# Card Games and the Monty Hall Problem

A standard deck contains 52 playing cards. All cards have one of two colors, red or black. Each card belongs to one of four suits, described as Hearts (red), Diamonds (red), Spades (black), and Clubs (black). There are 13 cards assigned to each suit, denoted as Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, and King.

Now select one card from a complete deck. The probability of drawing the Ace of Spades is 1/52. The probability of drawing a red card equals the probability of drawing a black card and equals 26/52 = ½.

Now suppose I want to calculate the odds of drawing two aces in two successive draws. On the first draw I have a 4/52 chance of drawing an ace. On the second draw, I have a 3/51 chance of drawing an ace, because they’re only three aces out of 51 cards left from which to draw. So the probability of drawing two aces is just the product of the two probabilities, or 4/52 * 3/51 = 12/2652 = 1/221.

Now suppose I want to calculate the odds of drawing one ace and one 2. I could draw the ace first and the 2 second or vice versa. So I could have the ace and then the 2 or vice versa. We’re going to write down all the ways this can occur:

Chose Ace of Hearts, Diamonds, Spades, Clubs and then 2 of Hearts, Diamonds, Spades, Clubs

Chose 2 of Hearts, Diamonds, Spades, Clubs and then Ace of Hearts, Diamonds, Spades, Clubs

The probability of drawing an ace first is 4/52, and of drawing a 2 second is 4/51, so the probability of (ace,2) = 4/52 * 4/51 = 16/2652 = 4/663. Similarly, the probability of drawing a 2 first and an ace second is 4/663. By accounting for both permutations, we see the probability of drawing an ace and a 2 is 4/663 + 4/663 = 8/663.

Now suppose I deal three cards from the diamond suit to you face down. The object of the game is to select the card with the highest value without looking. You choose one card, but don’t look at it. I turn over one card which is not the winning card. Should you keep your original card, or switch to the other card?

In fact, this is just a card-based version of a famous probability challenge known as the Monty Hall problem:

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

We’re going to use probability to determine whether you should switch. Your odds on the first choice of picking the correct door are 3/9 =1/3 (column 2). If you don’t switch, your odds remain the same (column 4). If you do switch, your odds of winning rise to 2/3 (column 6). So it is better to switch!

Car located behind door: A, A, A, B, B, B, C, C, C

The contestant chooses: A, B, C, A, B, C, A, B, C

Then host opens door: B or C, C, B, C, A or C, A, B, A, A or B

Result if doesn’t switch: wins, loses, loses, loses, wins, loses, loses, loses, wins

If contestant switches to: A for C or B, B for A, C for A, A for B, B for C or A, C for B, A for C, B for C, C for B or A

Then result becomes: loses, wins, wins, wins, loses, wins, wins, wins, loses

It remains true that all probability problems rely on counting and reasoning. But sometimes, like in this problem, the results can be very counterintuitive. Intuition fails here because we overlooked something. The host already knows which door is the correct door, so the door he chooses to open is not random. He will never open the correct door or the door chosen by the contestant. So, in point of fact, the probability remains 1/3 that the contestant’s first guess was right, and the probability of the switched-to choice being right is therefore 2/3.

So maybe this makes your brain hurt and you don’t believe this solution. In fact, there was a considerable dispute about these results for years. The great thing is, you can confirm a theoretical result like this with experiments. Experimental results use not only probability, but the field of statistics, which we will explore more in next week’s blog.