Card Counting

Card Counting: Simple Math

If Ben and Professor Rosen’s (Kevin Spacey) classroom banter threw you off, that’s probably because your high school teachers weren’t doing their jobs when it came to teaching you probability. Actually, their whole spiel is really just “simple math,” and you can pick up these card-counting principles, as well as counting itself, just as easily as the next guy. First, what you have to understand is that the law of probability is only determining what your odds are based on a limited number of possibilities. Basically, you take the total number of desired outcomes over the total number of potential outcomes.

The tricky part to all this, though, is that you have to be sure of what the proposition actually is. For example, in the case of Ben and Professor Rosen’s chalkboard game, it might seem that Ben is only picking a winner out of three random possibilities. Fair enough, you say, that means he has a 1:3 chance of choosing right.

The problem with thinking this way is that whether or not Ben chooses the winning board the first time is neither here nor there. Either way, Professor Rosen is actually the one doing the choosing, with the stipulation that he can’t slide down the chalkboard Ben picked. Since there is only one prize in this problem and it is only behind one of the doors, there is a 2:3 chance that Rosen will select a losing chalkboard. And because Rosen is only selecting one chalkboard, there is a 2:3 chance that Ben will select a non-losing one. As you can see, then, the chances are really in Ben’s favor the first round; he has a 2:3 chance of not selecting a losing board, and he has a 2:3 chance that Rosen will.

Assuming, then, that Rosen does reveal a non-winner, in the second round Ben’s chances now become dependent on the probability that he would have chosen the winner in the first round. If the game were “pick a board, and we’ll immediately reveal the other unpicked boards,” Ben would have a 1:3 chance of choosing correctly and, like Professor Rosen, a 2:3 chance of choosing incorrectly. Meanwhile, the probability that either of the two boards, which Ben doesn’t pick, conceal the prize is 2:3. To put it to you another way: since one of the other two boards is now eliminated, his original selection has only a 1:3 chance to the remaining unselected board’s 2:3. From here, then, the answer is simple. When Rosen asks him if he’d like to switch, he does—and wins an all-expenses paid trip to Vegas.

This little game is actually called “The Monty Hall Problem”—named for the famous “Let’s Make a Deal” host who asked contestants to pick between three doors—and it’s the kind of situation card counters deal with all the time. Like the possibility of getting certain cards out of an entire deck, the individual choices players made between doors in Monty Hall’s door game were not always “mutually exclusive.” That is, choosing one door in the first round did not mean that all the other options were permanently ruled out.

A good way to think of this is to compare dice and cards. Say, for instance, you roll a die and are shooting for a 6 and a 2. You can’t actually roll both numbers on the six-sided die without re-rolling it and, therefore, resetting the odds. Now say you try the same experiment with a deck of cards. Unless you reshuffle the deck before every draw, you can easily pull both numbers without either of them eliminating the possibility of drawing the other.