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Probability - Monty Hall - Very Tricky

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Probability - Monty Hall - Very Tricky

by MartyMurray » Tue Dec 29, 2015 8:06 am
On a game show, there are three closed doors. Behind one door is a prize. Behind the other two is nothing.

The contestant is instructed to choose one of the three doors.

After the contestant chooses a door, the game show host opens one of the two unchosen doors, behind which there is nothing, as the game show host was already aware. The host then offers to the contestant to choose to either look behind the still closed door the contestant had already chosen or to look behind the third, also still closed, door, the one chosen by neither the contestant or the host.

If the contestant is aware that the host always chooses a door that has nothing behind it and if all the probabilities of the contestant's choices are based on random selections, if the contestant were to change his choice to the third door, what is the probability of his finding the prize behind it?

(A) 1/4

(B) 1/3

(C) 1/2

(D) 2/3

(E) 3/4
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by Amrabdelnaby » Sat Jan 02, 2016 6:26 am
I think it's 1/4

Here is how i thought about it

first his probability was 1/3, but after the show host opens for him one empty door, the contestant's probability of finding the prize becomes 1/2

now to the second stage, the contestant decides to change doors so we multiply his first probability and his second probability of finding the prize --> 1.2 x 1/2 = 1/4

I personally would have stuck to the same door :)
Marty Murray wrote:On a game show, there are three closed doors. Behind one door is a prize. Behind the other two is nothing.

The contestant is instructed to choose one of the three doors.

After the contestant chooses a door, the game show host opens one of the two unchosen doors, behind which there is nothing, as the game show host was already aware. The host then offers to the contestant to choose to either look behind the still closed door the contestant had already chosen or to look behind the third, also still closed, door, the one chosen by neither the contestant or the host.

If the contestant is aware that the host always chooses a door that has nothing behind it and if all the probabilities of the contestant's choices are based on random selections, if the contestant were to change his choice to the third door, what is the probability of his finding the prize behind it?

(A) 1/4

(B) 1/3

(C) 1/2

(D) 2/3

(E) 3/4
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by MartyMurray » Sat Jan 02, 2016 6:52 am
Amrabdelnaby wrote:I think it's 1/4

Here is how i thought about it

first his probability was 1/3, but after the show host opens for him one empty door, the contestant's probability of finding the prize becomes 1/2

now to the second stage, the contestant decides to change doors so we multiply his first probability and his second probability of finding the prize --> 1.2 x 1/2 = 1/4

I personally would have stuck to the same door :)
Think about it. Why would you multiply the probabilities that way? He didn't really choose two doors. He only actually chose one, and you said that the probability of choosing the right one out of the two is 1/2. In other words, according to what you said, the probability of either door being the right door is 1/2.

Then again, 1/2 is not right either. That's the trick. Part of the reason I posted this is because someone brought it up over the holidays and two people who are into math and engineering type of stuff both insisted, and as far I am aware are still convinced, that 1/2 is correct. LOL
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by GMATGuruNY » Sat Jan 02, 2016 8:55 am
Marty Murray wrote:On a game show, there are three closed doors. Behind one door is a prize. Behind the other two is nothing.

The contestant is instructed to choose one of the three doors.

After the contestant chooses a door, the game show host opens one of the two unchosen doors, behind which there is nothing, as the game show host was already aware. The host then offers to the contestant to choose to either look behind the still closed door the contestant had already chosen or to look behind the third, also still closed, door, the one chosen by neither the contestant or the host.

If the contestant is aware that the host always chooses a door that has nothing behind it and if all the probabilities of the contestant's choices are based on random selections, if the contestant were to change his choice to the third door, what is the probability of his finding the prize behind it?

(A) 1/4

(B) 1/3

(C) 1/2

(D) 2/3

(E) 3/4
To switch to the WINNING door, the contestant must first select a LOSING door.

P(the contestant's 1st selection is a losing door) = 2/3. (Of the 3 doors, 2 are losing doors.)

P(the host's selection is a losing door) = 1. (The host must choose a losing door.)

When the contestant switches, P(the contestant's 2nd selection is the winning door) = 1. (Since the first 2 doors are losing doors, the remaining door must be the winning door.)

To combine these probabilities, we multiply:
2/3 * 1 * 1 = 2/3.

The correct answer is D.
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by Amrabdelnaby » Sat Jan 02, 2016 10:22 am
Now after giving it a second thought

I actually think that the probability should be 1/2 since the guy is only left with two doors.

although the contestant chose the door, he didn't open it and since one of the three doors were opened by the host, now the contestant is left with 2 doors only. regardless of which one he chose, the probability of finding the prize now is equal to the probability of not finding it behind either doors. so pls explain
GMATGuruNY wrote:
Marty Murray wrote:On a game show, there are three closed doors. Behind one door is a prize. Behind the other two is nothing.

The contestant is instructed to choose one of the three doors.

After the contestant chooses a door, the game show host opens one of the two unchosen doors, behind which there is nothing, as the game show host was already aware. The host then offers to the contestant to choose to either look behind the still closed door the contestant had already chosen or to look behind the third, also still closed, door, the one chosen by neither the contestant or the host.

If the contestant is aware that the host always chooses a door that has nothing behind it and if all the probabilities of the contestant's choices are based on random selections, if the contestant were to change his choice to the third door, what is the probability of his finding the prize behind it?

(A) 1/4

(B) 1/3

(C) 1/2

(D) 2/3

(E) 3/4
To switch to the WINNING door, the contestant must first select a LOSING door.

P(the contestant's 1st selection is a losing door) = 2/3. (Of the 3 doors, 2 are losing doors.)

P(the host's selection is a losing door) = 1. (The host must choose a losing door.)

When the contestant switches, P(the contestant's 2nd selection is the winning door) = 1. (Since the first 2 doors are losing doors, the remaining door must be the winning door.)

To combine these probabilities, we multiply:
2/3 * 1 * 1 = 2/3.

The correct answer is D.
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by GMATGuruNY » Sat Jan 02, 2016 10:37 am
Amrabdelnaby wrote:Now after giving it a second thought

I actually think that the probability should be 1/2 since the guy is only left with two doors.

although the contestant chose the door, he didn't open it and since one of the three doors were opened by the host, now the contestant is left with 2 doors only. regardless of which one he chose, the probability of finding the prize now is equal to the probability of not finding it behind either doors. so pls explain
Question:
If the contestant switches from his first selection to the remaining door, what is the probability that he switches to the winning door?

As I mentioned in my solution above, a favorable outcome will be yielded only if the following three events happen:
1. The contestant first selects a losing door.
2. The host then selects a losing door.
3. When the contestant switches, the remaining door is the winning door.

As shown in my solution:
The probability of the first event = 2/3.
The probability of the second event = 1.
The probability of the third event = 1.
Multiplying these probabilities, we get:
P(the contestant switches to the winning door) = 2/3 * 1 * 1 = 2/3.
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by DavidG@VeritasPrep » Sat Jan 02, 2016 10:39 am
Now after giving it a second thought

I actually think that the probability should be 1/2 since the guy is only left with two doors.

although the contestant chose the door, he didn't open it and since one of the three doors were opened by the host, now the contestant is left with 2 doors only. regardless of which one he chose, the probability of finding the prize now is equal to the probability of not finding it behind either doors. so pls explain
You're not alone: this problem became infamous when it was posted in Parade magazine some years ago. When it was revealed that there was a 2/3 probability of the contestant winning the prize should he choose to switch doors there was an uproar. PhD's in mathematics wrote in insisting that the answer was 1/2. (The PhD's were wrong.) This is all to say that this question feels counterintuitive to virtually everyone who comes across it.

Think of it this way. Say there were 100 doors and the prize is behind 1 of them. You choose 1 door. So there is a 1/100 probability that you selected the correct door. And a 99/100 probability that you did not select the correct door. Now say I, knowing where the prize is, open 98 doors and reveal that the prize is not behind any of the doors I open. (The key is that I'm not opening these doors randomly. I'm knowingly opening doors that don't have the prize.) If you stayed with the original selection, there's still a 1/100 probability of success. And there's still a 99/100 chance that your initial selection was a dud. Which means that there's a 99/100 chance that the prize is behind the other door, not a 1/2 chance, despite the fact that there are only 2 doors remaining. If given the option to switch, you definitely should.

Put another way, if you were given a choice between staying with your first choice or getting to open all 99 other doors at the same time, you'd obviously open all 99 other doors. But because I'm only opening 98 doors that I know don't have the prize, and then asking if you want to switch, this is effectively what I'm offering you.
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by Amrabdelnaby » Sat Jan 02, 2016 3:22 pm
Guru,

I see your logic. one final question though

The probability of the third event is one only if the event is sure to happen.

this means that when he switches doors, he knows for a fact that he will find the prize, which is not the case.

so for example

Bear with me and let me know your thoughts.

Thank you
GMATGuruNY wrote:
Amrabdelnaby wrote:Now after giving it a second thought

I actually think that the probability should be 1/2 since the guy is only left with two doors.

although the contestant chose the door, he didn't open it and since one of the three doors were opened by the host, now the contestant is left with 2 doors only. regardless of which one he chose, the probability of finding the prize now is equal to the probability of not finding it behind either doors. so pls explain
Question:
If the contestant switches from his first selection to the remaining door, what is the probability that he switches to the winning door?

As I mentioned in my solution above, a favorable outcome will be yielded only if the following three events happen:
1. The contestant first selects a losing door.
2. The host then selects a losing door.
3. When the contestant switches, the remaining door is the winning door.

As shown in my solution:
The probability of the first event = 2/3.
The probability of the second event = 1.
The probability of the third event = 1.
Multiplying these probabilities, we get:
P(the contestant switches to the winning door) = 2/3 * 1 * 1 = 2/3.
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by GMATGuruNY » Sat Jan 02, 2016 10:35 pm
Amrabdelnaby wrote:Guru,

I see your logic. one final question though

The probability of the third event is one only if the event is sure to happen.

this means that when he switches doors, he knows for a fact that he will find the prize, which is not the case.
The portion is red is a misstatement.
The contestant is not required to KNOW anything.
When the contestant switches doors, the two losing doors have already been chosen.
As a result, he is CERTAIN to switch to the winning door, since the winning door is the ONLY door that remains.
Thus, the probability that he switches to the winning door = 1.
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by regor60 » Mon Jan 04, 2016 6:51 am
Amrabdelnaby wrote:
I actually think that the probability should be 1/2 since the guy is only left with two doors.
Something I've learned that has been helpful to me is that just because there may be two alternatives doesn't necessarily mean that both are equally likely.

For example, the sun may come up tomorrow or it may not. Two alternatives. However, we would agree they are not equally likely.
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by Matt@VeritasPrep » Fri Jan 08, 2016 1:01 pm
regor60 wrote:For example, the sun may come up tomorrow or it may not. Two alternatives. However, we would agree they are not equally likely.
This is a really nice analogy. The naive definition of probability is called "naive" partly because it makes this assumption: if I have x events, the probability of any one of them is 1/x, since they all must be equally likely. But that, of course, is a huge assumption! You'd need to know that about the problem in question before using the naive definition to approach it.
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