Back to Probability Theory

### Problem

- You’re a contestant on a game show and asked to choose one of three closed doors. Behind one door is the car of your dreams; behind the others, goats. You pick a door, Door 1, say, and the host, knowing what’s behind each door, opens one of the other doors, Door 3 for example, revealing a goat. (He always opens a door with a goat.) He then gives you the option of changing your selection to Door 2. Do you have a better chance of winning if you switch; or are the odds the same?

### Solution

- This is the famous
*Monty Hall*problem, which made national news in 1991. It’s tricky, with hundreds of PhDs getting it wrong. There are two plausible solutions:*The Simple Solution*: For each door the initial probability is ⅓ the car’s behind it. But the situation changes when Door 3 is opened. For then only two doors are unopened, a car behind one, a goat behind the other. It’s as though a new game had begun, with two doors rather than three. For each door, therefore, the probability is ½ the car’s behind it.*The Complex Solution:*Having selected Door 1, there are three equally likely scenarios.- The car’s behind Door 1, in which case the host opens either of the other doors. In this scenario you win if you stick with Door 1.
- The car’s behind Door 2, in which case the host opens Door 3. In this scenario you win if you switch.
- The car’s behind Door 3, in which case the host opens Door 2. In this scenario you also win if you switch.
- The probability of each scenario is ⅓, the scenario’s being equally likely. Therefore, since you win if you switch in two of the three scenarios, the probability you win if you switch is ⅔.

- There are thus two plausible arguments. Happily, there’s a way of deciding the matter: play the game lots of times. Perhaps not the actual game, since we’d need doors, goats, and a cool car. But we can simulate the game by playing cards. The
*host*randomly places three cards face down, an ace and two deuces. Peeking at the cards, he knows which is which. The*contestant*picks a card. The host turns over one of the other cards, revealing a deuce. The cards are then examined to see whether sticking or switching wins and the result tallied. The game is played hundreds or thousands of times. Here are the results of a computer simulation that plays the game thousands of times using a random number generator:

- If the probability of winning-if-you-switch were ½, the more games played the closer the fractions in the right-most column would be to 0.5. If the probability were ⅔, the more games played the closer the fractions would be to 0.66666666… The probability of winning-if-you-switch is therefore ⅔. The underlying principle is the Law of Large Numbers, roughly the following:
- if P(A|B) = n, A / B approaches n, as the number of Bs increases.

- The Law of Large Numbers keeps casinos in business.

### Backstory

- Now the story behind the problem. Listed in the
*Guinness Book of Records*as having the highest I.Q., Marilyn vos Savant writes the*Ask Marilyn*column for*Parade*magazine, part of many Sunday newspapers. Readers write her with questions and puzzles. In 1991, replying to a letter that had posed the Monty Hall problem, she argued by analogy that the probability of winning-if you-switch was ⅔:- “Suppose there are a
*million*doors, and you pick door number 1. Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door number 777,777. You’d switch to that door pretty fast, wouldn’t you?”

- “Suppose there are a
- The letters poured in, about 10,000, most critical. Typical comments: “You blew it and you blew it big!”; “May I suggest that you obtain and refer to a standard textbook on probability before you try to answer a question of this type again.”; “Maybe women look at math problems differently than men”; “You are the goat.” A thousand or so critical letters were from PhDs, many in math and science.
- Ms. vos Savant wrote a second column setting forth the three-scenario argument. Still more letters. Finally, in a third column she asked readers to test her theory by conducting a “nationwide experiment” involving a simulation using three paper cups and a penny. The results were what you’d expect.
- Many readers got the problem wrong because the answer was obvious to them: if there are two doors and one car, it’s “obvious” the probability’s ½. But if there are plausible arguments that the probability is ⅔ rather than ½, the arguments need to be examined, not dismissed with snide comments. The fact vos Savant’s arguments had no apparent logical flaws should have raised red flags in the minds of her critics, making them question their beliefs, however “obvious” to them.

### Mathematica Code for Simulation

- sticker = 0; switcher = 0;
- Do[car = RandomInteger[{1, 3}];
- guess = RandomInteger[{1, 3}];
- sum_cargas = car + guess;
- If[car == guess, Switch[sum_cargas, 2, open = RandomInteger[{2, 3}], 4, RandomChoice[{1, 3}], 6, open = RandomInteger[{1, 2}]], Switch[sum_cargas, 3, open = 3, 4, open = 2, 5, open = 1]];
- If[car == guess, sticker = sticker + 1, switcher = switcher + 1];,

- {10000000}];