More probability - the Monty Hall Problem

[T_R_Oglodyte]

Here's a classic probability problem.

You are the contestant on a game show hosted by Monty Hall. You are brought up to the stage, where there are three doors. Monty tells you that behind one of the doors is the deed to a fixed week 52, 3-bedroom upper floor oceanfront unit at the new Westin Ka`anapali (or substitute Atlantis if you prefer). Plus all annual fees and assessments have been prepaid for the next 25 years. Behind the other two doors are timeshare salesmen from the Royal Sunset.

Monty tells you to pick a door. If you pck the door that has the deed to the Westin (or Atlantis) it's yours. If you pick a door with a timeshare salesman, you are obliigated to sit through a four-hour sales presentation and not receive any freebies.

You pick a door. After you make your selection, Monty opens one of the doors you didn't pick, revealing a timeshare salesman. Monty now asks you if you want to stay with your original selection, or if you want to switch to the other unopened door.

What should you do? Does it make a difference if you switch doors?

 

[chemteach]

Steve,
I have to laugh when I see this problem. I teach high school AP Chemistry. Another woman at the school teaches AP Biology. She has a Ph.D. in Geophysics from Princeton. I have a Ph.D. in Chemical Engineering from Caltech. The AP Physics teacher has Masters in Physics from Caltech. I was trying to explain to them why you should always switch, and my AP Chem students got into competitions with her AP Biology students over which teacher was correct. After a day of discussions back and forth and an internet search where I found a great graphic showing the problem, she finally conceded. The AP Physics teacher was still stuck in "It shouldn't matter." Alas, he wouldn't accept that Monty knew where the good prize was located, and that this affected the outcome.

If you actually do the simulation, you find that it really does work to always switch. It's a very fun problem to debate. (If you enjoy these types of debates, of course...)

Edye

 

[awsherm]

I've always loved this problem. Especially how adamantly people will argue that switching shouldn't matter. I also like that choosing incorrectly with the first choice, guarantees a win with a switch. One of those rare instances where it pays to be wrong.

Alex

 

[T_R_Oglodyte]

I also like that choosing incorrectly with the first choice, guarantees a win with a switch.

That's a great way to analyze the problem that I hadn't previously seen!!

When you pick your door initially, the odds are 2:1 against you. By switching after the door is opened, you get to flip those odds, so the odds are now 2:1 in your favor instead of 2:1 against you.

 

[teachingmyown]

So what about "Deal or No Deal?" I saw an episode where it went all the way down to two cases (from beginning with 26) and before opening the chosen case the contestant was offered the opportunity to switch for the one remaining unchosen. It is always asserted that "NO ONE" knows what value prize is hidden in any of the cases, so the "Monty always knows" factor is not there. Should the contestant switch or not switch?

Steve,

The AP Physics teacher was still stuck in "It shouldn't matter." Alas, he wouldn't accept that Monty knew where the good prize was located, and that this affected the outcome.


Edye

 

[T_R_Oglodyte]

... the "Monty always knows" factor is not there. Should the contestant switch or not switch?

But Monty does always know - that's key.

If Monty didn't know what door had the prize, then one-thrid of the time he would pick the door that had the prize. But whenever Monty gives you the chance to switch, he never picks a door that has the prize. He always picks a door that does not have the prize. That means that he is not picking a random door. That difference is critical to the problem.

 

[teachingmyown]

But Monty does always know - that's key.

If Monty didn't know what door had the prize, then one-thrid of the time he would pick the door that had the prize. But whenever Monty gives you the chance to switch, he never picks a door that has the prize. He always picks a door that does not have the prize. That means that he is not picking a random door. That difference is critical to the problem.

I believe you misread my question, which might be my fault since I was going off on a tangent to the original post. I was referencing a different game show..."Deal or No Deal" not "Let's Make a Deal". In DoND, the host is Howie, not Monty and --according to what he is constantly saying-- NO ONE knows which case is the best. Therefore, the "Monty knows" factor IS removed, so does it then simply become a 50/50 problem?

 

[T_R_Oglodyte]

I believe you misread my question, which might be my fault since I was going off on a tangent to the original post. I was referencing a different game show..."Deal or No Deal" not "Let's Make a Deal". In DoND, the host is Howie, not Monty and --according to what he is constantly saying-- NO ONE knows which case is the best. Therefore, the "Monty knows" factor IS removed, so does it then simply become a 50/50 problem?

I don't know the Deal or No Deal game, but I'm assuming the setup is similar to the Monty Hall problem - i.e.,

• you select from among three choices
• After you choose, one of the choices you didn't make is revealed. If the prize is revealed, you lose and the game is over.
• If the revealed choice is not the prize, the game continues and you are offered the option to switch to the remaining choice you didn't select.

You are still better off switching if given the opportunity.

Your overall odds of winning the game at the time you start the game are one-third. But after you survive the door opening hurdle, the situation is the same as the Monty Hall problem.

****

This is an example of conditional probability.

Your overall odds of winning the game are one-third.

Your odds of winning the game, given that you now know that one of the choices you didn't make is not a winner becomes two-thrids if you switch; one-third if you don't switch.

The odds change in the middle of the game because at the time you are offered the chance to switch, you have more information about what is behind the doors than you had at the start of the game.

 

[teachingmyown]

I don't know the Deal or No Deal game,

The odds change in the middle of the game because at the time you are offered the chance to switch, you have more information about what is behind the doors than you had at the start of the game.

I suspect you'd enjoy it...at least for a show or two.

It begins with 26 choices ranging in values from $0.01 to $1million; the contestant picks one case from the 26 to call his own. Then the elimination begins as the values of the unchosen cases are revealed one by one, in an order chosen by the contestant. (Insert dramatic pauses and cute quips by host) If the contestant is able to maintain their nerve and not sell out (give up) he will eventually find himself choosing between two cases, the one originally chosen and the last unchosen remaining to be opened.

So, in the beginning, his chances are 1 in 26 of choosing the correct case. But at the end, his chances are 1 in 2. Since the Monty factor is not there, is it simply a 50/50 choice?

 

[swift]

Okay, I don't get it. Why does it matter whether or not Monti knows. When it comes down to the last two doors you are picking the door and no matter which one you choose you have a 50/50 shot at being correct.

 

[BocaBum99]

Steve,
I have to laugh when I see this problem. I teach high school AP Chemistry. Another woman at the school teaches AP Biology. She has a Ph.D. in Geophysics from Princeton. I have a Ph.D. in Chemical Engineering from Caltech. The AP Physics teacher has Masters in Physics from Caltech. I was trying to explain to them why you should always switch, and my AP Chem students got into competitions with her AP Biology students over which teacher was correct. After a day of discussions back and forth and an internet search where I found a great graphic showing the problem, she finally conceded. The AP Physics teacher was still stuck in "It shouldn't matter." Alas, he wouldn't accept that Monty knew where the good prize was located, and that this affected the outcome.

If you actually do the simulation, you find that it really does work to always switch. It's a very fun problem to debate. (If you enjoy these types of debates, of course...)

Edye

Edye,

So, you are clearly off the charts in terms of analytical capabilities. I've always been told that engineers are the hardest to sell on timesharing. Is that true?

What process did you use to buy into timesharing? Did you buy from the developer or on the resale market first?

Just curious.

 

[Dave M]

swift -

You are correct that it would be 50-50 if the contestant picked his door - the first time - after[i/]Manty opened one of the doors.

When the contestant picks one of the three doors, there is a 1 in 3 chance that the pick was for the door with the most valuable prize. The chances are 2 in 3 that the prize is behind one of the other two doors.

Monty will then show the contestant what's behind one of the two remaining doors. However, Monty always picked a door other than the one with the big prize, based on his knowledge of where the big prize was or wasn't.

Now there are two doors left. But the odds don't change. The contestant's odds are still the same 1 in 3 that the chosen door has the big prize. Why? Because there was always at least one losing door that Monty could pick to show. Thus, the odds are still 2 out of 3 that the big prize is behind the other door. Thus, the contestant should accept the offer to switch.

Here's another example. Suppose I place coins under 100 napkins and, without showing them to you, ask you to pick the only coin that is heads-up (versus tails-up). You pick one, but I don't let you see if you guessed correctly. Then one by one, I uncover 98 of the coins, all of which are tails-up. Do you think your chances of having picked the only heads-up coin have changed from 1 in 100? No, they haven't. Why? Because I knew which losing (tails) coins to show you, making you erroneously believe you had a better chance of winning than 1 in 100. The theory of Monty's doors is the same.

 

[T_R_Oglodyte]

Okay, I don't get it. Why does it matter whether or not Monti knows. When it comes down to the last two doors you are picking the door and no matter which one you choose you have a 50/50 shot at being correct.

No you don't. If you switch you have a 2/3 chance of being right. If you don't switch, you have a one-third chance of being right.

Let's work it out. The prize can be behind any one of three doors, call them Doors A, B, and C. The prize is equally likely to be behind any of the three doors. You can pick any of the three doors to start the game. So there are nine combinations of doors to pick and choices you make.

Those nine combinations define all of the playing options, and there is an equal probability of winning with any of the nine combinations.

Case 1. Prize is behind Door A.

• You pick door A. Monty opens either Door B or Door C (doesn't matter which one).
  • Switch = lose
  • Hold = win
• You pick door B. Monty opens Door C
  • Switch = win
  • Hold = lose
• You pick door C. Monty opens Door B
  • Switch = win
  • Hold = lose

Case 2. Prize is behind Door B.

• You pick door A. Monty opens Door C
  • Switch = win
  • Hold = lose
• You pick door B. Monty opens either Door A or Door C (doesn't matter which one).
  • Switch = lose
  • Hold = win
• You pick door C. Monty opens Door A
  • Switch = win
  • Hold = lose

Case 3. Prize is behind Door C.

• You pick door A. Monty opens Door B
  • Switch = win
  • Hold = lose
• You pick door B. Monty opens Door A
  • Switch = win
  • Hold = lose
• You pick door C. Monty opens Door A
  • Switch = lose
  • Hold = win

Summary:
Out of the nine equally possible combinations to start the game, the switching strategy results in winning the prize in six of them. Holding is the winning strategy in only three of them.

It works out this way because at the start of the game 2/3 of the time the prize will be behind one of the doors you didn’t pick. When Monty opens the door, it doesn’t change the 2/3 probability that you picked the wrong door to start. What is different is that now you know that if you did pick incorrectly, there is a 100% chance that the prize is behind the other door. Since the chances that you picked the wrong door are 2/3, there is 2/3 chance the prize is behind the remaining door you didn’t pick.

**************

Another way to look at this is to recognize that at the start of the game, there was a 2/3 chance that you picked the wrong door. Switching after one of the other doors has been opened is the wrong decision only if your original selection was correct. Note that there was only a 1/3 chance your initial selection was correct. Ergo, the decision to switch is correct 2/3 of the time.

 

[Elan]

Steve,
I have to laugh when I see this problem. I teach high school AP Chemistry. Another woman at the school teaches AP Biology. She has a Ph.D. in Geophysics from Princeton. I have a Ph.D. in Chemical Engineering from Caltech. The AP Physics teacher has Masters in Physics from Caltech. I was trying to explain to them why you should always switch, and my AP Chem students got into competitions with her AP Biology students over which teacher was correct. After a day of discussions back and forth and an internet search where I found a great graphic showing the problem, she finally conceded. The AP Physics teacher was still stuck in "It shouldn't matter." Alas, he wouldn't accept that Monty knew where the good prize was located, and that this affected the outcome.

If you actually do the simulation, you find that it really does work to always switch. It's a very fun problem to debate. (If you enjoy these types of debates, of course...)

Edye

I hope the AP Biology teacher has a substitute come in to teach genetics..........

 

[awsherm]

Here's another example. Suppose I place coins under 100 napkins and, without showing them to you, ask you to pick the only coin that is heads-up (versus tails-up). You pick one, but I don't let you see if you guessed correctly. Then one by one, I uncover 98 of the coins, all of which are tails-up. Do you think your chances of having picked the only heads-up coin have changed from 1 in 100? No, they haven't.

This is an excellent extension of the problem to illustrate why a switch is in your best interest. You can't change the fact that the original pick was 1 in 100. When there are two coins left (yours and the one under the napkin), the chance that one of them is heads-up is 100 in 100.

Your Choice (1/100) + Coin under napkin (99/100) = 100/100

SWITCH!!!

 

[T_R_Oglodyte]

I suspect you'd enjoy it...at least for a show or two.

It begins with 26 choices ranging in values from $0.01 to $1million; the contestant picks one case from the 26 to call his own. Then the elimination begins as the values of the unchosen cases are revealed one by one, in an order chosen by the contestant. (Insert dramatic pauses and cute quips by host) If the contestant is able to maintain their nerve and not sell out (give up) he will eventually find himself choosing between two cases, the one originally chosen and the last unchosen remaining to be opened.

So, in the beginning, his chances are 1 in 26 of choosing the correct case. But at the end, his chances are 1 in 2. Since the Monty factor is not there, is it simply a 50/50 choice?

At the end the chances are not 1 in 2. It depends on what the remaining prizes are. For example, if the big $1 million prize is still in play at the end of the game, the chances of winning the $1 million is 25/26 if you switch, 1 in 26 if you hold.

I don't have time to work out the logic for a more complete answer - it will depend on the rankings of the remaining choices and the payout table for th echoices.

 

[Dave M]

I don't come up with that 25/26 answer, Steve, primarily because the contestant, not the game show host, is randomly picking the cases.

Example: With three cases left, one of which the contestant picked at the start of the game, assume that the prizes remaining are $1, $1,000 and $1,000,000. Using your logic, when the contestant next picks a case to eliminate, the odds would be extremely high that his one choice of the other two cases will hold the $1,000,000. In fact, any of the remaining prizes has an equal chance of being uncovered every time the contestant chooses a case.

Phrased differently, when a contestant picks the first case of 26 (his case), the chances that it holds any particular prize is 1 in 26. Then he opens and eliminates one case/prize, leaving 25 cases and 25 prizes. There was an equal chance (1 in 26) that any particular prize was eliminated when that first case was opened. When he next opens a case, the chance that it holds any particular prize is 1 in 25. Etc. When there are two cases left, the two remaining prizes will have been determined by random draw - the contestant's random draw.

Thus, the chances that the contest's case holds the larger of the two prizes should be exactly 1 in 2. The chances are only 2 in 26 that the top prize ($1,000,000) will be one of the two remaining prizes. But if it is still there, it should have an equal chance of being in either case.

In fact, the "banker's offer" to the contestant to cash out at various points during the show (e.g., after 5 and 10 and 15, etc. cases have been opened) takes into account that random calculation.

On the other hand, if someone asked what are the chances my case that I pick at the start of the game holds $1,000,000, I would confidently state 1 in 26. The chances are 24 in 26 that I will eliminate that as a prize in getting down to the last two cases. But if it's still there....

Or try it this way. Assume the last two prizes of the 26 are B and G (of the 26 prize letters). Thus, I know that my case holds either B or G. The chance that I might have originally picked B was 1/26. The same for G. Now that I'm down to B and G (regardless of what the actual prize amounts are), I have a 50-50 chance that my case holds B.

 

[#1 Cowboys Fan]

OK---back to the Monty Hall thing.
I simply am one that does not see whether switching or keeping makes any difference.

So, let me pose this:

What is Monty had FOUR doors, and offered you one of them.

Let's say you take #4, and he then reveals #1 as being worthless, and offers you to switch.

So, given your advice, let's say I switch to #3.

Now let's say he reveals #2 as being worthless, and asks if you want to switch again.

Are you advising that I switch back to #4, which I already exchanged?????!!!!!!!

I still say it doesn't matter---what do you think?

Pat

 

[RonaldCol]

Let's get to the chase. What is the right answer? Thank you.

 

[Dave M]

Pat -

Look carefully at Steve's post #13 in this thread. He goes through each of the nine possible combinations and shows that in six of the nine situations it pays to switch. In other words, switching wins in six situations and loses in three. Thus, it's two-to-one in favor of switching.

Look carefully at each example in that post and then add them up. If you don't understand his explanation (and, specifically, the nine situations) in that particular post, come back and ask about it.

 

[T_R_Oglodyte]

OK---back to the Monty Hall thing.
I simply am one that does not see whether switching or keeping makes any difference.

So, let me pose this:

What is Monty had FOUR doors, and offered you one of them.

Let's say you take #4, and he then reveals #1 as being worthless, and offers you to switch.

So, given your advice, let's say I switch to #3.

Now let's say he reveals #2 as being worthless, and asks if you want to switch again.

Are you advising that I switch back to #4, which I already exchanged?????!!!!!!!

I still say it doesn't matter---what do you think?

Pat

Pat - go back to my post #13 and work through that.

Now to the situation you posit with four doors.

First, at the start of the game you pick door 4. You have a 25% chance of being right. Also, there is a 75% chance that the prize will be behind one of doors 1 through 3. Those two probabilities never change during the progress of the game..

Before Monty opens a door, there is a 25% chance of the prize being behind any of the doors. As soon as Monty opens the first door (#1 in your example), you now know that door #1 did not have the prize.

Now remember that there is still a 75% chance that the prize is behind doors 1, 2, and 3. But you also know that door 1 did ot have the prize. So now you know there is a 75% chance that the prize is behind door 2 and door 3. There's an equal chance that the prize is behind either of those doors, so now the odds for door 2 and door 3 are 37.5% each. (Note they were originally 25% each, but with door 1 eliminated the odds that they have the prize increases.

So now you switch to door 3, having increased your odds of winning from 25% to 75%.

Continuing with your scenario, Monty now opens door 2. What's the situation now?

Again, remember that at the start of the game there was a combined 75% chance that the prize was behind doors 1, 2 or 3. Since doors 1 and 2 are now revealed to be worthless, door 3 has a 75% chance of being a winner. So when Monty offers you the chance to switch back to door 4, you decline.

You can extend it further, as Dave suggested. Imagine that there are 100 doors to choose from. Then there only a 1% chance that the prize will be behind the door you picked. If Monty then shows you 98 doors that don't have prizes behind them, don't you think it's pretty certain that the prize is going to be behind the one Monty didn't show you? Or do you believe that there was actually a 50% chance you guessed right in the first place??

 

[T_R_Oglodyte]

I don't come up with that 25/26 answer, Steve, primarily because the contestant, not the game show host, is randomly picking the cases. ...

Dave - I think we're looking at the problem slightly differently.

I am presuming that the player has already arrived at the end of the game, and there are only two cases left, one of which has $1 million, and the contestant hasn't switched cases. At that point, if the contestant should switch cases because there's a 25/26 chance that the case the contestant didn't select has the $1 million.

I'm not familiar with that particular game so I may be missing something in the play of the game.

 

["Roger"]

OK---back to the Monty Hall thing.
I simply am one that does not see whether switching or keeping makes any difference....

Pat

See if this helps (it is actually Dave M's problem, explained differently)...

Suppose there are 100 doors with a prize behind just one of them. You get to pick. What are you're chances of picking the correct door?

1/100

That means that that the chances of the prize being behind one of the other doors is

99/100

Certainly you would be better off if you could pick all 99 of those doors rather than the one that you did pick.

Now Monty, knowing which door the prize is behind, shows you every one of those 99 doors except one. Should you switch?

Yes, you should. The odds that it was behind one of those 99 doors was

99/100

The fact that Monty picked out 98 of those doors (knowing that he was not picking the right one) and shown you that the prize was not behind those doors has not changed the fact that the odds were overwhelmingly in favor of the prize being behind one of the 99 doors.

So, the odds that the prize is behind that one remaining door is

99/100

You should switch.

The same holds true when you only start with three doors.

 

[Icarus]

Dave - I think we're looking at the problem slightly differently.

I am presuming that the player has already arrived at the end of the game, and there are only two cases left, one of which has $1 million, and the contestant hasn't switched cases. At that point, if the contestant should switch cases because there's a 25/26 chance that the case the contestant didn't select has the $1 million.

I'm not familiar with that particular game so I may be missing something in the play of the game.

You would be wrong

There's a 50/50 chance that either case holds either of the two remaining dollar amounts at the end of DoND.

I thought about applying the Monty Hall problem to DoND when it first started, but it doesn't apply. In fact, we talked about it on BARGE.

The primary differences are:

- The host is not eliminating the booby prizes (lower amounts) on DoND. The host doesn't know what's in each case. The host does not eliminate a case with a low dollar amount for you, so you don't have additional information.

- It doesn't matter what amounts are left at the end of the game. There's one higher amount and one lower amount.

- The only additional information you get by opening a case is the knowledge that the amount in that case cannot be in your case. But since that amout AND that choice is eliminated, you still have no new information about what's in the remaining cases. Your odds improve slightly because the new N is smaller by 1. But your odds never ever improve greater than 1 in N, and the smallest N can be is 2.

Let's say at the end of the game, you have two cases left, 1 case holds $1 and the other case holds $1,000,000. That's still all the information you have.

If your case holds $1.
Possibility 1: Keep it and you win $1
Possibility 2: Switch it and you win $1,000,000

If your case holds $1,000,000
Possibility 3: Keep it and you win $1,000,000
Possibility 4: Switch it and you win $1

2 out of 4 possibilities nets you $1
2 out of 4 possibilities nets you $1,000,000
There's no advantage in keeping or switching.

If we apply this to the MH problem, each curtain has a 1/3 chance of being the winning curtain on LMaD. But, when Monty eliminates one of the curtains, after you've chosen one, he will always eliminate a loser. So, there's still your curtain with that same 1 in 3 possibility of being the winner because you chose it randomly prior to him revealing new information, but the other curtain that you didn't choose has gone from the same 1 in 3 to a 2 in 3 possibility because of the new information that Monty gave us when he revealed the third curtain. You proved that in your post with the truth table you posted earlier in this thread.

I hope you can see how this is different from DoND. In DoND, there's no free information that can be used to improve the odds of the other choices.

ok, here's another one ...

Let's say you are playing "Who wants to be a Millionaire" and you get down to a question that you have no clue as to what the correct answer is. So you take the 50/50, in which the computer randomly eliminates 2 out of 3 incorrect answers, leaving you with one correct answer and one incorrect answer. What's your strategy?

-David

 

[Big Matt]

It depends on whether you have any more life lines and what stage you are in the game.

If I am trying for 64k and have no life lines then you just guess. You can't lose any money and have no other options.

If I'm going for a million and have no life lines I can lose a ton of money and my pay back odds are much worse than 50/50. In that case I'd take the 500 thousand and walk away.

You would be wrong

ok, here's another one ...

Let's say you are playing "Who wants to be a Millionaire" and you get down to a question that you have no clue as to what the correct answer is. So you take the 50/50, in which the computer randomly eliminates 2 out of 3 incorrect answers, leaving you with one correct answer and one incorrect answer. What's your strategy?

-David