Ask Marilyn

Quote: Wizard

You can also play this Monty knows game.

I picked door #1, and stuck with it. I won a car!

I'm not sure how the calculations work in their report.

Quote: Monty Knows

CONGRATULATIONS! You're a winner!
RECAP: You originally picked door 1 and then stayed with that door.
Here is a summary of how previous contestants have fared.

Switched

# of Players	Winners	Percent Winners
141256	1326	0.9

Didn't Switch

# of Players	Winners	Percent Winners
-25551013	-12840718	50.3

Someone on here years ago gave the best way to convince people they should switch.

Just add doors. There’s 100 doors. 98 are opened. Should you switch? Of course. What’s good for 100 is good for 10, 5, 3...

Quote: billryan

But you aren't really picking one of three. No matter which one you originally choose, one of the others is eliminated. Your original choice has a 50-50 chance, just like the unlocked box.

While it is true that the unlocked box has a 1 in 2 shot of having the prize, so does your original pick.

Just try the experiment, then.

It is important that you are choosing before any wrong choices have been revealed, and that the winning prize will not be revealed when non-winning doors are opened.

When you pick one door, there is a 1/3 chance that you were right, and a 2/3 chance that you were wrong.
1/3 door 1
1/3 door 2
1/3 door 3

Say you pick door one. If door two is opened, there is still a 1/3 chance than you were right, and a 2/3 chance that you were wrong.
1/3 door 1
0/3 door 2
2/3 door 3

Like someone else mentioned, if you start with 100 doors, there is only a 1% chance that you were right.

Anyway, I am not going to try to explain it in different ways. Try the experiment, in the way that it is described, and see for yourself.

Here is that link netzer was trying to post: The American Statistician, Volume 29, 1975 - Issue 1. It has the same problem as Marilyn's, that it isn't clear what Monty's behavior is. We don't even know if Monty knows where the keys are in the wording of this one. My personal interpretation of the wording is Monty doesn't know and it is a Deal or No Deal kind of situation.

Quote: Ayecarumba

Quote: Wizard

You can also play this Monty knows game.

I picked door #1, and stuck with it. I won a car!

I'm not sure how the calculations work in their report.

Quote: Monty Knows

CONGRATULATIONS! You're a winner!
RECAP: You originally picked door 1 and then stayed with that door.
Here is a summary of how previous contestants have fared.

Switched

# of Players	Winners	Percent Winners
141256	1326	0.9

Didn't Switch

# of Players	Winners	Percent Winners
-25551013	-12840718	50.3

It must have gotten fixed overnight -- it seems to be working better this morning. So far, I have won 5 cars and 3 goats!

Quote: Monty Knows

Switched

# of Players	Winners	Percent Winners
42	28	66.7

Didn't Switch

# of Players	Winners	Percent Winners
38	9	23.7

I've always wondered if the contestants who got zonked got any sort of compensation. Livestock does have a cash value. Maybe in the rules it says that if you get zonked, you don't get anything, regardless of the value of the zonk?

Quote: FinsRule

Someone on here years ago gave the best way to convince people they should switch.

Just add doors. There’s 100 doors. 98 are opened. Should you switch? Of course. What’s good for 100 is good for 10, 5, 3...

That is the first argument Marilyn used to defend her answer. She had a page on her personal web site explaining the responses and her struggles with them. She finally settled the question by asking school teachers to have their classes conduct simulations. Her web site appears to be down, so try "The Time Everyone “Corrected” the World's Smartest Woman" on Priceonomics.

Quote: Joeman

I've always wondered if the contestants who got zonked got any sort of compensation. Livestock does have a cash value. Maybe in the rules it says that if you get zonked, you don't get anything, regardless of the value of the zonk?

I think with any prize they make a cash offer after the show to buy it back, which most players take. For a goat, I imagine they offer you $100 or so, but I'm not sure.

If you read Steve Selvin's first letter to The American Statistician (link provided by the Wizard) you should pay special attention to Monty's last quoted remark;

"I'll do you a favor and open one of the remaining boxes on the table (he opens Box A) It's empty! Now either Box B of your Box A has the car keys. Since there are two boxes left, the probability of your box containing the keys is now 1/2. I'll give you $1,000 cash for your box."

At this point Selvin stops to question Monty's conclusion. He then goes on to show that the probability for the contestant's box remains at 1/3 but that the probability for the remaining box is now 2/3. He sent a second letter to the publication in August reporting the responses he had received. Here it is:

On the Mony Hall Problem

I have received a number of letters commenting on my "Letters to the Editor" in The American Statistician of February, 1975, entitled "A Problem in Probability." Several correspondents claim my answer is incorrect. The basis to my solution is that Monty Hall knows which box contains the keys and when he can open either of two boxes without exposing the keys, he chooses between them at random. An alternative solution to enumerating the mutually exclusive and equally likely outcomes is as follows:

A = event that keys are contained in box B
B = event that contestant chooses box B
C = event that Monty Hall opens box A

Then

P(keys in box B | contestant selects B and Monty opens A)

= P(A | BC) = P(ABC)/P(BC)
= P(C | AB)P(AB)/P(C | B)P(B)
= P(C | AB)P(B | A)P(A)/P(C | B)P(B)
= (1/2)(1/3)(1/3)(1/2)(1/3)
1/3

If the contestant trades his box B for the unopened box on the table, his probability of winning the card is 2/3.

D.L. Ferguson presented a generalization of this problem for the case of n boxes, in which Monty Hall opens p boxes. In this situation, the probability the contestant wins when he switches boxes is (n-1)/[n(n-p-1)].

Benjamin King pointed out the critical assumpitons about Monty Hall's behavior that are necessary to solve the problem, and emphasized that "the prior distribution is not the only part of the probabilistic side of a decision problem that is subjective."

Monty Hall wrote and expressed that he was not "a student of statistics problems" but "the big hole in your argument is that once the first box is seen to be emprty, the contestant cannot exchange his box." He continues to say, "Oh and incedentally, after one [box] is seen to be empty, his chances are no longer 50/50 but remain what they were in the first place, one out of three. It just seems to the contestant that one box having been eliminated, he stands a better chance. Not so."

I could not have said it better myself.

Steve Selvin
School of Public Health
University of California
Berkley, CA

You have three boxes. one of them has a million dollars ,the others have nothing. We know one box will be opened and it will have nothing. The box you pick has a 1/3 chance of being correct. The box you didn't pick has a 1/3 chance of being correct. How does switching your choice change that? Explain it in simple terms please, because I really want to see the fault in my logic.

Quote: billryan

You have three boxes. one of them has a million dollars ,the others have nothing. We know one box will be opened and it will have nothing. The box you pick has a 1/3 chance of being correct. The box you didn't pick has a 1/3 chance of being correct. How does switching your choice change that? Explain it in simple terms please, because I really want to see the fault in my logic.

The box you originally pick has a 1/3 chance of being correct

That means there’s a 2/3 chance it’s in one of the other boxes.

No matter what the host opens or doesnt open , shows or doesn’t show , those odds above do not change. You can’t “change” odds simply by moving curtains or opening doors.

So no matter what , there’s always a 1/3 chance you picked right originally .. and a 2/3 chance the million dollars is elsewhere. By opening that second losing box , elsewhere has been narrowed down to one place. Elsewhere is still 2/3 likely to be the winner

billryan:

You didn't state it correctly. It's like this:

The box you pick has a 1/3 chance of being correct.
The two boxes you didn't pick taken together have a 2/3 chance of being correct.
Eliminate one of these and the 2/3 passes to the other.

Lets say there are a million boxes.

You pick 1. Odds= 1 in a million.

Monty opens all but one of the unpicked boxes. He knows where the prize is.

Is the last box the same 1 in a million. Would you switch? The last box is 100% guaranteed to be the winner, minus the 1 in a million shot that you happened to pick the winner in the first place.

Quote: netzer

billryan:

You didn't state it correctly. It's like this:

The box you pick has a 1/3 chance of being correct.
The two boxes you didn't pick taken together have a 2/3 chance of being correct.
Eliminate one of these and the 2/3 passes to the other.

Why? Label the boxes A,B, and C.
I choose Box A. I have a 1 in 3 chance of winning. Box C has a one in three chance of winning. Box B is opened and is empty. Why does Box C now go from 1/3 chance to 2/3 chance instead of Box A and C both being 50-50? If I had chosen C in the first place, how does that affect the chance of Box A having it?
I really don't get this.

Same answer as michael99000 but with different wording:

Suppose you are the contestant and then call the box/door/curtain you chose #1. Then suppose that the host makes this offer: "You can stay with box #1, or instead I will let you have the contents of both box #2 and box #3, while guaranteeing that at least one of them contains nothing of value."

Do you prefer box #1 or boxes #2 and #3 combined?

Is that not the same? How does anything differ between the host guaranteeing that at least one unchosen box is "empty" and the host showing you that one unchosen box is empty? Do you really prefer your 1/3 chance that box #1 is the right one vs. your 2/3 chance that the right one is either #2 or #3?

I say it isn't a 1/3 choice because you know one box that isn't the big prize will be opened so my chance of hitting it is 50-50 all along. Does anyone dispute that?
It doesn't matter what box I chose first, because I get a second chance and at that point, the two remaining boxes are 50-50.
If the opened door was random and might show the big prize, it changes everything but it seems to be agreed by all that the opened door will not have the big prize.
Lets suppose you managed to see a cheat sheet and knew in advance that door #2 was going to be opened if you didn't chose it. That tells you the prize is behind door 1 or door 3. 50-50 right? Now, suppose you didn't see the cheat sheet, picked a door and then they opened #2. How is that any different? In both cases, the prize has an equal chance to be behind either door.

Quote: billryan

Why? Label the boxes A,B, and C.
I choose Box A. I have a 1 in 3 chance of winning. Box C has a one in three chance of winning. Box B is opened and is empty. Why does Box C now go from 1/3 chance to 2/3 chance instead of Box A and C both being 50-50? If I had chosen C in the first place, how does that affect the chance of Box A having it?
I really don't get this.

What would you rather have as your choice from the beginning ?

Box A
Or
Box B AND Box C

Obviously you’d pick the two box option. You have a 66% shot at being right vs a 33% shot.

Here’s the part you’re having trouble with .. just because one of those other two boxes was opened by the host , that does not change the fact that the Box B and Box C choice was better from the start and still is better , because it’s 66% likely to be there

How did you get a 50% chance when choosing among 3 boxes? Suppose the contest was that you get to choose one box first, then your "opponent" gets both of the other boxes. Who has a better shot of winning? If after you choose your single box and your "opponent" gets the other two, does it change something if the host shows your opponent that one -- just one -- of his boxes is empty?

Quote: michael99000

What would you rather have as your choice from the beginning ?

Box A
Or
Box B AND Box C

Obviously you’d pick the two box option. You have a 66% shot at being right vs a 33% shot.

Here’s the part you’re having trouble with .. just because one of those other two boxes was opened by the host , that does not change the fact that the Box B and Box C choice was better from the start and still is better , because it’s 66% likely to be there

But that's not an option. Obviously being able to open two doors increases your chances. To me, it comes down to you have to pick one of two boxes. Its 50-50.
Lets go back to the 100 door option. You pick one so its 1 out of 100 that you are right. All the remaining boxes are opened but #72. When I choose, the chance I was right was one percent. The chance door 72 was right was one percent. With two boxes left, are my chances still 1% or 50%? If my original box only had a 1% chance, does that mean box 72 now has a 99% chance? Or am I as likely to win by switching to #72, assuming that the host knew the money was not in any opened box.
Does door 72 have a better chance than door 1 of being the one in 100 with the prize? If Door 72 and Door 1 had the same chance of being the magic box when I have 100 choices, how do they not have the equal chance when they are two out of 50, or two out of five?

Quote: billryan

But that's not an option. Obviously being able to open two doors increases your chances. To me, it comes down to you have to pick one of two boxes. Its 50-50.
Lets go back to the 100 door option. You pick one so its 1 out of 100 that you are right. All the remaining boxes are opened but #72. Ipicked Door 1, I only had a 1 in 100 chance of being right.
Does door 72 have a better chance than door 1 of being the one in 100 with the prize? If Door 72 and Door 1 had the same chance of being the magic box when I have 100 choices, how do they not have the equal chance when they are two out of 50, or two out of five?

With the 100 box version of this ..

Let’s say you can keep playing the game as many times as you want. Every single time you play the game ,the host will have the ability to open 98 losing boxes after you make your choice.

So you think that every single time you play a game where you’re choosing 1 Box out of 100 trying to find the one winning box .. that somehow your odds of having picked that winning box can be narrowed to 50/50 ? Which also means you’re saying if you played the game 100 times, you’d have selected the right box originally 50 times

Quote: Doc

How did you get a 50% chance when choosing among 3 boxes? Suppose the contest was that you get to choose one box first, then your "opponent" gets both of the other boxes. Who has a better shot of winning? If after you choose your single box and your "opponent" gets the other two, does it change something if the host shows your opponent that one -- just one -- of his boxes is empty?

You have a choice of three, but you know the host will open a door that contains the non winning prize, so it really is you picking one of two unopened doors. Either your original or the other. That is 50-50.
Suppose your opponent gets to go first and choses the first two boxes. You get #3. He opens #1 and it is empty. Do you switch with him? Your chances of winning have gone from 1/3 to 1/2.
What I see in this problem is a lot of smoke and distraction that avoids the relative simplicity of the problem. You have a 50-50 shot at winning, while it appears you should only have a one in three.

Quote: billryan

You have a choice of three, but you know the host will open a door that contains the non winning prize, so it really is you picking one of two unopened doors. Either your original or the other. That is 50-50.
Suppose your opponent gets to go first and choses the first two boxes. You get #3. He opens #1 and it is empty. Do you switch with him? Your chances of winning have gone from 1/3 to 1/2.
What I see in this problem is a lot of smoke and distraction that avoids the relative simplicity of the problem. You have a 50-50 shot at winning, while it appears you should only have a one in three.

I would love to book this bet. I’ll take your money all day long at 50/50 odds if you promise never to switch.

Quote: billryan

You have a choice of three, but you know the host will open a door that contains the non winning prize, so it really is you picking one of two unopened doors. Either your original or the other. That is 50-50.
Suppose your opponent gets to go first and choses the first two boxes. You get #3. He opens #1 and it is empty. Do you switch with him? Your chances of winning have gone from 1/3 to 1/2.
What I see in this problem is a lot of smoke and distraction that avoids the relative simplicity of the problem. You have a 50-50 shot at winning, while it appears you should only have a one in three.

I would love to book this bet to. Here are the conditions I suggest:

1. You will be the host.
2. We'll use three cups. Under two will be a representation of the "prize." Under the other two will be nothing.
3. You, of course, can choose where to hide the prize.
4. I will pick one of the cups.
5. You must lift one of the cups showing nothing.
6. You must give me the option to switch, which I always will.
7. If you are concerned about me looking for tells, you may insist I randomize my cup picking choices. I, too, may do that if I wish.
8. I'm willing to pay 60% of the value of "prize" each time we play.
9. Since you feel I have only a 50% chance of winning, you will have a 16.67% advantage, the same as on the worst bet in craps, the "any 7."
10. You may quit at anytime.

Any takers?

Quote: Wizard

I would love to book this bet to. Here are the conditions I suggest:

1. You will be the host.
2. We'll use three cups. Under two will be a representation of the "prize." Under the other two will be nothing.
3. You, of course, can choose where to hide the prize.
4. I will pick one of the cups.
5. You must lift one of the cups showing nothing.
6. You must give me the option to switch, which I always will.
7. If you are concerned about me looking for tells, you may insist I randomize my cup picking choices. I, too, may do that if I wish.
8. I'm willing to pay 60% of the value of "prize" each time we play.
9. Since you feel I have only a 50% chance of winning, you will have a 16.67% advantage, the same as on the worst bet in craps, the "any 7."
10. You may quit at anytime.

Any takers?

The truest test of how strongly someone backs their stance , offering them a real money wager.

People. I am not saying I am right. I am laying out my thinking and asking where I go wrong.
1)Three boxes. Each box has an equal chance of having a prize, so longterm each box is worth one third of the prize.
2)I pick box A and have a 1/3 chance to win. On this we all agree. At this point, each box is still worth one third of a prize. Am I correct?
3)So now the host opens box B, which means it has a value of zero. Still correct, I think.
Now, this is where I go astray. Two boxes left. Why doesn't the value of Box B( one third of the prize) get divided equally into the two remaining boxes, bringing each box to a value of one half the prize, 50%.
Since your final decision is one of two, I don't understand why one choice has a value of one third of the prize and the other has two thirds.
I'm trying to understand how this is. I accept that I'm incorrect. What I don't understand is where the flaw in my thinking is that brings me to the incorrect answer.

Quote: billryan

People. I am not saying I am right. I am laying out my thinking and asking where I go wrong.
1)Three boxes. Each box has an equal chance of having a prize, so longterm each box is worth one third of the prize.
2)I pick box A and have a 1/3 chance to win. On this we all agree. At this point, each box is still worth one third of a prize. Am I correct?
3)So now the host opens box B, which means it has a value of zero. Still correct, I think.
Now, this is where I go astray. Two boxes left. Why doesn't the value of Box B( one third of the prize) get divided equally into the two remaining boxes, bringing each box to a value of one half the prize, 50%.
Since your final decision is one of two, I don't understand why one choice has a value of one third of the prize and the other has two thirds.
I'm trying to understand how this is. I accept that I'm incorrect. What I don't understand is where the flaw in my thinking is that brings me to the incorrect answer.

The host opening an empty door provides you with no new information, so you can’t update the chance that door A is right 1/3 of the time. You know that of doors B and C, at least one is empty. You also know the host will only open empty doors. So if the prize is in B, the host opens C. If the prize is in C, the host opens B. In the 1/3 chance the prize is in A, the host can open either B or C.

So when the host opens a door and it’s empty, you learned absolutely nothing. So your door still has 1/3 chance of being right. The other closed door (which could be B or C) has a 2/3 chance of being right. It’s because the host avoids opening any door with a prize.

Re-run the experiment with 1,000,000 doors. You pick A. The host then opens up 999,998 empty doors and leaves one closed. Do you want to switch?

Quote: billryan

People. I am not saying I am right. I am laying out my thinking and asking where I go wrong.
1)Three boxes. Each box has an equal chance of having a prize, so longterm each box is worth one third of the prize.
2)I pick box A and have a 1/3 chance to win. On this we all agree. At this point, each box is still worth one third of a prize. Am I correct?
3)So now the host opens box B, which means it has a value of zero. Still correct, I think.
Now, this is where I go astray. Two boxes left. Why doesn't the value of Box B( one third of the prize) get divided equally into the two remaining boxes, bringing each box to a value of one half the prize, 50%.
Since your final decision is one of two, I don't understand why one choice has a value of one third of the prize and the other has two thirds.
I'm trying to understand how this is. I accept that I'm incorrect. What I don't understand is where the flaw in my thinking is that brings me to the incorrect answer.

Your flaw is that you are thinking of this like a Deal or No Deal kind of game, which it isn't. It is mathematically equivalent to the host letting you swap your door for BOTH other doors, assuming you put no value on a goat. Since the host knows where the car is, and we know at least one of the other doors contains a goat, he isn't giving up any relevant information by revealing the goat.

Here is another way of wording the same problem. What is your answer to it?

There are three prisoners on death row, Adam, Bob, and Charlie. The warden says to Adam, "Tomorrow somebody will be executed, but I won't say who yet. The choice was made randomly." Adam replies, can you at least tell me one of the other two prisoners who won't be executed, since at least one of them must be safe? The warden says, "Why not, Bob will not be executed tomorrow." What are the chances Adam is executed?

Nice, my answer in before the Wiz and pretty close to Wiz in substance. Feeling math smart!

Quote: billryan

People. I am not saying I am right. I am laying out my thinking and asking where I go wrong.
1)Three boxes. Each box has an equal chance of having a prize, so longterm each box is worth one third of the prize.
2)I pick box A and have a 1/3 chance to win. On this we all agree. At this point, each box is still worth one third of a prize. Am I correct?
3)So now the host opens box B, which means it has a value of zero. Still correct, I think.
Now, this is where I go astray. Two boxes left. Why doesn't the value of Box B( one third of the prize) get divided equally into the two remaining boxes, bringing each box to a value of one half the prize, 50%.
Since your final decision is one of two, I don't understand why one choice has a value of one third of the prize and the other has two thirds.
I'm trying to understand how this is. I accept that I'm incorrect. What I don't understand is where the flaw in my thinking is that brings me to the incorrect answer.

Your mistake is in thinking that the host opening any doors or boxes changes the odds.

This puzzle is small enough to count every possible outcome.

There are three possible prize arrangements
G P P

P G P

P P G

Count the number of times you win the prize, for all three possible arrangements, by picking door 1, and eliminating a non-winning prize from either of door 2 or 3, and sticking with your initial choice of door 1
Then do the same by starting with door 2, and finally door 3
Add up the number of times you have won, after all nine possible arrangements of prizes and door selections.

You can then do the same thing all over again, but with switching.
For three prize arrangements, start with door one
Then remove one of the non-winning prizes from either door 2 or 3
Now switch to the other door. Record how many times out of the three possible starting prize arrangements that you win.

Then start with door 2 and switch,
Then start with door 3 and switch.

Again, how many times out of the 9 possible arrangements of prizes and starting door selections did you win?

Quote: unJon

The host opening an empty door provides you with no new information, so you can’t update the chance that door A is right 1/3 of the time. You know that of doors B and C, at least one is empty. You also know the host will only open empty doors. So if the prize is in B, the host opens C. If the prize is in C, the host opens B. In the 1/3 chance the prize is in A, the host can open either B or C.

So when the host opens a door and it’s empty, you learned absolutely nothing. So your door still has 1/3 chance of being right. The other closed door (which could be B or C) has a 2/3 chance of being right. It’s because the host avoids opening any door with a prize.

Re-run the experiment with 1,000,000 doors. You pick A. The host then opens up 999,998 empty doors and leaves one closed. Do you want to switch?

Let’s say Contestant 1 is in a room with Monty Hall playing the boxes game. Monty tells him that two of the boxes have a goat and one has a car, and to choose one. Contestant 1 chooses box A, and Monty opens Box B to show a goat. He then offers Contestant 1 the chance to switch to Box C ..

At that exact moment another contestant , Contestant 2 , enters the room, not having seen any of what just occurred. Monty tells Contestant 2 that under one of the two remaining boxes (A and C) is a car and under the other is a goat, and to simply choose one.

From Contestant 2’s standpoint , does he have a 50% chance of choosing the car ?

If so , how can two people both be staring at the same two covered boxes , yet to one of them the odds are 33/66 as to which has the car, and to the other it’s 50/50?

Quote: michael99000

Let’s say Contestant 1 is in a room with Monty Hall playing the boxes game. Monty tells him that two of the boxes have a goat and one has a car, and to choose one. Contestant 1 chooses box A, and Monty opens Box B to show a goat. He then offers Contestant 1 the chance to switch to Box C ..

At that exact moment another contestant , Contestant 2 , enters the room, not having seen any of what just occurred. Monty tells Contestant 2 that under one of the two remaining boxes (A and C) is a car and under the other is a goat, and to simply choose one.

From Contestant 2’s standpoint , does he have a 50% chance of choosing the car ?

If so , how can two people both be staring at the same two covered boxes , yet to one of them the odds are 33/66 as to which has the car, and to the other it’s 50/50?

I’ll see your two people and raise you a third person. Monty himself is looking at the same two doors and is unexpectedly told by the announcer that it is “Hosts Can Win Day” and Monty can select one of the two remaining doors and win what’s behind it. Does Monty have a 100% chance of getting the car?

How can three people looking at the same two doors have three different chances of choosing the winning door?

Quote: unJon

I’ll see your two people and raise you a third person. Monty himself is looking at the same two doors and is unexpectedly told by the announcer that it is “Hosts Can Win Day” and Monty can select one of the two remaining doors and win what’s behind it. Does Monty have a 100% chance of getting the car?

How can three people looking at the same two doors have three different chances of choosing the winning door?

Exactly. What I was getting at is odds are based on information known.

Quote: Joeman

NSFW. You Have Been Warned!

Here we go again...

You B45t***d! Now I'll have to go back into therapy.

The good news is that billryan at least understands he’s wrong.

He just doesn’t know why.

That’s better than most. Most would be calling us all idiots or drones.

Quote: billryan

People. I am not saying I am right. I am laying out my thinking and asking where I go wrong.
1)Three boxes. Each box has an equal chance of having a prize, so longterm each box is worth one third of the prize.
2)I pick box A and have a 1/3 chance to win. On this we all agree. At this point, each box is still worth one third of a prize. Am I correct?

Correct so far.

Quote:

3)So now the host opens box B, which means it has a value of zero. Still correct, I think.

Indeed BUT
We know the probability of the prize NOT being in box B or box C is 2/3. You agree there?
Monty adds more info. He knows for sure that the probability of it being in a certain one of those boxes is zero.
There is No contradiction or conflict here. Nothing has changed.
Probability prize is in A = 1/3
Probability it's in B or C =2/3
Probability that it's in the box that Monty is about to open =0
He opens box C Still nothing changes. You know exactly what you knew before, plus a little extra knowledge.
p(B or C)=2/3
p(C)=0
therefore p(B)=2/3
That's where I showed that the 2/3 probability got fully assigned to box B.
Now. You show otherwise.

Quote:

Now, this is where I go astray. Two boxes left. Why doesn't the value of Box B( one third of the prize) get divided equally into the two remaining boxes, bringing each box to a value of one half the prize, 50%.

Because that's not how maths works. You give your working out if you can. MONTY KNEW THE PROBABILITY of it being in the box he opened. He simply revealed that extra bit of data. Nothing else changed.

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Since your final decision is one of two, I don't understand why one choice has a value of one third of the prize and the other has two thirds.

It always did. Nothing changed.

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I'm trying to understand how this is. I accept that I'm incorrect. What I don't understand is where the flaw in my thinking is that brings me to the incorrect answer.

Quote: OnceDear

Quote: Joeman

NSFW. You Have Been Warned!

Here we go again...

You B45t***d! Now I'll have to go back into therapy.

Hey, I put it under a spoiler tag and warning. This one is all on you, OD! ;)

Here's another way of looking at it.

You make your choice.

Then the host, without revealing anything, offers you to switch to BOTH of the other choices.

You KNOW that at least one of the other two contains a goat, but you'll still switch, right?

So what difference does it make if the host shows you which of those other two contain the goat before making the offer to switch?

Quote: Wizard

...What is your answer to it?

There are three prisoners on death row, Adam, Bob, and Charlie. The warden says to Adam, "Tomorrow somebody will be executed, but I won't say who yet. The choice was made randomly." Adam replies, can you at least tell me one of the other two prisoners who won't be executed, since at least one of them must be safe? The warden says, "Why not, Bob will not be executed tomorrow." What are the chances Adam is executed?

Adam now has a 50% chance of being executed given the new information. The warden knows who will be executed. Here's the breakdown of the 3 possible situations before the warden reveals who is safe (the capital letter represents who is to be executed):
abC
aBc
Abc
Given that we now know that aBc is not possible, we can eliminate that choice, and the possible situations are:
aC
Ac

This is different than Monte Hall in that there is a distinction between Bob & Charlie. If we pose the question as "One prisoner will be executed, Adam or one of two others," we can now breakdown the possible situations as:
apP
aPp
App
Since we don't know Bob from Charlie here, any of the three situations is still possible, and Adam's chances would remain 1/3.
Similarly, if we re-worded the Monty Hall problem to "Goat, Pig, or Car", and a Pig was revealed, we can eliminate the possibility that a Pig was initially chosen, and it's now 50/50 that you have a car.

Quote: Chuckleberry

Adam now has a 50% chance of being executed given the new information.

Sorry, you're wrong. Adam still has a 1 in 3 chance to be executed, for the same reason that staying in the Monty Hall problem has a 1 in 3 chance of winning the car.

Quote: Wizard

Sorry, you're wrong.

That stings. Not sure where my logic was flawed.

Quote: Chuckleberry

That stings. Not sure where my logic was flawed.

Distinguishing things a A, B and C doesn’t change the information. As long as one is the good thing (car) and two are the bad things (goats or pig and goat) it’s the same problem because Monty always has a bad door he can and will open.

Quote: unJon

Distinguishing things a A, B and C doesn’t change the information.

It does change the situation. If it was Goat/Pig/Car and a Pig is revealed, your odds of having each are 50%-Goat, 0%-Pig, or 50%-Car.

If it was Goat/Goat/Car and a Goat is revealed, your odds of having each are 33%-First Goat, 33%-Second Goat, or 33%-Car. And since the Goats are indistinguishable, your odds of having each are 67%-Goat, or 33%-Car.

Quote: Wizard

Your flaw is that you are thinking of this like a Deal or No Deal kind of game, which it isn't. It is mathematically equivalent to the host letting you swap your door for BOTH other doors, assuming you put no value on a goat. Since the host knows where the car is, and we know at least one of the other doors contains a goat, he isn't giving up any relevant information by revealing the goat.

Here is another way of wording the same problem. What is your answer to it?

There are three prisoners on death row, Adam, Bob, and Charlie. The warden says to Adam, "Tomorrow somebody will be executed, but I won't say who yet. The choice was made randomly." Adam replies, can you at least tell me one of the other two prisoners who won't be executed, since at least one of them must be safe? The warden says, "Why not, Bob will not be executed tomorrow." What are the chances Adam is executed?

What if Adam asked the warden to tell him the name of one prisoners who won’t be executed, and Adam himself was one of the possible answers the warden could give ? In other words the warden chooses his answer completely randomly among the 2 he knows won’t be killed.

Then if the warden says bob or Charlie , Adam would be 50/50 to be killed

Quote: Chuckleberry

It does change the situation. If it was Goat/Pig/Car and a Pig is revealed, your odds of having each are 50%-Goat, 0%-Pig, or 50%-Car.

If it was Goat/Goat/Car and a Goat is revealed, your odds of having each are 33%-First Goat, 33%-Second Goat, or 33%-Car. And since the Goats are indistinguishable, your odds of having each are 67%-Goat, or 33%-Car.

That’s the flaw in your logic. Half the time Monty will show a Goat and half the time he will show a Pig. Of the times he shows a Goat, you will have picked the Pig 2/3 of those times and the car 1/3 of those times. Of the times he shows a Pig, you will have picked a Goat 2/3 of the time and a car 1/3 of the time. This is because he shows the Pig 100% of the time you picked the Goat, but only 50% of the time you picked the car (the other 50% of the time you picked the car Monty shows the Goat and not the Pig).

Quote: unJon

That’s the flaw in your logic.

You are correct. 1/3, same as the original Monty Hall problem.

Quote: michael99000

What if Adam asked the warden to tell him the name of one prisoners who won’t be executed, and Adam himself was one of the possible answers the warden could give ? In other words the warden chooses his answer completely randomly among the 2 he knows won’t be killed.

Then if the warden says bob or Charlie , Adam would be 50/50 to be killed

Adam asked for the name of another inmate who would be spared. If Adam were not to be executed, the warden would have said the other safe name.

Let's say Monty, after revealing a goat to player A (call him Adam) behind curtain 2, then asks another player (call him Bob) to stand up and join Adam. Monty tells Bob that he will get the prize behind the remaining curtain that Adam does not choose.

It has been established that if Adam switches, his odds of winning the car go from 1/3 to 2/3. What are Bob's odds of winning the car? If they are different than Adam's, why?

Quote: Ayecarumba

Let's say Monty, after revealing a goat to player A (call him Adam) behind curtain 2, then asks another player (call him Bob) to stand up and join Adam. Monty tells Bob that he will get the prize behind the remaining curtain that Adam does not choose.

It has been established that if Adam switches, his odds of winning the car go from 1/3 to 2/3. What are Bob's odds of winning the car? If they are different than Adam's, why?

If Adam does not know the math , and is equally likely to switch or not switch , then I’d say Bobs chances are 50/50

If Adam knows the math, then he’s definitely going to switch , and bobs chances are 1/3 because he’s stuck with Adams’s original choice.

Quote: unJon

That's the flaw in your logic.

Quote: Chuckleberry

You are correct. 1/3, same as the original Monty Hall problem.

After a (very) long time, and looking at it from every angle I could, I think I figured out my error.
If we look at the situations:
1) Adam is to be executed, Bob will be named 50% of the time
2) Charlie is to be executed, Bob will be named 100%, and
3) (this was my error in logic) Bob is to be executed, Charlie would be named 100%.

(I originally thought that since Bob was named, we would have to eliminate this situation when calculating the odds going forward. If we ran one million simulations, and eliminated all the occurrences where Bob was to be executed, Adam would get the axe 50%. But I failed to consider that just because Bob was deemed safe doesn't mean that he was never in danger of being executed.)

In order for me to be right, the warden would have had to walk into a room with all 3 of them, and said "one of you was randomly chosen to be executed, and it's not Bob!"