# Can a mathematically, statistically inclined person please explain this to me?

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Suppose...

...You have the choice of 3 doors - Door #1, Door #2 and Door #3 and you should guess behind which door a prize is hidden.

You say Door #1 (it is just a guess)

The game show moderator then opens Door # 3 and shows that the prize is NOT hidden behind Door # 3.

Now you can stick with your initial choice of Door # 1 or you can switch to Door # 2.

Statistically, you should switch to Door # 2 to double your chances of picking the correct door behind which is the prize.

Your chances in the beginning were 1/3 - 3 doors and you choose one.

After one door is opened and you know this is not the correct door, the chances for Door # 2 are now (supposedly) 2/3 while the chance for Door # 1 remains at 1/3.

In my feeble mind, there are still two doors and the prize could be behind either one.

What do I not understand here?

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???

You're hurting my brain.

:)

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It's called the Monty Hall problem. If you google that, you'll find lots of info, interactive explanations, and you tube videos.

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The problem is that the one door has been opened by the game master who knows behind which door the prize is hidden. He does not open a random door, but one where he knows it will not reveal the prize. That introduces the change. The three doors are no longer equivalent.

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Yes, this is based on the old Let's Make a Deal program. Monty Hall was the host, hence the problem is named after him.

Here's a link to a column in which this has been discussed at length, with some dissenting opinions, and what I think are good explanations of the correct reasoning. Anything I have to say would be redundant, I think.

http://marilynvossavant.com/game-show-problem/

Here's a link to a site where you can run a simulation:

http://nlvm.usu.edu/en/nav/frames_asid_117_g_4_t_5.html?from=category_g_4_t_5.html

Or, you could always try it yourself. Use three index cards, two with zonker "prizes" and one with a good prize written on them. You will need a partner who knows which is where. You choose one at random, your partner shows you one of the zonkers. Keep track of how many times you get the good prize if you stick with your original choice, and how many times you get the good prize if you switch.

Remember, you must run it many times to be statistically significant.

HTH

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Three doors, two goats, one car. When you make your initial pick, you only have a 1/3 chance of picking the car right off the bat. You have a 2/3 chance of picking a goat. Since it's most likely you picked a goat, when the host picks the only other goat door, he's pretty much revealing that the final door hides the car.

This all presumes that you picked a goat door in the first place, which is most likely since there are two goats and only one car.

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Your chances in the beginning were 1/3 - 3 doors and you choose one.

After one door is opened and you know this is not the correct door, the chances for Door # 2 are now (supposedly) 2/3 while the chance for Door # 1 remains at 1/3.

In my feeble mind, there are still two doors and the prize could be behind either one.

What do I not understand here?

Think of it this way: your chances of getting the "good" door were 1/3 to begin with; or, to put it another way 2/3 that you chose the wrong door. Just because one of the wrong doors was revealed, doesn't change the fact that you had a 2/3 chance of having chosen the wrong door to begin with. So if you switch when given the chance, there is a 2/3 chance that you will now have chosen the good door.

Of course, this doesn't take into account the psychological factor of "going with your gut".

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Think of it this way: your chances of getting the "good" door were 1/3 to begin with; or, to put it another way 2/3 that you chose the wrong door. Just because one of the wrong doors was revealed, doesn't change the fact that you had a 2/3 chance of having chosen the wrong door to begin with. So if you switch when given the chance, there is a 2/3 chance that you will now have chosen the good door.

Of course, this doesn't take into account the psychological factor of "going with your gut".

I get this math - the part where we start out with 1/3 chance, then after one door is opened there are 2 doors left. I am thinking now the chances are 50/50 or 1/2. This supposes that most people initially choose the wrong door.

I will now go to the links. I am curious about the explanations.

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I get this math - the part where we start out with 1/3 chance, then after one door is opened there are 2 doors left. I am thinking now the chances are 50/50 or 1/2. This supposes that most people initially choose the wrong door.

I will now go to the links. I am curious about the explanations.

It doesn't "suppose" most people would initially choose the wrong door. Two out of three times, the person chooses the wrong door initially.

The chances would be 50-50 if there were only two choices to begin with. But think about it for a minute: how would revealing a bad door change the odds of your having chosen the correct door to begin with? It simply doesn't.

It's similar to this: you have a fair coin. On any given flip, the odds of getting a tail are 50-50. Let's say you've gotten 4 tails in a row. What are the odds that the next flip will be a tail? Still 50-50! If it's a fair coin, it doesn't matter what happened before, as far as a single flip goes. However, if you asked what are the odds that you would get 5 tails in 5 flips, that's a whole different question, and would be very low odds.

You're in good company! If you read that column I linked you to, you will see that there were numerous guys with PhDs who were wrong in their thinking initially. Some of them were big enough to come back later and admit it. One teacher, however, even after her students ran the experiment and proved the outcome, still refused to believe it.

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It might help to imagine it this way:

* Instead of 3 doors, there are 1 million doors. You choose door #1. The odds that door #1 is the correct door is 1 in 1,000,000. The odds that the prize is behind one of the doors you did not pick is 999,999 in 1,000,000.

* Monty knows where the prize is. Monty opens 999,998 doors, leaving you with door #1 and door #2 still closed.

* The odds that the prize is behind door #1 is still 1 in 1 million, because you originally chose it out of 1 million possibilities. Monty's reveal did nothing to affect that.

* The odds that the prize is behind door #2 is still 999,999 out of 1,000,000, because that was the odds that the prize was not behind door #1 and Monty - who knew where the prize is - eliminated all of those other doors.

ETA: I think I finally got it right! LOL

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It doesn't "suppose" most people would initially choose the wrong door. Two out of three times, the person chooses the wrong door initially.

The chances would be 50-50 if there were only two choices to begin with. But think about it for a minute: how would revealing a bad door change the odds of your having chosen the correct door to begin with? It simply doesn't.

It's similar to this: you have a fair coin. On any given flip, the odds of getting a tail are 50-50. Let's say you've gotten 4 tails in a row. What are the odds that the next flip will be a tail? Still 50-50! If it's a fair coin, it doesn't matter what happened before, as far as a single flip goes. However, if you asked what are the odds that you would get 5 tails in 5 flips, that's a whole different question, and would be very low odds.

You're in good company! If you read that column I linked you to, you will see that there were numerous guys with PhDs who were wrong in their thinking initially. Some of them were big enough to come back later and admit it. One teacher, however, even after her students ran the experiment and proved the outcome, still refused to believe it.

I did read the column and it was quite amazing that PhD's evidently could not quite grasp it either.

Here is something you said: By revealing one door as being the wrong one, it does not change my chances.

But now I know that #3 is not the correct door, therefore I am still in the running with door #1. Of course, it could also be #2. But my chances have just gotten better, have they not? If the prize is indeed behind "my" door, then the moderator could have opened either of the other doors but chose #3.

Now y'all know why I am not a math major. :lol:

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I did read the column and it was quite amazing that PhD's evidently could not quite grasp it either.

Here is something you said: By revealing one door as being the wrong one, it does not change my chances.

But now I know that #3 is not the correct door, therefore I am still in the running with door #1. Of course, it could also be #2. But my chances have just gotten better, have they not? If the prize is indeed behind "my" door, then the moderator could have opened either of the other doors but chose #3.

Now y'all know why I am not a math major. :lol:

Ah, but the moderator knew that he was opening a "bad" door -- he didn't randomly choose which of the other two doors to open. If he did that, there would be a (good) chance that he would reveal the "good" prize.

Yes, you are still "in the running" with your original door, but the odds are still only 1 in 3 that you chose the "good" door, or 2 in 3 that you chose one of the "bad" doors. Since you wouldn't choose to switch to the door that's been revealed to be bad, that means there is a 2 in 3 chance that the door you didn't pick originally is the "good" door. Strictly playing the odds, you ought to switch.

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Your first pick will determine which door will be opened. I diagrammed it the following way:

#1 *p

#2

#3

#1*

#2p

#3 ( this door needs to be opened after first pick)

#1*

#2 (this door needs to be opened after first pick)

#3p

There are three doors for you to pick. The * stands for the door that has the prize, in my example always #1. The p stands for the door that you picked. In the first example door #2 or #3 can be opened after your initial pick. In example number two and three however, the person who opens the door after your initial pick needs to open the door that does not have the prize. If you look at all three possibilities, in 2/3 the prize is hidden behind the other door. In only 1/3 the prize is hidden in your first initial pick.

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I think I will never understand this one. Once that 3rd door is opened, it becomes irrelevant. We know that the prize isn't there so it's no longer a chance. In my mind it becomes 50/50 - there are only 2 doors - or 2 chances - at that point. I aced my stats class & am very mathematical. But this makes no sense to me whatsoever.

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It's not that the 3rd door becomes irrelevant. It's that your odds are locked in when you make your original guess.

You choose 1 door out of 3, so your odds are 1/3 that you picked the right door and 2/3 that you picked the wrong door.

So long as all 3 doors remain closed, each door has a 1/3 chance of holding the prize.

When the host (who knows where the prize is) eliminates one of the doors that you did not choose from the competition, it doesn't change your original odds. You still have a 1/3 chance that you picked the right door, and a 2/3 chance that you picked the wrong door. But now instead of that 2/3 chance being spread across 2 doors, there is only one door remaining that embodies that 2/3 chance.

Your odds are better if you now switch to that door with the 2/3 chance.

The only way you would get 50/50 odds is if, after the host eliminates one door from the competition, the location of the prize is reassigned randomly, and you get to guess again (i.e., the offer to switch your choice). At that point, you'd be choosing between 2 doors that each have a 50% chance of containing the prize.

The whole psychological aspect of the game is that it feels like your odds should go up when the host opens that door and there is no prize there, and you're now down to two doors. But no matter what door you pick originally, the host can always open one door and reveal a room with no prize. His opening the door does not provide you with any new information - you always knew that one door that you did not pick would definitely not contain the prize.

Contrast that to the situation where you have two contestants playing and neither knows the location of the prize. You (contestant #1) choose door 1. Contestant #2 chooses door 3, which could contain the prize or not. If they open door 3 and you see that it does not contain the prize, that does give you new information and so it does change your odds. Before any door was opened, each door had a 1/3 chance of containing the prize. But after one door is chosen randomly and opened and shown not to contain the prize, the other two doors now have a 50/50 chance of containing the prize.

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So you switch....and the car could still be behind the door you originally chose. That's why the whole concept of probability makes my head hurt.

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. You still have a 1/3 chance that you picked the right door, and a 2/3 chance that you picked the wrong door. But now instead of that 2/3 chance being spread across 2 doors, there is only one door remaining that embodies that 2/3 chance.

Your odds are better if you now switch to that door with the 2/3 chance.

.

This is the way I have heard it explained that actually made sense to me.

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See if this helps There are 9 outcomes. You and a friend are working together. There is one prize and two goats. You pick first but your friend gets to help. After you pick, she can peek at the two remaining choices and show you one of the goats, before you finalize your decision.

There are 9 possible outcomes if you have no help:

you pick the prize and stick with it. WIN

you pick goat 1 and stick with it.

you pick goat 2 and stick with it

you pick the prize but switch to a goat 1.

you pick the prize but switch to goat 2

you pick goat 1 but switch to a prize. WIN

you pick goat 1 but switch to goat 2

you pick goat 2 but switch to goat 1

you pick goat 2 but switch to prize WIN

Without interference you will win 3 out of 9 times, or 1/3 of the time. Common math. Makes sense.

Lets say door 1 is the prize and goats are behind 2 and 3.

you pick door 1 and stick with it YOU WIN

you pick door 1 and switch to door 2

you pick door 1 and would have switched to door 3 but friend eliminated this option by showing the goat

you pick door 2 and switch to door 1 YOU WIN

you pick door 2 and stick with it

you pick door 2 and would have switched to door 3 but friend eliminated this option by showing the goat

you pick door 3 and switch to door 1 YOU WIN

you pick door 3 and would have switched to door 2 but friend eliminated this option by showing the goat

you pick door 3 and stick with it

Yes, each time you have a remaining 50/50 chance of a goat. BUT by your friend showing you one goat, by SWITCHING you win 2/3 times. It is the words 'remaining 50/50 chance' vs 'switching' that causes the mental problem. They are not representing the same odds, because:

Each time she peeks and finds 2 goats left, she is no help to you. She simply leaves one goat, either goat 1 or 2. You now have a 50/50 chance of getting the prize vs goat.

Where your friend helps you, is when she sees one goat and one prize. By her telling you, don't pick door 2, or don't pick door 3, she is altering your odds to your advantage by taking away one of the bad options but only when you switch your answer based on her advice.

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So you switch....and the car could still be behind the door you originally chose. That's why the whole concept of probability makes my head hurt.

Yes, it could be behind the door you originally chose. And it would be about 1/3 of the time.

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Contrast that to the situation where you have two contestants playing and neither knows the location of the prize. You (contestant #1) choose door 1. Contestant #2 chooses door 3, which could contain the prize or not. If they open door 3 and you see that it does not contain the prize, that does give you new information and so it does change your odds. Before any door was opened, each door had a 1/3 chance of containing the prize. But after one door is chosen randomly and opened and shown not to contain the prize, the other two doors now have a 50/50 chance of containing the prize.

Wait a minute!!! This isn't true. I read this too quickly last night.

Just because there is another player involved, it doesn't change the odds as to whether you picked the winning door originally. The scenario described here is no different than the original, except the second player didn't get the prize.

There is, however, a 2/3 chance that one of you picked the correct door.

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Wait a minute!!! This isn't true. I read this too quickly last night.

Just because there is another player involved, it doesn't change the odds as to whether you picked the winning door originally. The scenario described here is no different than the original, except the second player didn't get the prize.

There is, however, a 2/3 chance that one of you picked the correct door.

By having another door chosen and revealed randomly (by another contestant) rather than non-randomly (by the host who knows where the prize is) it does change your odds. When door #3 is opened, your odds drop to 0% if the prize is revealed. Your odds increase to 50% if the prize is not revealed. The odds change because when door #3 is chosen randomly and revealed to contain nothing, you now have new information. There now exist 2 doors that randomly contain 1 prize, so each has 50% odds of containing the prize. You can stick with your first choice or change, but either way you have 50% odds.

This is the equivalent of if, after the host shows you one empty door, the location of the prize is randomly re-assigned. You now have 2 doors and 1 prize randomly assigned behind them, so each door has a 50/50 chance.

This is different than when the host reveals a door that he knows did not contain the prize (non-random). Since the host knows where the prize is, he will reveal a door with nothing behind it. This does not change your odds. There is no chance that your odds could drop to 0% because the prize will definitely not be behind the door that the host opens.

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But you don't have a 50/50 chance. Another person who is asked to pick whether or not you have the prize would have a 50/50 chance. You're still part of the original problem and the original problem is what determines your chances.

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But you don't have a 50/50 chance. Another person who is asked to pick whether or not you have the prize would have a 50/50 chance. You're still part of the original problem and the original problem is what determines your chances.

OK, I'll give it more thought - maybe I have the 2nd scenario wrong (with 2 players both with no knowledge). In any event, it's a different scenario than the original one posited by the op, where 1 player has no knowledge and the host has complete knowledge.

ETA: Alright, I concede that I my understanding of the 2nd scenario was wrong and Michelle and MyThreeSons were right. Your odds don't change after you've made your guess unless the prize is randomly re-assigned (shuffled) and then you have the option to switch your choice.

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But doesn't the fact that you're making another choice after the one door has been opened change your odds? Now you're being asked to make a second choice, choosing between two remaining doors, one of which holds the prize.

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The real choice is choosing the right door. You have a 1/3 chance to pick the right door right off the bat and a 2/3 chance of picking the wrong door. That is always your chance when you pick the first door. When the host opens a wrong door, your chances of having picked the right or wrong one are still the same. The difference is that because the host has removed one of the 2/3 doors from the equation, it's better to switch because it was more likely that you picked a wrong door in the first place. You're not changing based on new odds; you're changing based on the odds of the original problem.

It's easier (for me) to think of it as percents. You have a (roughly) 33% chance of picking the prize and 66% chance of not picking the prize. One door is labeled 33% and the other two are both labeled 66%. You randomly pick a door without looking at the labels. You've most likely chosen a 66% door because there are two of them. The host, who looks at the labels, opens a 66% door. Now, you're left with choosing a 33% door or a 66% door. You know in your head that the one you picked first was most likely a 66% door and you know the host has already opened the other 66% door, so it's better to change your choice to the door you didn't pick first, which is most likely the 33% door.

I think, and I may be completely off, that an outside observer coming in at this point and choosing whether or not you will win the game would have a 50/50 chance because their choices would be a) you win or b) you don't win. But that would be a different gamble. You would still be stuck in the original problem of choosing the correct door out of three.

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But doesn't the fact that you're making another choice after the one door has been opened change your odds? Now you're being asked to make a second choice, choosing between two remaining doors, one of which holds the prize.

This would be true if they took what was behind these two doors and shuffled them, and then you chose.

AHA! I think I may have figured out a different way to make sense of this:

Go back to the beginning of the game. Forget about the door you did choose -- instead think about the two you didn't choose. You know that at least one of those is a bad prize. But what are the odds that they are both bad? Only 1 in 3, which could only happen if you chose the good door. So when the host reveals one of the unchosen doors as bad, you know that there is only a 1 in 3 chance that the other unchosen door is also bad. In other words, there is a 2 in 3 chance that the other unchosen door is good. Therefore, you ought to switch.

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