You are on a game show and there are three closed doors.

Behind one of them is the grand prize and nothing behind the others.

You choose door #1 so the host opens door #3 knowing it is empty.

The host then asks you if you would like to change your selection.

Should you?

Answer below.

If you change your choice to door #2 your odds of winning will actually increase from 1/3 to 2/3!

Confused? Yea, I was too.

A good way to understand how this is possible is to increase the total number of doors in the problem to one million.

Pick a door from the million? Do you like your odds?

The game show host then opens all losing doors leaving only yours and one other.

Do you want to change your choice now?

Put another way, you are very likely to pick the wrong door the first time (2/3) and then the game show host does you a favor by revealing the other losing door.

Thank you to the theproblemsite.com for explaining this to me!

## Monday, April 14, 2008

### Game Show Host Problem

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## 5 comments:

What in God's name is this supposed to mean.

Hmmm, well firstly i would like to ask - How is it only door number 2 gets that increased probability of winning? If you are in fact talking in terms of probability,

Probability = No. of possible outcomes DIVIDED BY no. of favorable outcomes.

In the first case he has a one third chance of winning cause the favorable outcome being that the door he chooses is the right door is one, and number of possible outcomes is three because there are three doors with three different outcomes. Now after one door is eliminated, He has a 1/2 chance of winning meaning his odds are increased to 50%, because the number of possible outcomes has changed and so has everything else. His favorable outcome is still that the door he chooses is the right door and the possible outcomes are hes right or hes wrong. How can you suppose that after eliminating door number three, the odds of door number two increase?

OK. So everyone I've talked to agrees that the first choice has a 33% chance of winning. The problem comes with the second choice (after

the game show host (GSH) changes the problem. Are the improved odds given in the movie (67%) correct or are the odds only 50/50?

I think of it this way. For the first offer, there are 3 possible ways the doors can be configured (C=car; G=goat):

Door: 1 2 3

Prize: C G G

G C G

G G C

So, if we choose any set of outcomes (a group of 3 possibilities vertically above, defined by a door choice) then there is one car and two goats in each column, so the chances of winning, regardless of the choice is 1 in 3 (33%).

Clearly then, the first choice is a complete gamble - any group chosen has an equal chance of a positive outcome.

The situation after choosing Door #1 is:

Door: 1 | 2 3

Prize: C | G G

G | C G

G | G C

Where the vertical line divides the the set of choices you made (Door 1) from the set not chosen (Door 2 + Door3).

Now, keep in mind the following requirements for the next part of the problem:

1. The GSH must open exactly one door.

2. The GSH cannot open your chosen door.

3. The GSH cannot show you a prize.

Number 3 here is key. If the GSH opens a door with a prize, the game is over and you lose. Therefore, you won't be around to make the second choice and, we presume that the producers don't

want that outcome on their show. Requirement 3 is the result of the information in the movie that the GSH "knows what's behind the other doors."

So, what is the situation after the GSH opens another door? (O=open)

Door: 1 | 2 3

Prize: C | G O

G | C O

G | O C

Note that in row 3 above, the GSH must open door #2, not door #3 by requirement 3 above. Now, the set not chosen (door 2 + door 3) has

2 cars and 1 goat, making the chance of winning 2 in 3 or 67% should you choose to change from one set to the other; meaning that

you change your choice from door #1 to the remaining available door (which may be door #2 or door #3 depending upon the prize/goat setup). This is the outcome given in the movie.

Now, I can hear a lot of sputtering about why didn't the GSH always open door #3?

If the GSH had no knowledge of the game i.e. requirement 3 above were relaxed, then there would be a 1 chance in 3 (33%) possibility that he would open a door with a prize and you would lose outright. In such a case, you would not be around to make the second choice. The outcome in that case is

Door: 1 | 2 3

Prize: C | G O

G | C O

X | X X

Where X is 'game over.' Now, guess what? The alternative set has one goat and one car, giving you a 1 chance in 2 (50%) probability

of winning if you change to door #2. The set chosen has the same probability (50%), since you're still around to play. So, in this case, wer're back to gambling;

change your door or don't, it doesn't make any difference in this case.

The key point in this problem is that the GSH is not carrying out random, independent trials as if this were roulette. Because he has knowledge of the game, he can change the likelihood of future success.

There's a possiblity that you left out of your table:

C | O G

Using your logic, that would give a 2/3 probably that staying with your original choice is a winner, except the numbers have changed and it's really 2/4 (or 1/2) if you keep your original choice and 2/4 (or 1/2) if you change your mind.

Now, you may say "that's fine if your original choice had a car behind the door, but what if it had a goat?" I did a set of tables for that scenario and it still came out the same.

I don't know if trying it in real life makes a difference. Doing a series of coin tosses in real life sometimes yeilds a much greater discrepency than I'd expect. I guess it matters how many times you try it.

If your strategy is to always switch, you have a 2/3 chance of initially picking a goat, which will leave you with the car after switching EVERY time. A 'switch' strategy therefore offers 67% chance of success.

For anyone who disagrees, let's set up a real-life game where we each play it 25 times, with $100 to play forr each time. You will soon realise.

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