[bonanova]

You know the drill.

Only one of the doors is a winner; but this time there are four doors.

Behind three of them stand bill collectors.

Behind the other is your high-school sweetheart.

Monty Hall gives you the rules:

1. You point to a door.
   
2. Of the other three doors, at least two have bill collectors. I will open one of them.
   
3. You decide whether to stay with your choice or pick another door.
   
4. I will then open another bill collector door that is not your current pick.
   
5. You then decide whether to stay with your choice or pick the only other unopened door.
   

What are the best odds for kissing once again your HS sweetie, and how do you get them?

Should you

• stick / stick,
  
• stick / switch,
  
• switch / stick, or
  
• switch / switch?
  
  1. You pick door 1.
     
  2. MH opens door 2: Bill Collector.
     
  3. You switch your pick to door 3.
     
  4. MH opens door 1: Bill Collector.
     
  5. You decide whether to stick with door 3 or switch to door 4.
     
  Or:

[*]You pick door 4.

[*]MH opens door 1: BC.

[*]You stick with door 4.

[*]MH opens door 2: BC.

[*]You decide whether to stick with door 4 or switch to door 3.

[Guest]

This is incorrect. The probability of A being the winner is still 1/4. Opening door D doesn't change that, because it wasn't part of the set of possible choices for Monty when he opened door D.

That statement follows the previous statement, "If B is not the winner, then the winner must be A or C. " I was only considering cases where B is not the winner, to show that the door that gets opened gives you some information about where the sweetheart is.

I'll try to explain how I'm seeing this by comparing it to another situation - suppose two people are playing darts, and that you know one of them is better than the other. Specifically, one will hit a bullseye 3/5 of the time, and one will hit a bullseye 1/2 of the time. The first player steps up to throw, but you don't know whether he is the good player or the not as good player. If he hits the bullseye, you can say it's more likely he's the good player. His hitting the bullseye gave you more information.

This is similar - if the sweetheart is not behind door B then you know the probabilities of which door Monty will open (since he HAS to open the one with the bill collector, and doors A and C have different probabilities). If the sweetheart is behind door B, then Monty gets to choose. If he picks doors A and C with equal probabilities, then that is a difference from the case where the sweetheart is not behind door B. Of course, you don't know for sure how Monty chooses the door to open, and even if he was equally likely to choose each door you would still be better off switching. It would give you more information though.

[Prime]

That statement follows the previous statement, "If B is not the winner, then the winner must be A or C. " I was only considering cases where B is not the winner, to show that the door that gets opened gives you some information about where the sweetheart is.

I'll try to explain how I'm seeing this by comparing it to another situation - suppose two people are playing darts, and that you know one of them is better than the other. Specifically, one will hit a bullseye 3/5 of the time, and one will hit a bullseye 1/2 of the time. The first player steps up to throw, but you don't know whether he is the good player or the not as good player. If he hits the bullseye, you can say it's more likely he's the good player. His hitting the bullseye gave you more information.

This is similar - if the sweetheart is not behind door B then you know the probabilities of which door Monty will open (since he HAS to open the one with the bill collector, and doors A and C have different probabilities). If the sweetheart is behind door B, then Monty gets to choose. If he picks doors A and C with equal probabilities, then that is a difference from the case where the sweetheart is not behind door B. Of course, you don't know for sure how Monty chooses the door to open, and even if he was equally likely to choose each door you would still be better off switching. It would give you more information though.

In your scenario (SWITCH/?) the probability of Monty's last remaining door to be A (the door initially selected by player) is 7/16, and the probability of the last remaining Monty's door to be door C is 9/16. But that does not help your final selection at all. For the probabilities of a particular door to be the last unopened Monty's door and probability of HS hiding behind it can not be mixed or combined.

You set a scenario where Monty has possession of HS and therefore the probabilities for the doors A and C are 2/5 and 3/5 respectively. But overall probability of that scenario is 5/8, so overall probabilities of the doors are (2/5)*5/8=1/4 and (3/5)*5/8=3/8 -- something that we know already. That is before Monty opened his last door. When he does, his choice is FORCED NOT RANDOM in your scenario. Therefore remaining door simply inherits the probability of 5/8 of hiding HS.

[Guest]

I'd rather hang out with the bill collectors.

[Guest]

I have seen a lot of incorrect information posted about this and the Monty Hall puzzle. The correct answer is that when there are two choices left and one of them is right and the other is wrong, the odds are 50-50 that you will pick correctly. The previous choices make no difference at this point. For this particular puzzle, you started with 4 choices one of which was correct, so your probablity of a correct choice at that point were .25. You chose, and one choice was taken away. You then had three choices, one of which was correct, so your probablity of a correct choice at that point was .333. Another choice was then taken away. You finally have 2 choices, one of which is correct, so your probablity of correct choice is .5.

Another way to think of it at this point is there are four situations, two of which are winners and two are losers.

1) You keep the door you have selected and it is the correct door

2) You keep the door you have selected and it is the wrong door

3) You choose the other door and it is the correct door

4) You choose the other door and it is the wrong door.

You know the drill.

Only one of the doors is a winner; but this time there are four doors.

Behind three of them stand bill collectors.

Behind the other is your high-school sweetheart.

Monty Hall gives you the rules:

1. You point to a door.
   
2. Of the other three doors, at least two have bill collectors. I will open one of them.
   
3. You decide whether to stay with your choice or pick another door.
   
4. I will then open another bill collector door that is not your current pick.
   
5. You then decide whether to stay with your choice or pick the only other unopened door.
   

What are the best odds for kissing once again your HS sweetie, and how do you get them?

Should you

• stick / stick,
  
• stick / switch,
  
• switch / stick, or
  
• switch / switch?
  
  1. You pick door 1.
     
  2. MH opens door 2: Bill Collector.
     
  3. You switch your pick to door 3.
     
  4. MH opens door 1: Bill Collector.
     
  5. You decide whether to stick with door 3 or switch to door 4.
     
  Or:

[*]You pick door 4.

[*]MH opens door 1: BC.

[*]You stick with door 4.

[*]MH opens door 2: BC.

[*]You decide whether to stick with door 4 or switch to door 3.

[unreality]

autonav: I have two doors. One has a llama behind it, the other has an alpaca. You pick a random door and open it. What are the chances of it being a llama?

Just because there are two options, it doesn't mean they are 50-50. In my llama-alpaca example, the chances are 50-50 though because you don't have any further information. Information changes probability. If you are told that there is twice as likely chance of picking an alpaca, the chances of the llama are 1/3. If you are told the llama went sick and had to go home, the chances of an alpaca are still 1/2, but the llama is 0 ("nothing" is the other 1/2). If there are 7 doors, with 6 alpacas and 1 llama, and three doors are opened to show alpacas, you have a 1/4 of picking a llama now as opposed to 1/7 before

now if there are those same 7 doors, but you pick one, and three doors are opened (all alpacas, and the host specifically opened alpaca doors), the door you picked has a 1/7 chance of being a llama. The other three doors must sum to 6/7, and thus are worth 2/7 each, which is why, if you wanted a llama, you should switch to a different door, because you have new information.

Information changes probability

[Guest]

In your scenario (SWITCH/?) the probability of Monty's last remaining door to be A (the door initially selected by player) is 7/16, and the probability of the last remaining Monty's door to be door C is 9/16. But that does not help your final selection at all. For the probabilities of a particular door to be the last unopened Monty's door and probability of HS hiding behind it can not be mixed or combined.

You set a scenario where Monty has possession of HS and therefore the probabilities for the doors A and C are 2/5 and 3/5 respectively. But overall probability of that scenario is 5/8, so overall probabilities of the doors are (2/5)*5/8=1/4 and (3/5)*5/8=3/8 -- something that we know already. That is before Monty opened his last door. When he does, his choice is FORCED NOT RANDOM in your scenario. Therefore remaining door simply inherits the probability of 5/8 of hiding HS.

What I'm saying is that his choice is forced some times, and not forced other times. Specifically, when HS is behind door B, his choice is not forced - he can open either A or C. When HS is not behind B, his choice is forced - he has to open whichever of A or C does not have HS.

What I'm saying is that the way Monty makes his choice when he is not forced may give you some information. There are limitless ways he could choose which door to open when the door the player has chosen hides the HS. I'll give you an extreme example so you can see that this is true - imagine that he will always open door C whenever he has a choice between A and C. Then, if he opens door A, you know that door C hides the HS, otherwise he would've opened that.

Now imagine that he will open door C 90% of the time when Monty has a choice. If door A opens, you no longer know that door C has the HS - since door A will open 10% of the time when door B is the winner - but you can be much more confident than you were before either door was open.

So if Monty chooses the door when has a choice using any probability besides what he is forced to choose one door or the other (which comes from the probabilities of each door being the winner), you gain information from which door is opened.

[Prime]

What I'm saying is that his choice is forced some times, and not forced other times. Specifically, when HS is behind door B, his choice is not forced - he can open either A or C. When HS is not behind B, his choice is forced - he has to open whichever of A or C does not have HS.

What I'm saying is that the way Monty makes his choice when he is not forced may give you some information. There are limitless ways he could choose which door to open when the door the player has chosen hides the HS. I'll give you an extreme example so you can see that this is true - imagine that he will always open door C whenever he has a choice between A and C. Then, if he opens door A, you know that door C hides the HS, otherwise he would've opened that.

Now imagine that he will open door C 90% of the time when Monty has a choice. If door A opens, you no longer know that door C has the HS - since door A will open 10% of the time when door B is the winner - but you can be much more confident than you were before either door was open.

So if Monty chooses the door when has a choice using any probability besides what he is forced to choose one door or the other (which comes from the probabilities of each door being the winner), you gain information from which door is opened.

CR,

I commend your attempts to confuse the issue. They are clever and of some complexity. But they will not work.

Once again your line of reasoning mixes up probabilities of Monty having some particular door unopened in the end and probability of HS hiding behind that door.

When you switch your choice on the first opportunity, probability that Monty leaves your originally picked door unopened is 7/16, probability for the other door to be the last unopened is 9/16; whereas probability of the HS the behind last unopened door is 5/8 in either case. Do you disagree?

Note, how the first two probabilities add up to 1 with each other, whereas 5/8 adds up to 1 with the probability of your second pick (3/8).

When you break down your variation into two sub-variations, where in one case you have HS and in another – Monty does, no longer can you use the previous probabilities (1/4 and 3/8) of having HS for any of Monty’s doors. Since in one case you have already presumed that probability to be zero; and in the other case Monty’s choice is not random, but forced.

Your “extreme example” changes the statement of the problem, making it different from what OP meant. I have already done that humorously, drawing some skepticism from Bonanova. (See posts 19, 20, and 21).

Think of it this way:

When you chose a door of probability P of hiding HS, all the remaining doors in Monty’s possession have total probability of 1-P. Monty’s manipulations change probabilities of his individual doors, but their total remains the same. Until you make another switch, that is.

See my general solution in post #18.

[Guest]

I'm not disagreeing with your solution in #18. I actually think we agree on just about everything.

To review the situation, so we're on the same page: contestant originally chose door A. Monty opened door D, revealed a bill collector. Contestant then switched to door B. You calculated the probabilities of each door being opened correctly - A will be opened 9/16 of the time, C will be opened 7/16 of the time (you phrased it as "which door remains unopened," but that's equivalent). I calculated the probability of door B being the winner, given that door A had been opened and given that door C had been opened. I found p=1/3 when door A was opened, and p=3/7 when door C was opened. Note that all of these assume that Monty chooses each door equally often in situations where he is not forced to open one or the other (ie, when B is the HS).

Here's something interesting though (red numbers are yours, blue numbers are mine, the "|" sign means "given"):

P(B winner)=P(A opened)*P(B winner|A opened)+P(C opened)*P(B winner|C opened)

P(B winner)=9/16*1/3+7/16*3/7=3/16+3/16=3/8

You'll notice the answer, 3/8, is exactly what we both agree it should be - before the second door is opened. After the second door is opened, Monty's choice of door has given you more information. Depending on which door opened, the probability of your choice being the winner has increased or decreased slightly.

I want to stress again that (1) the above numbers assume Monty chooses A and C each with equal probability when he is not forced to choose one or the other, (2) either way, you are still better off switching again, and (3) you're even better off going with a stay-switch strategy from the beginning.

I apologize if this is still unclear, but rest assured I'm not trying to confuse you.

Edited for clarity (hopefully)

Edited August 13, 2008 by Chuck Rampart

[Prime]

I'm not disagreeing with your solution in #18. I actually think we agree on just about everything.

To review the situation, so we're on the same page: contestant originally chose door A. Monty opened door D, revealed a bill collector. Contestant then switched to door B. You calculated the probabilities of each door being opened correctly - A will be opened 9/16 of the time, C will be opened 7/16 of the time (you phrased it as "which door remains unopened," but that's equivalent). I calculated the probability of door B being the winner, given that door A had been opened and given that door C had been opened. I found p=1/3 when door A was opened, and p=3/7 when door C was opened. Note that all of these assume that Monty chooses each door equally often in situations where he is not forced to open one or the other (ie, when B is the HS).

Here's something interesting though (red numbers are yours, blue numbers are mine, the "|" sign means "given"):

P(B winner)=P(A opened)*P(B winner|A opened)+P(C opened)*P(B winner|C opened)

P(B winner)=9/16*1/3+7/16*3/7=3/16+3/16=3/8

You'll notice the answer, 3/8, is exactly what we both agree it should be - before the second door is opened. After the second door is opened, Monty's choice of door has given you more information. Depending on which door opened, the probability of your choice being the winner has increased or decreased slightly.

I want to stress again that (1) the above numbers assume Monty chooses A and C each with equal probability when he is not forced to choose one or the other, (2) either way, you are still better off switching again, and (3) you're even better off going with a stay-switch strategy from the beginning.

I apologize if this is still unclear, but rest assured I'm not trying to confuse you.

Edited for clarity (hopefully)

Sorry, CR.

But the formula that you show appear to be invalid and the numerical values for probabilities incorrect. The formula uses P(B winner) both on the left and the right side of the equation. And they seem to disagree. Both P(B winner|A opened) and P(B winner|C opened) are equal to 3/16 each -- not 1/3 and 3/7 as you stated. Or if you consider the probabilities of A or C open inside the case when B is the winner -- then it is 1/2 each. Your P(A open) and P(C open) on the right side use overall probability including the case when B is not a winner. Therefore, multiplication of that by the probability of the cases when B can only be a winner makes no sense.

Your formula should be: P(Bw) = P(Bw)*P(Ao) + P(Bw)*P(Co), or substituting 3/8 = (3/8)*(1/2) + (3/8)*(1/2). And of course, it is not anything useful, because all it is saying: 1 = P(Ao) + P(Co) and we know that Monty will surely open one of his doors without any calculations.

After checking the other posts, I found that "general solution" (N-1)/N, although without proof, has been posted already as far back as post #4.

Edited August 13, 2008 by Prime

[bonanova]

I found that "general solution" (N-1)/N, although without proof, has been posted already as far back as post #4.

Proof goes like this.

If you stick with your choice until the final losing door is opened, two closed doors remain:

the one you initially picked [prob = 1/N] and the only other door that it could be.

But you knew that, right?

[Guest]

Sorry, CR.

But the formula that you show appear to be invalid and the numerical values for probabilities incorrect. The formula uses P(B winner) both on the left and the right side of the equation. And they seem to disagree. Both P(B winner|A opened) and P(B winner|C opened) are equal to 3/16 each -- not 1/3 and 3/7 as you stated. Or if you consider the probabilities of A or C open inside the case when B is the winner -- then it is 1/2 each. Your P(A open) and P(C open) on the right side use overall probability including the case when B is not a winner. Therefore, multiplication of that by the probability of the cases when B can only be a winner makes no sense.

Your formula should be: P(Bw) = P(Bw)*P(Ao) + P(Bw)*P(Co), or substituting 3/8 = (3/8)*(1/2) + (3/8)*(1/2). And of course, it is not anything useful, because all it is saying: 1 = P(Ao) + P(Co) and we know that Monty will surely open one of his doors without any calculations.

After checking the other posts, I found that "general solution" (N-1)/N, although without proof, has been posted already as far back as post #4.

I think you're misunderstanding the notation. See http://mathworld.wolfram.com/TotalProbabilityTheorem.html and equation (8) from http://mathworld.wolfram.com/BayesTheorem.html

Again, I'm not disagreeing with the general solution. Stick-Switch is still your best bet.

[Prime]

I think you're misunderstanding the notation. See http://mathworld.wolfram.com/TotalProbabilityTheorem.html and equation (8) from http://mathworld.wolfram.com/BayesTheorem.html

Again, I'm not disagreeing with the general solution. Stick-Switch is still your best bet.

I understand the notation. In your scenario SWITCH/? the probability of your door to hide the prize is 3/8 and the probability of Monty's last remaining door is 5/8 regardless which door he opened 2nd time (A or C). If any equations/and or reasoning show otherwise, they are erroneus. (Like the equation you showed in the previous post.) And I am convinced at this point that you are quite capable of finding errors in your equation.

They say you can't add apples and oranges. Well, sometimes you can't multiply them either.

[Prime]

Proof goes like this.

If you stick with your choice until the final losing door is opened, two closed doors remain:

the one you initially picked [prob = 1/N] and the only other door that it could be.

But you knew that, right?

I posted a proof that you get (N-1)/N chance to win in this game by sticking to your first choice and switching when Monty has only one door left AND that you cannot do better than (N-1)/N with any other strategy.

Here is a derivative from this problem:

Say there are N doors, N>4. Suppose, you knew that if you switched your original choice, Monty would keep that door closed to tease you, except when he has to open it (he is down to two doors and has to open your originally chosen door, because the other one hides the prize.) What would be your chances and best strategy in that situation?

[Guest]

In your scenario SWITCH/? the probability of your door to hide the prize is 3/8 and the probability of Monty's last remaining door is 5/8 regardless which door he opened 2nd time (A or C). If any equations/and or reasoning show otherwise, they are erroneus. (Like the equation you showed in the previous post.)

That's not true, and you know it. If you reread the OP, you'll see there's no mention of how Monty chooses which bill collector door to open when he is not forced. This means that he could open door A 100% of the time, 0% of the time, or anywhere in between, in cases where the HS is behind door B. It would help me see your objection if you answer the follow questions:

1. If you know he opens door A 0% of the time when he is not forced, and he opens door A, are the probabilities still 3/8 and 5/8?

2. If you know he opens door A 1% of the time when he is not forced, and he opens door A, are the probabilities still 3/8 and 5/8?

3. What number or numbers x would make the following statement true: "You know he opens door A x% of the time when he is not forced, and he opens door A. The probabilities after the second door is opened are still 3/8 and 5/8."

If you object to the word "know" in the above statements, please keep in mind that all my previous reasoning was explicitly based on the assumption that he would open each door with equal probability when he was not forced.

[Prime]

That's not true, and you know it. If you reread the OP, you'll see there's no mention of how Monty chooses which bill collector door to open when he is not forced. This means that he could open door A 100% of the time, 0% of the time, or anywhere in between, in cases where the HS is behind door B. It would help me see your objection if you answer the follow questions:

1. If you know he opens door A 0% of the time when he is not forced, and he opens door A, are the probabilities still 3/8 and 5/8?

2. If you know he opens door A 1% of the time when he is not forced, and he opens door A, are the probabilities still 3/8 and 5/8?

3. What number or numbers x would make the following statement true: "You know he opens door A x% of the time when he is not forced, and he opens door A. The probabilities after the second door is opened are still 3/8 and 5/8."

If you object to the word "know" in the above statements, please keep in mind that all my previous reasoning was explicitly based on the assumption that he would open each door with equal probability when he was not forced.

I already posted a derivative problem, where Monty teases player who swithched the door. (See my previous post.)

Whereas, OP did not mention any criteria for Monty to send one or the other bill collector home, we must assume his choice is random in that situation. And for that case the formula that you showed is still incorrect. As well as it is incorrect for any other case with Monty's non-random decision between two bill collectors. (I have already explained how several times.)

If I explained how Monty's partiality with respect to bill collectors would affect probabilities, I'd be giving away clues on solving my derivative problem.

[Guest]

Going back a few posts, you said

Once again your line of reasoning mixes up probabilities of Monty having some particular door unopened in the end and probability of HS hiding behind that door.

Is that your objection to my reasoning in post 24? Because those two probabilities most certainly are linked, in a way that doesn't mix them up. You know that, because you calculated the 9/16 and 7/16 probabilities earlier for doors A and C being opened.

Look at it this way - when Monty chooses the door, he picks door A with prob 1/2 (presumably). When he is forced to open one door, he opens door A 3/5 of the time, because it is more likely that HS is behind door C than door A. That means that if he opens door A, it is more likely to be an occasion where HS is not behind door B.

[Guest]

step1:

chance on succes = 1/4, (3/4 against) => SWITCH

step 2:

change on success = 1/2 * 3/4 = 3/8 (5/8 against) => SWITCH

you must always switch

*edit:

forget above => STICK, SWITCH will give a 3/4

Edited August 14, 2008 by dfhwze

[Prime]

Going back a few posts, you said Is that your objection to my reasoning in post 24? Because those two probabilities most certainly are linked, in a way that doesn't mix them up. You know that, because you calculated the 9/16 and 7/16 probabilities earlier for doors A and C being opened.

Look at it this way - when Monty chooses the door, he picks door A with prob 1/2 (presumably). When he is forced to open one door, he opens door A 3/5 of the time, because it is more likely that HS is behind door C than door A. That means that if he opens door A, it is more likely to be an occasion where HS is not behind door B.

I am under the impression that you suggest that if you switch the door at your first opportunity, then probability of your newly chosen door (B) to have the prize would vary depending on which door Monty opened second time around -- your original choice, or the other one (A or C).

I disagree with that assessment and state as following:

In your scenario, SWITCH/?, the probability of your second choice (B) will remain 3/8 regardless of Monty's consequent manipulations. Whereas, the probability of Monty opening door A second time around is 9/16, versus door C -- 7/16.

If you disagree with that, or have calculated different values, I suggest check your line of reasoning/calculations for error.

(Note, that I gave exact numbers for probabilities as I calculate them.)

[Guest]

I am under the impression that you suggest that if you switch the door at your first opportunity, then probability of your newly chosen door (B) to have the prize would vary depending on which door Monty opened second time around -- your original choice, or the other one (A or C).

I disagree with that assessment and state as following:

In your scenario, SWITCH/?, the probability of your second choice (B) will remain 3/8 regardless of Monty's consequent manipulations. Whereas, the probability of Monty opening door A second time around is 9/16, versus door C -- 7/16.

If you disagree with that, or have calculated different values, I suggest check your line of reasoning/calculations for error.

(Note, that I gave exact numbers for probabilities as I calculate them.)

So we're clear, I do think the statement "if you switch the door at your first opportunity, then probability of your newly chosen door (B) to have the prize would vary depending on which door Monty opened second time around -- your original choice, or the other one (A or C)" is true.

I go back to my darts player example: There are two people playing darts. You know one of them is better than the other one (he will hit a bullseye more often), but you do not know which is which. Before either player throws a dart, you can safely say the probability is 1/2 that either one is the better darts player. If the first one throws a dart and hits the bullseye, you can now say it is more likely that that player is the better darts player.

Do you think my darts example is wrong, or do you think it's not a suitable analogy for the problem we've been discussing?

[Prime]

So we're clear, I do think the statement "if you switch the door at your first opportunity, then probability of your newly chosen door (B) to have the prize would vary depending on which door Monty opened second time around -- your original choice, or the other one (A or C)" is true.

I go back to my darts player example: There are two people playing darts. You know one of them is better than the other one (he will hit a bullseye more often), but you do not know which is which. Before either player throws a dart, you can safely say the probability is 1/2 that either one is the better darts player. If the first one throws a dart and hits the bullseye, you can now say it is more likely that that player is the better darts player.

Do you think my darts example is wrong, or do you think it's not a suitable analogy for the problem we've been discussing?

Darts example is not a suitable analogy. Each dart player can either hit or miss, whereas in case of Monty -- he must always open non-winning door. Until he opens the very last one and shows your high school sweetheart hiding behind it. But at that point it's too late for the player to make calculations.

With probability problems, when in doubt, go through all possible permutations. And there are but 4 in your scenario.

We agreed that after you make the initial switch from door A to B, the probabilities for the doors to hide the prize are A: 1/4, B: 3/8, and C: 3/8.

Then there are 4 mutually exclusive and collectively exhaustive cases:

1. A has the prize/Monty opens C (forced move). --------------------> P1=(1/4)*1 = 1/4.

2. B has the prize/Monty opens C (random choice from A and C).--> P2=(3/8)*1/2=3/16

3. B has the prize/Monty opens A (random choice from A and C).--> P3=(3/8)*1/2=3/16

4. C has the prize/Monty opens A (forced). ---------------------------> P4=(3/8)*1=3/8

Now case 2 and case 3 are the only two where door B is the winner. The probability P2 + P3 = 3/16 + 3/16 = 3/8. And that's the total probability for B to be the winner after Monty opens his second door.

Cases 1 and 2 are the only cases when door C is opened. P1 + P2 = 1/4 + 3/16 = 7/16.

Cases 3 and 4 are the only cases when door A is opened. P3 + P4 = 3/16 + 3/8 = 9/16.

(Note how the four cases above add up to 1).

I believe you can figure out where your formula is wrong/inapplicable without my help. When you multiply probabilities of two events, those must be dependent. When you add two probabilities -- their events must be mutually exclusive. E.g., you cannot multiply probability of door A being open second time around by probability of door B being a winner. Such operation has no meaning in this problem.

[Guest]

I'd rather hang out with the bill collectors.

haha, you know, I wouldn't take a 25% chance/risk of kissing a bill collector. But there's got to be a 25% chance of the bill collector being a woman, and if it's a game show, your 100% guaranteed to be ripped off.

[Prime]

And let me preempt possibility for further confusion.

Let's say Monty opened door C second time around. That excludes case 3 described above and case 2 inherets its probability, becoming 3/16 + 3/16 = 3/8. The choice has been made and no longer its probability is 1/2. Similarly, after door C is opened, case 4 gets excluded and case 1 inherits its probability becoming 1/4 + 3/8 = 5/8. Since those two cases were mutually exclusive.

[Guest]

You should always stick with your original door until the last opportunity to switch, and then switch.

In general, for N doors, the chance your sweetie is behind the door you originally picked is

1/N,

and the chance she's not is

1 - (1 / N), originally spread out over N-1 other doors.

By staying with your original pick, your chance of having picked correctly is the same, 1/N, but the 1 - (1 / N) probably that she is not is now only spread out across one other door. In the classic 3-door Monty Hall problem, you have a 1 - (1/3) = 66% chance of picking correctly by switching; as N goes up, your odds only get better. N=4, chance for last switch is right = 1 - (1/4) = 75%, N=10 chance = 1 - (1/10) = 90%

[Guest]

Darts example is not a suitable analogy. Each dart player can either hit or miss, whereas in case of Monty -- he must always open non-winning door. Until he opens the very last one and shows your high school sweetheart hiding behind it. But at that point it's too late for the player to make calculations.

With probability problems, when in doubt, go through all possible permutations. And there are but 4 in your scenario.

We agreed that after you make the initial switch from door A to B, the probabilities for the doors to hide the prize are A: 1/4, B: 3/8, and C: 3/8.

Then there are 4 mutually exclusive and collectively exhaustive cases:

1. A has the prize/Monty opens C (forced move). --------------------> P1=(1/4)*1 = 1/4.

2. B has the prize/Monty opens C (random choice from A and C).--> P2=(3/8)*1/2=3/16

3. B has the prize/Monty opens A (random choice from A and C).--> P3=(3/8)*1/2=3/16

4. C has the prize/Monty opens A (forced). ---------------------------> P4=(3/8)*1=3/8

Now case 2 and case 3 are the only two where door B is the winner. The probability P2 + P3 = 3/16 + 3/16 = 3/8. And that's the total probability for B to be the winner after Monty opens his second door.

Cases 1 and 2 are the only cases when door C is opened. P1 + P2 = 1/4 + 3/16 = 7/16.

Cases 3 and 4 are the only cases when door A is opened. P3 + P4 = 3/16 + 3/8 = 9/16.

(Note how the four cases above add up to 1).

I believe you can figure out where your formula is wrong/inapplicable without my help. When you multiply probabilities of two events, those must be dependent. When you add two probabilities -- their events must be mutually exclusive. E.g., you cannot multiply probability of door A being open second time around by probability of door B being a winner. Such operation has no meaning in this problem.

At this point I've become convinced that I'll never be able to change your mind, at least over the internet (give me a whiteboard and it might be a different story). I suspect you're thinking something similar. It's probably in both of our best interests to give up this discussion and go about our merry ways for the rest of our lives, happily convinced that we were right about this problem.

[Prime]

At this point I've become convinced that I'll never be able to change your mind, at least over the internet (give me a whiteboard and it might be a different story). I suspect you're thinking something similar. It's probably in both of our best interests to give up this discussion and go about our merry ways for the rest of our lives, happily convinced that we were right about this problem.

Are you offering to beat me with your whiteboard? I decline. Don't take me wrong, I love violence, but it's not the most effective way to solve probability problems.

Bearing the whiteboard threat in mind, I have to concede my point. After you switch your first choice and AFTER Monty opens second door, the probability of your door to hide the prize becomes 1/3 if Monty opened your originally chosen door second time around and 3/7 if he opened another one.

See my post of another probability problem, which I derived from our exchange.