| Monty Hall Problem |
Article Index for Monty |
Website Links For Monty Hall |
Information AboutMonty Hall Problem |
| CATEGORIES ABOUT MONTY HALL PROBLEM | |
| probability theory paradoxes | |
| lets make a deal | |
|
The Monty Hall problem is a puzzle involving Probability loosely based on the American game show '' Let's Make A Deal ''. The name comes from the show's host, Monty Hall . A widely known statement of the problem is from Craig F. Whitaker of Columbia, Maryland in a letter to Marilyn Vos Savant 's September 9 , 1990 , column in '' Parade Magazine '' (as quoted by Bohl, Liberatore, and Nydick). ''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?'' The problem is also called the Monty Hall paradox, in the sense that the solution is counterintuitive, although the problem does not yield a logical contradiction. Many people incorrectly assume since there are only two doors left after the host opens one, that the remaining doors must have equal probability. However, assuming the host knows what is behind each door, always opens a door revealing a goat, and always makes the offer to switch, the host's actions do not affect the probability of the player initially choosing the car (1/3). This means the answer is '''yes''' — switching results in the chances of winning the car improving from 1/3 to 2/3. PROBLEM AND SOLUTION The problem, with all constraints explicit The version of the problem popularized by ''Parade'' unfortunately leaves room for possible misunderstandings, some of which may affect the correctness of the answer. An unambiguous statement of the problem, with explicit constraints on the host, was described by Mueser and Granberg: :''A thoroughly honest game-show host has placed a car behind one of three doors. There is a goat behind each of the other doors. You have no prior knowledge that allows you to distinguish among the doors. "First you point toward a door," he says. "Then I'll open one of the other doors to reveal a goat. After I've shown you the goat, you make your final choice whether to stick with your initial choice of doors, or to switch to the remaining door. You win whatever is behind the door." You begin by pointing to door number 1. The host shows you that door number 3 has a goat. Do the player's odds of getting the car increase by switching? The solution The answer to the problem is yes; the chance of winning the car is doubled when the player switches to another door rather than sticking with the original choice. At the point the player is asked whether to switch there are three possible situations corresponding to the player's initial choice, each with equal probability (1/3):
If the player chooses to switch, the car is won in the first two cases. A player choosing to stay with the initial choice wins in only the third case. Since in two out of three equally likely cases switching wins, the odds of winning by switching are 2/3. In other words, a player who has a policy of always switching will win the car on average two times out of the three. Another way of thinking about this explanation is by observing that the player begins the game with a 2/3 chance of picking a door with a goat behind it. In this case the host must show the other goat, so switching turns the initial 2/3 chance of picking a goat into an equivalent chance of winning the car. Switching loses only if the player picks the car first (1/3 chance). The total expected outcome for the player who always switches is therefore a 2/3 chance of winning the car. The problem would be different if the game host were permitted to make the offer to switch more often (or only) depending on knowledge of the player's original choice or if the host does not know what is behind each door. Some statements of the problem, notably the one in ''Parade Magazine'', do not explicitly exclude these possibilities. For example, if the game host only offers the opportunity to switch if the contestant originally chooses the car, the odds of winning by switching are 0%. In the problem as stated above, it is because the host must reveal a goat and must make the offer to switch that the player has a 2/3 chance of winning by switching. AIDS TO UNDERSTANDING The most common objection to the solution is the idea that, for various reasons, the past can be ignored when assessing the probability. Thus, the first door choice and the host's forced response are ignored. Because there are two doors to choose from, many people jump to the conclusion that there must be a fifty-fifty chance of choosing the right one. Although ignoring the past works fine for some games, like Coin Flipping , it doesn't work for all games. In this case what should be ignored is the ''opening of the door''. The player's choice is between the originally picked door and the ''other two'' — opening one is simply a distraction. There's only one car. The original choice divides the possible locations of the car between the one door the player picks (1/3 chance) and the other two (2/3 chance). It is already known that at least one of the two doors contains a goat. The revealing of the goat therefore gives the player no additional information about his own door. It doesn't change the 2/3 probability that the car is still in the block of two doors. Another possible reason for confusion is that the problem is often stated as though the host takes the player by surprise with his opening of the door. This tends to give the impression that the host is trying to confuse a player who has chosen correctly. This is why the rules are so strictly formulated in the previous section. The player may still not know these rules, but that does not alter the probability; it just means the player is discouraged from rationally considering the options and making the optimal choice. Increasing the number of doors It may be easier to appreciate the result by considering a hundred doors instead of just three. In this case there are 99 doors with goats behind them and 1 door with a prize. The player picks a door; 99% of the time, the player will pick a door with a goat. Thus, the chances of picking the winning door at first are very small: only 1%. The game host then opens 98 of the other doors revealing 98 goats and offers the player the chance to switch to the only other unopened door. On 99 out of 100 occasions the other door will contain the prize, as 99 out of 100 times the player first picked a door with a goat. At this point a rational player should always switch. To drive the point home, one only has to imagine the host having to start with the first door and go down the line, opening the doors but skipping over only the player's door and one other door. The player can then more easily appreciate the randomness of his first choice and the large amount of information he has gained since he made that choice and then see the wisdom in switching. There is a reasonable question to this logic: if we increase the number of doors, why does this explanation assume the host would open 98 doors to make the problem similar to the original? Why doesn't the host open 33 doors instead? That's actually an essential ingredient for the counter-intuitiveness of the original problem. It's correct to assume the host would open 98 doors in this alternate game because in the 3 door game the player has only one switching option — so the player in the 100 door game must also be presented with a single switching option. The 3 door game is misleading because the player is always presented with 1/3 proportions: he has a 1/3 chance of winning, the host reveals 1/3 of the mystery, and he is allowed to switch to the other 1/3 option — therefore all options seem equal. Venn diagram The probability that the car is behind the remaining door can be calculated using Venn Diagram s. After choosing door 1, for example, the player has a 1/3 chance of having selected the door with the car, leaving a 2/3 chance between the other two doors. Note that there is a 100% chance of finding a ''goat'' behind at least one of the two unchosen doors because there is only one car. The host now opens door 3. Since the host must always open a door revealing a goat, opening this door does not affect the chance that the car is behind the originally chosen door which remains 1/3. The car is not behind door 3, so the entire 2/3 probability of the two unchosen doors is now carried only by door 2, as shown below. Another way to state this is that if the car is behind either door 2 or 3, by opening door 3 the host has revealed it must be behind door 2. Decision tree More formally, the scenario can be depicted in a decision tree. In the first two cases, wherein the player has first chosen a goat, switching will yield the car. In the third and fourth cases, since the player has chosen the car initially, a switch will lead to a goat. The total probability that switching wins is equal to the sum of the first two events, . Likewise, the probability that staying wins is . Combining doors Instead of one losing door being opened (and thus eliminated from the possible array of choices), it may equivalently be regarded as combining two doors into one (i.e. reducing the two doors which were not chosen into a single option, since the player can't, and won't choose the opened, losing door anymore). In essence, this means the player has the choice of either sticking with the original choice of door (1/3 chances), or choosing the sum of the contents of the two other doors (2/3 chances). Notice how the game assumptions play a role here — the switching is equivalent to taking the combined contents ''because'' the game host is required to open a door with a goat. Bayes' theorem Analysis of the problem using Bayes' Theorem has the least reliance on verbiage and the most on formal mathematics. It also makes explicit the effect of the assumptions given earlier. Consider the position when door 3 has been chosen and no door has been opened; the probabilities P(C1) that the car is behind door 1, P(C2) (car behind door 2), and P(C3) (car behind door 3), are all plainly 1/3. The probability that the game host will open door 1, P(O1), is 1/2: : |
|
|