[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]

Stick/switch. This gives a 3/4 chance, better than the 5/8 given by switch/switch. Stick/switch makes this just like the original problem, when Monty opens up all but one of the other doors (i.e. he opens up one, you stick, he opens up the other one, you switch). At that point, if you switch, you will win if you were wrong to begin with.

In general, with N doors, with this problem, you should pick a door and stick with it until your last chance to switch. This will give you an (N-1)/N chance of winning.

[Guest]

I pick a door. I have a 1 in 4 chance of having picked the right door. Therefore there is a 1/4 chance of all the rest of the doors containing bill collectors.

When Monty picks a door with a bill collector in it, each door has a 13/24 chance of being picked: 1/4 chance having a 1/3 chance of being picked, and a 3/4 chance of having a 1/2 chance of being picked.

Afterwards, the doors that were not picked would end up having a 11/24 chance of containing HS (?) which is higher than the 1/3 chance that my current door has, so I switch.

When Monty picks that next bill collector door, the leftover doors each have a 37/48 chance of being picked: 13/24 chance of definitely being picked and a 11/24 chance of a 1/2 chance of being picked.

Then Monty will open one of the doors, and the leftover door will hence have a 11/48 chance (?) of having HS, which is less than the 1/2 chance that my current door has, so I stay.

I haven't tried the original Monty problem yet, so I'm not sure how to solve this riddle.

[unreality]

if you switch both times, you have a 5/8 chance of getting the girl as opposed to smaller chances

[Guest]

Stick/switch. This gives a 3/4 chance, better than the 5/8 given by switch/switch. Stick/switch makes this just like the original problem, when Monty opens up all but one of the other doors (i.e. he opens up one, you stick, he opens up the other one, you switch). At that point, if you switch, you will win if you were wrong to begin with.

In general, with N doors, with this problem, you should pick a door and stick with it until your last chance to switch. This will give you an (N-1)/N chance of winning.

I'm with chuck...

Stick then Switch then kiss Daniel again because he was so cute!!!

[Guest]

hm, isn't it when you have probability divided into 4- like 25% each- then each time he opens the door- you change- so finally you have 75% to kiss sweetheart comparing with 25% to loose.....tell me if i'm stupid

[Guest]

75% make perfect sense when staying/switching. However, how do you get 5/8 from switching/switching?

[Guest]

75% make perfect sense when staying/switching. However, how do you get 5/8 from switching/switching?

After the first door closes:

Your door has sweetheart: 25% (you picked it independently)

The other two have evenly matched probabilities so 75%/2 = 37.5% or 3/8.

If you switch now, your door will have that 37.5% chance of containing the sweetheart. Host closes another door -> you switch and have a 62.5% or 5/8 chance of winning, the chance that your initial switch did NOT land on the sweetheart.

[brainz]

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.

ill say door 4 did i win?

[Guest]

if you switch both times, you have a 5/8 chance of getting the girl as opposed to smaller chances

i think switch/switch too because of how in the original MH problem, the probability that the right door is one you haven't picked is bigger than the probability that the right door is the one you have picked.

[Guest]

switch/switch

[Prof. Templeton]

Regardless of what happens before, when you get to the last decision you have 2/3 chance to win if you switch.

[Guest]

Regardless of what happens before, when you get to the last decision you have 2/3 chance to win if you switch.

That's not true in the four door problem. The 2/3 calculation in the three door problem results from the fact that all 3 doors originally had a prob of 1/3. In this case, though, the probability of the various doors depends on what you did in the first step. The probabilities work out like this:

Stick - Switch: 3/4

Switch - Switch: 5/8

Switch - Stick: 3/8

Stick - Stick - 1/4

[Prof. Templeton]

That's not true in the four door problem. The 2/3 calculation in the three door problem results from the fact that all 3 doors originally had a prob of 1/3. In this case, though, the probability of the various doors depends on what you did in the first step. The probabilities work out like this:

Stick - Switch: 3/4

Switch - Switch: 5/8

Switch - Stick: 3/8

Stick - Stick - 1/4

You are right.

[Guest]

A question regarding switch/switch.

You pick a door. The probability you picked the one with your high-school sweetheart behind it is 1/4. The probability she's behind one of the remaining doors is 3/4 (1/4 each). If a bill collector is removed from those three, that group still has a 3/4 chance of being the group with the sweetheart behind it (each door now has a 3/8 probability of your high-school sweetheart behind it). You decide to switch.

Now you have a door with a 3/8 probability of having your high-school sweetheart behind it. The probability of her being behind one of the other doors is 5/8. The one you originally picked still has a 1/4 probability of having your high-school sweetheart behind it and the other door 3/8.

You have a 3/8 probability of having the favorable door. The probability that the favorable door is among the other two is 5/8, but the probability is not distributed equally among the two of them. One has a 1/4 probability of being the favorable door and the other a 3/8 probability. After he reveals one of the others, if you switch, does your probability of picking the favorable door depend on which door reveals a bill collector?

[Guest]

A question regarding switch/switch.

You pick a door. The probability you picked the one with your high-school sweetheart behind it is 1/4. The probability she's behind one of the remaining doors is 3/4 (1/4 each). If a bill collector is removed from those three, that group still has a 3/4 chance of being the group with the sweetheart behind it (each door now has a 3/8 probability of your high-school sweetheart behind it). You decide to switch.

Now you have a door with a 3/8 probability of having your high-school sweetheart behind it. The probability of her being behind one of the other doors is 5/8. The one you originally picked still has a 1/4 probability of having your high-school sweetheart behind it and the other door 3/8.

You have a 3/8 probability of having the favorable door. The probability that the favorable door is among the other two is 5/8, but the probability is not distributed equally among the two of them. One has a 1/4 probability of being the favorable door and the other a 3/8 probability. After he reveals one of the others, if you switch, does your probability of picking the favorable door depend on which door reveals a bill collector?

That's really interesting, and I think it might make a small difference. Suppose you originally picked A, then D was opened, and you switched to B. If B is not the winner, they are more likely to open door A next (since C will be the winner 3/5 times when it isn't B). Specifically, they would open door A 21/40 times and door C 19/40 times if I did the math right. If B is the winner, they would presumably open A and C 1/2 of the times each. This means that, if door C gets opened, it is slightly more likely that you have the winner. There are two problems with this:

1. The 1/40 difference is not big enough to make this option preferable to the stick-switch method, and

2. It could be overcome by the selector adjusting the way they select the doors to match your expected probability. There's nothing in the OP that says they have to choose the next door randomly.

[bonanova]

That's really interesting, and I think it might make a small difference. Suppose you originally picked A, then D was opened, and you switched to B. If B is not the winner, they are more likely to open door A next (since C will be the winner 3/5 times when it isn't B). Specifically, they would open door A 21/40 times and door C 19/40 times if I did the math right. If B is the winner, they would presumably open A and C 1/2 of the times each. This means that, if door C gets opened, it is slightly more likely that you have the winner. There are two problems with this:

1. The 1/40 difference is not big enough to make this option preferable to the stick-switch method, and

2. It could be overcome by the selector adjusting the way they select the doors to match your expected probability. There's nothing in the OP that says they have to choose the next door randomly.

Hi CR,

You say

If B is not the winner, they are more likely to open door A next (since C will be the winner 3/5 times when it isn't B).

That isn't necessarily true.

Monty Hall is always certain to open a non-winning door (since he knows the winning door).

Are you saying that his choice of non-winning door affects your chances of winning?

[Prime]

Here is a general solution for the problem of this type. And it hardly uses any math at all. And no calculations, I promise.

It’s a stick/switch.

For N doors and only one prize, in general you always must stick to your original choice until your last pick. See it this way:

When it comes to your last choice, if the probability that your door hides your sweetheart is X/Y, then probability of Monty’s door is complement of that: (1-X)/Y. The smallest possible probability that you can get is your original choice: 1/N and that would leave (N-1)/N for Monty’s last door, which you should surely choose. If at any stage of the game you’d be switching your choice, you always end up with a door of greater probability than 1/N, but lesser than (N-1)/N. And thus you’d never get a choice as good as (N-1)/N.

[Prime]

A question regarding switch/switch.

You pick a door. The probability you picked the one with your high-school sweetheart behind it is 1/4. The probability she's behind one of the remaining doors is 3/4 (1/4 each). If a bill collector is removed from those three, that group still has a 3/4 chance of being the group with the sweetheart behind it (each door now has a 3/8 probability of your high-school sweetheart behind it). You decide to switch.

Now you have a door with a 3/8 probability of having your high-school sweetheart behind it. The probability of her being behind one of the other doors is 5/8. The one you originally picked still has a 1/4 probability of having your high-school sweetheart behind it and the other door 3/8.

You have a 3/8 probability of having the favorable door. The probability that the favorable door is among the other two is 5/8, but the probability is not distributed equally among the two of them. One has a 1/4 probability of being the favorable door and the other a 3/8 probability. After he reveals one of the others, if you switch, does your probability of picking the favorable door depend on which door reveals a bill collector?

Monty's possibility for random decision destroys the uneven distribution among his two doors. Unless, you know of his teasing nature, and expect that if he has two doors with bill collectors, he will leave your original choice unrevealed -- just to tease you.

[bonanova]

Monty's possibility for random decision destroys the uneven distribution among his two doors. Unless, you know of his teasing nature, and expect that if he has two doors with bill collectors, he will leave your original choice unrevealed -- just to tease you.

Are you saying that his choice of non-winning door affects your chances of winning?

[Prime]

Are you saying that his choice of non-winning door affects your chances of winning?

Do you mean which one of many bill collectors he sends home? If he chooses at random -- then his choice gives you no additional iformation. But if he wants to tease you, or have some other criteria that you can guess -- then yes!

Or do you mean the fact that he chooses bill collector -- not the prize? That is the very thing that allows you to improve your chance over the original choice.

[Guest]

Are you saying that his choice of non-winning door affects your chances of winning?

I'm saying it could, depending on how the door he opens is chosen. If it is chosen from a uniform distribution over all the non-winning non-chosen doors left, then I think it does. Follow me on this

Here is the situation: you started with door A. Monty opened door D. You then switched to door B. At this point, you know the probability of each door being a winner. If B is not the winner, then the winner must be A or C. Further, you know the probability of A being the winner is 1/4/(1/4+3/8)=2/5, and the prob of C being the winner is 3/5. This means that if B is not the winner, door A will have to be opened 3/5 of the time, since C can't be opened.

It's been a while since I've gone through Bayes' theorem, so I might have messed something up here. Let's give it a shot though:

To find P(B wins|Door A opened) you need: P(A opened|B wins), P(A opened|C wins), P(A opened|A wins), and P(A wins), P(B wins), and P(C wins)

P(A opened|B wins)=1/2 (assuming Monty chooses randomly).

P(A opened|C wins)=1

P(A opened|A wins)=0

P(A wins)=1/4

P(B wins)=3/8

P(C wins)=3/8

Going through Bayes' theorem, this gives us P(B wins|Door A opened)=(1/2*3/8)/(1/2*3/8+1*3/8+0*1/4)=1/3

Now let's calculate P(B wins|Door C opened). We now have

P(C opened|B wins)=1/2 (assuming Monty chooses randomly).

P(C opened|C wins)=0

P(C opened|A wins)=1

P(A wins)=1/4

P(B wins)=3/8

P(C wins)=3/8

Going through Bayes' theorem, this gives us P(B wins|Door C opened)=(1/2*3/8)/(1/2*3/8+0*3/8+1*1/4)=3/7>1/3

So your chances of winning with door B are better if Door C is opened (although still not as good as if you switch). Please let me know if I screwed something up here, as that is well within the realm of possibility. Having gone over it again, though, it looks pretty solid.

[Prime]

I'm saying it could, depending on how the door he opens is chosen. If it is chosen from a uniform distribution over all the non-winning non-chosen doors left, then I think it does. Follow me on this

Here is the situation: you started with door A. Monty opened door D. You then switched to door B. At this point, you know the probability of each door being a winner. If B is not the winner, then the winner must be A or C. Further, you know the probability of A being the winner is 1/4/(1/4+3/8)=2/5, and the prob of C being the winner is 3/5....

Actually, the probability of A is still 1/4 at that point, and probability of B and C is 3/8 each.

[Guest]

For clarity, I've gone back and color-coded the equations. Didn't get back to it in time to edit

To find P(B wins|Door A opened) you need: P(A opened|B wins), P(A opened|C wins), P(A opened|A wins), and P(A wins), P(B wins), and P(C wins)

P(A opened|B wins)=1/2 (assuming Monty chooses randomly).

P(A opened|C wins)=1

P(A opened|A wins)=0

P(A wins)=1/4

P(B wins)=3/8

P(C wins)=3/8

Going through Bayes' theorem, this gives us P(B wins|Door A opened)=(1/2*3/8)/(1/2*3/8+1*3/8+0*1/4)=1/3

Now let's calculate P(B wins|Door C opened). We now have

P(C opened|B wins)=1/2 (assuming Monty chooses randomly).

P(C opened|C wins)=0

P(C opened|A wins)=1

P(A wins)=1/4

P(B wins)=3/8

P(C wins)=3/8

Going through Bayes' theorem, this gives us P(B wins|Door C opened)=(1/2*3/8)/(1/2*3/8+0*3/8+1*1/4)=3/7

Edited for error

Edited August 11, 2008 by Chuck Rampart

[Prime]

For clarity, I've gone back and color-coded the equations. Didn't get back to it in time to edit

Edited for error

Color code is difficult to follow. However, is it possible that Bayes' theorem is not applicable here (whatever it is)? Especially, if it yields an incorrect result.

The two doors that the host has after the initial switch in your scenario have collective probability of 1/4 + 3/8 = 5/8. And Monty's consequent manipulations do not change that probability for his remainin doors. If you break that situation into 3 uneven scenarious, then in one of them (when B hides the prize) the probability for B is 100% -- no further calculations are needed. Whereas, in the other 2 cases, the probability for B is 0 and the probability for Monty's remaining door to hide the prize is 100% (and he did not choose randomly in those cases) -- no theorems need to be applied here either.

[Guest]

you know the probability of A being the winner is 1/4/(1/4+3/8)=2/5, and the prob of C being the winner is 3/5. This means that if B is not the winner, door A will have to be opened 3/5 of the time, since C can't be opened.

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.

Edited August 12, 2008 by itsclueless