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First, the Monty Hall Problem is very different. Both depend on how poor most people's understanding of probability is. But how each problem uses this ignorance is very different.
The flaw in the Wikipedia solution is in this statement:
To see this, imagine that your benefactor started with two identical, empty envelopes, one $20 bill, one $10 bill, and nine $5 bills.
He puts the $10 bill in one envelope, and seals it. Then he randomly selects one of the other bills, and seals it in the second envelope.
It is true that the chances are 1/2 that you pick the larger envelope, and 1/2 that you pick the smaller. But if you try to associate the random variable A with your envelope, it depends on what A is.
What the misunderstanding is, is that a random variable, like this A, represents every possible value at the same time. But each value has a weight, called the probability.
There is a 45% chance that A=$5, a 50% chance that A=$10, and a 5% chance that A=$20. This makes the expected value
Exp(A) = $5*0.45+$10*0.50+$20*0.20 = $8.25.
But what about the other envelope (call it B)? If A=$5, there is a 100% chance that B=2A=$10. If A=$20, there is a 100% chance that B=A/2=$10. And if A=$10, there is a 90% chance that B=A/2=$5, and a 10% chance that B=2A=$20. Note that the total chance that B=A/2 is 50%, but it is never 50% for any particular value.
And the expected value of B is
Exp(B) = $5*0.9/2 + $10*0.50 + $20*0.10/2 = $8.25.
+++++
You can't calculate the expected value of the other envelope without knowing the distribution of values going into th envelopes. But you can say each has the same distribution now. If you open an envelope, the answer still depends on the distribution for A/2 and 2A, and you don't know that. (If you do, and they have equal probability? Then it is advantageous to switch.)
The flaw in the Wikipedia solution is in this statement:
quote:
Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
To see this, imagine that your benefactor started with two identical, empty envelopes, one $20 bill, one $10 bill, and nine $5 bills.
He puts the $10 bill in one envelope, and seals it. Then he randomly selects one of the other bills, and seals it in the second envelope.
It is true that the chances are 1/2 that you pick the larger envelope, and 1/2 that you pick the smaller. But if you try to associate the random variable A with your envelope, it depends on what A is.
What the misunderstanding is, is that a random variable, like this A, represents every possible value at the same time. But each value has a weight, called the probability.
There is a 45% chance that A=$5, a 50% chance that A=$10, and a 5% chance that A=$20. This makes the expected value
Exp(A) = $5*0.45+$10*0.50+$20*0.20 = $8.25.
But what about the other envelope (call it B)? If A=$5, there is a 100% chance that B=2A=$10. If A=$20, there is a 100% chance that B=A/2=$10. And if A=$10, there is a 90% chance that B=A/2=$5, and a 10% chance that B=2A=$20. Note that the total chance that B=A/2 is 50%, but it is never 50% for any particular value.
And the expected value of B is
Exp(B) = $5*0.9/2 + $10*0.50 + $20*0.10/2 = $8.25.
+++++
You can't calculate the expected value of the other envelope without knowing the distribution of values going into th envelopes. But you can say each has the same distribution now. If you open an envelope, the answer still depends on the distribution for A/2 and 2A, and you don't know that. (If you do, and they have equal probability? Then it is advantageous to switch.)
1
The MHP is misunderstood because it is almost always - including the few answers I've read here, even the ones that get the right answer - explained incorrectly.
Let's say you always pick door #3. At this point, there are three possibilities that I hope everybody can agree with:
(A) There is a 1/3 chance that the car is behind door #1.
(B) There is a 1/3 chance that the car is behind door #2.
(C) There is a 1/3 chance that the car is behind door #3.
What happens next is what everybody seems to get confused about. Monty Hall has to open a door other than door #1, and it has to have a goat. How he does this is different in one of the three cases, than the other two:
(A) There is a 1/3 chance that the car is behind door #1, and he must open door #2.
(B) There is a 1/3 chance that the car is behind door #2, and he must open door #1.
There is a 1/3 chance that the car is behind door #3, BUT NOW HE HAS A CHOICE. So case (C) must get split into two:
(C1) There is a 1/6 chance that the car is behind door #3 and he opens door #1.
(C2) There is a 1/6 chance that the car is behind door #3 and he opens door #2.
What everybody seems to miss, is that WE SEE WHICH DOOR HE OPENED. If it is #1, we know we are in either case (B) or case (C1). If it is #2, we know we are in either case (A) or case (C2).
Either way, the probability of arriving at the point where WE SEE WHAT DOOR IS OPEN and are asked if we want to switch is 1/3 if you picked a goat, and 1/6 if you picked the car. This is what makes it twice as likely that you picked a goat than the car, not the fact that you originally had a 2/3chance to pick a goat.
Most of the answers that get the right answer combine cases (A) and (B) into a single case, as well as cases (C1) and (C2). WITHOUT RECOGNIZING THAT THEY DO SO. As long as the split of (C) into (C1) and (C2) is even, the answer has to be the same, but that doesn't make it correct to combine them this way.
Let's say you always pick door #3. At this point, there are three possibilities that I hope everybody can agree with:
(A) There is a 1/3 chance that the car is behind door #1.
(B) There is a 1/3 chance that the car is behind door #2.
(C) There is a 1/3 chance that the car is behind door #3.
What happens next is what everybody seems to get confused about. Monty Hall has to open a door other than door #1, and it has to have a goat. How he does this is different in one of the three cases, than the other two:
(A) There is a 1/3 chance that the car is behind door #1, and he must open door #2.
(B) There is a 1/3 chance that the car is behind door #2, and he must open door #1.
There is a 1/3 chance that the car is behind door #3, BUT NOW HE HAS A CHOICE. So case (C) must get split into two:
(C1) There is a 1/6 chance that the car is behind door #3 and he opens door #1.
(C2) There is a 1/6 chance that the car is behind door #3 and he opens door #2.
What everybody seems to miss, is that WE SEE WHICH DOOR HE OPENED. If it is #1, we know we are in either case (B) or case (C1). If it is #2, we know we are in either case (A) or case (C2).
Either way, the probability of arriving at the point where WE SEE WHAT DOOR IS OPEN and are asked if we want to switch is 1/3 if you picked a goat, and 1/6 if you picked the car. This is what makes it twice as likely that you picked a goat than the car, not the fact that you originally had a 2/3chance to pick a goat.
Most of the answers that get the right answer combine cases (A) and (B) into a single case, as well as cases (C1) and (C2). WITHOUT RECOGNIZING THAT THEY DO SO. As long as the split of (C) into (C1) and (C2) is even, the answer has to be the same, but that doesn't make it correct to combine them this way.
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