my pile, then one more for your pile, then one more for mine. Finally, I'll deal whatever two cards remain---something you will have determined---the first one goes to you, the second to me."
Hold the next two cards of the packet in front of you, making sure that nobody can see their faces. "One of these cards will be added to your pile, the other will be tucked underneath the packet in my hand. You decide which you get." If the spectator selects the top card, deal it to appropriate pile, and tuck the next card underneath the remaining cards in your hand. If the spectator wants the second card, tuck the first card underneath and deal the second card to the pile. Repeat for the next two cards; the net result is that three cards are now in a pile in front of the spectator, and two have been tucked underneath. Next, give the spectator the same freedom of choice for the next two pairs of cards, only this time the cards selected are dealt into the second pile in front of you. Then, repeat for the next pair of cards in the packet, dealing the selected card onto the spectator's pile, and finally once more with the next pair of cards, dealing the selected card onto your own pile. "That's four cards for each of us," you comment, "We need one more apiece." Deal out the last two cards just as you promised you would; the top one to the spectator's pile, and the bottom one to your own pile.
"You had so many free choices, and I bet used them wisely! I think you'll find that you have a poker hand that will easily beat mine." Have the two piles turned over to verify your claim. (You can always control who has the better hand of course, but we've suggested that you let the spectator win here for a change.)
Now that know how to conspire to have winning (or losing) poker hands, under
conditions that appear quite fair, we should point out that none of this helps
you to win at real games of poker, where the rules generally do not permit the
kinds of peeking, shuffling, mixing and dealing considered above!
I beg your pardon, I never promised you a song, reader.
Colm Mulcahy (colm@spelman.edu) has taught
at Spelman College since 1988, where he is almost finished his three year term
as department chair. You can find more writing on mathematical card tricks at
http://www5.spelman.edu/~colm/cards.html.
"Roaming Nerds" is an anagram of "Gardner/Simon," and
"song, reader" is an anagram of "rose garden."