Should you switch doors?

[exreo]

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?

[Brian GC]

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?

I would assume a game show would do that to get you to change your door choice. The big question is, do they want you to win a car or lose?

[matt1124]

I took a statistics class that addressed this in a lecture. The conclusion was that you should pick another door. I wish I could remember the whole thing! It had something to do with at first having a 1 in 3 chance of picking the prize door, and once you know what one of the doors is, you now have a 50/50 chance of picking the prize door. Surely someone around here knows what I'm talking about because I sure don't!

[jtrammel]

i would listen for the other goat to baa then choose the door thats not baaaing

[exreo]

I took a statistics class that addressed this in a lecture. The conclusion was that you should pick another door. I wish I could remember the whole thing! It had something to do with at first having a 1 in 3 chance of picking the prize door, and once you know what one of the doors is, you now have a 50/50 chance of picking the prize door. Surely someone around here knows what I'm talking about because I sure don't!

You're close. If you switch doors, you have a 2 in 3 chance of winning. If you stay with the original door, you have a 1 in 3 chance of winning.

http://farmdev.com/thoughts/41/the-m...oat-or-a-car-/

[hearthman]

Nope, this is wrong. We ran this problem in statistics class, too. Originally, yes, you had a one in three chance of winning. Once one door was revealed, that reduced it to two chances remaining--not three. Therefore, if offered the opportunity to change your choice, you would be left with a one in two or 50/50 chance of winning.

One of the problems in your 2 in 3 model is the two. When changing or offered a change, you don't count the original plus a second, meaning you don't get to choose two doors at once. You get either/ or. A 2 in3 means they offer you to choose two doors and keep both choices.

[exreo]

Nope, this is wrong. We ran this problem in statistics class, too. Originally, yes, you had a one in three chance of winning. Once one door was revealed, that reduced it to two chances remaining--not three. Therefore, if offered the opportunity to change your choice, you would be left with a one in two or 50/50 chance of winning.

One of the problems in your 2 in 3 model is the two. When changing or offered a change, you don't count the original plus a second, meaning you don't get to choose two doors at once. You get either/ or. A 2 in3 means they offer you to choose two doors and keep both choices.

I don't think you have a 50/50 chance. If you did, you should not switch doors. The people who switch doors have a 66.67% chance of winning. Classrooms around the country did this test 100 or 200 times, and by switching doors they were able to win 66.67% of the time. If you had a 50/50 chance of winning, you would only win 50% of the time.

[dls2]

Yes Myth Busters ran this same test a while ago and got the same results always better to switch you choice.

[Brian GC]

Why would it be better to switch doors?

When the game starts you have a 1 in 3 chance of winning. With one door eliminated you now have a 50 /50 chance of winning, not a 66% of winning when there are only two doors left.

This same thing is done with roulette or blackjack. Just because you lost 10 hands in a row doesn’t make your chances of winning the next hand any better.

Another example: say you had ten doors to choose from. You eliminated eight of them and only two are left to be opened. Do you have an 80 -90% chance of getting that flip of the coin right? NO!

[samtheman]

the math can be hard to wrap your head around, but it is better to switch doors after one is revealed. DLS is right, the mythbusters did an episode on this, and the results clearly support switching doors. When first choosing, you are picking one door from three possible choices. Odds are 1 in 3 that you are choosing correctly. After one door is revealed as a loser, only two choices are left. If you stay with your original guess, it is still 1 in 3 odds that you are right, as that is what it was when the door was originally chosen. If you switch, you are now choosing between two doors, and your odds are 1 in 2 that you will be guessing right. seems counter-intuitive, i know. But the results after repeating the test many times over clearly support this. They also did another test on that episode about human nature, with the participants overwhelmingly staying with their original choice, and most of them losing

[Brian GC]

the math can be hard to wrap your head around, but it is better to switch doors after one is revealed. DLS is right, the mythbusters did an episode on this, and the results clearly support switching doors. When first choosing, you are picking one door from three possible choices. Odds are 1 in 3 that you are choosing correctly. After one door is revealed as a loser, only two choices are left. If you stay with your original guess, it is still 1 in 3 odds that you are right, as that is what it was when the door was originally chosen. If you switch, you are now choosing between two doors, and your odds are 1 in 2 that you will be guessing right. seems counter-intuitive, i know. But the results after repeating the test many times over clearly support this. They also did another test on that episode about human nature, with the participants overwhelmingly staying with their original choice, and most of them losing

I will venture to say I cannot wrap my head around it because it is wrong. As for Mythbusters, they have a good show but if you want to see them perform a bogus test watch them test pyramid power. It was a sham.

Prior to picking one out of three doors you have a choice to pick a door. That gives you a one out of three chance IF you open the other two doors at once. If you open just one and are given another chance to pick a door to open then your chances become 50/50. It can never be more than 50/50 with two doors remaining and you have a choice between them.

If this is wrong then it must be explained…not left up to Mythbusters. If they ran a few hundred tests on their method and came up with a 66% advantage, that test is also a sham. Anybody can go out in Vegas a beat the 47% to 53% odds against them for a night. But try doing that for a few million hands.

[samtheman]

i don't know where the 66% advantage came from, i said 1 in 2 (50-50) if you switch. I believe that is what they said on the show as well. That being said, i just looked at the link above, and maybe i am wrong. I do know that it is in your favor to switch after one door is revealed

[Brian GC]

i don't know where the 66% advantage came from, I do know that it is in your favor to switch after one door is revealed

Sorry, you didn’t say the 66% part. It was in post #7.

The only advantage comes from having the opportunity to switch doors after one has been revealed. No opportunity 33%, opportunity 50%.

Switching doors is like flipping a coin, it’s 50/50. Watching someone do this test manually is not the way to test it. Not enough information. It would be like following one person on a night in Vegas…not enough test data. Do it on a computer where you can run a few million tests.

[Brian GC]

Sam,

I was very wrong. Switching doors does give you a 66% chance of winning.

It is explained very clearly here: http://www.youtube.com/watch?v=mhlc7peGlGg

BTW - Mythbusters does a poor job of explaining it, probably because thy are more interesting in making a TV show than understanding what they are testing.

[samtheman]

thanks brian, that explaination makes a lot more sense. it is based on the probability of your first choice , not so much the swap.

[ArthurHagar]

I'm still not seeing their logic in this. Yes you have a 33% chance when you pick the first time around. But when they open a door, then instead of staying with your pick at 33%, what your actually doing is just re-picking your original door at a 50% chance. So either way your falling with 50/50 odds.

[Brian GC]

I'm still not seeing their logic in this. Yes you have a 33% chance when you pick the first time around. But when they open a door, then instead of staying with your pick at 33%, what your actually doing is just re-picking your original door at a 50% chance. So either way your falling with 50/50 odds.

That is what I thought too. I thought two doors means 50/50. But not so.

I’ll try putting it a different way. You have a 33% chance of picking the car. So let’s say you pick the car 33% of the time. They will always open and eliminate the other door with a goat behind it. Therefore, if you stick with your original pick and not switch doors you will win a car 33% of the time.

Done the other way: You have a 66% chance of picking the door with a goat behind it. Again they will open and eliminate the other door with a goat behind it. This will leave you with your pick of a goat and the other door with a car behind it 66% of the time. In that case would you rather stick with your goat or switch doors?

[hearthman]

incorrect logic applied. Here's why:
Once the first door is revealed, the percentages change. The original odds were 33% car/ 66% goat. However, once the first goat is revealed, you have reduced your choices from 3 to to. This brings it down to a 50/50 chance.

Now, you won't know whether your first pick was a car or goat until the end of the show. So, if you choose door 1, he reveals a goat behind door 2, that means there are only 2 doors left and one has the the car and one has the goat. Its the best you can get. This scenario is not accepting the fact the odds change once the first door is revealed.

[nismo1]

incorrect logic applied. Here's why:
Once the first door is revealed, the percentages change. The original odds were 33% car/ 66% goat. However, once the first goat is revealed, you have reduced your choices from 3 to to. This brings it down to a 50/50 chance.

Now, you won't know whether your first pick was a car or goat until the end of the show. So, if you choose door 1, he reveals a goat behind door 2, that means there are only 2 doors left and one has the the car and one has the goat. Its the best you can get. This scenario is not accepting the fact the odds change once the first door is revealed.

+1

[samtheman]

here's the way i think the logic works. when you make your first choice, there are two goats and one car, you have no idea which is which so you are picking a door at random. one door is as car, one door is a goat, one door is also a goat. so, choosing at random, you have a 66%chance of being wrong. The odd are 2 in 3 that you are choosing WRONG. after your choice, one door is revealed as a goat. your original choice is a 66% chance, still, that probably wrong. Since if you change, the only possible choice is the correct choice (since the other door was revealed as a goat, and you can't choose that door) you can turn your most likely wrong choice into the right choice 66% of the time.