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Monty Hall solution flawed?

The solution to the Monty Hall problem seems flawed to me.

The solution states that the probability of winning with a switch after one door is eliminated is 2/3.
I contend that as one door is ALWAYS eliminated, 3 doors should not be accounted for in calculating probability.
It doesn't matter what you choose from the beginning, you will always wind up with a 50/50 chance of winning, hence it doesn't matter if you switch or not, odds are 1/2 anyway.

I would like it if someone would explain to me why I am wrong. :)

R Jeppesen
Thursday, January 22, 2004

Monty chose to give you a choice to switch with knowledge of what door you chose and what door the prize is behind. Nowhere does it say that Monty had to give you that choice. You don't know what Monty's reasons for giving you a choice are. You don't know if he would have given you a choice had you picked another door.

The problem is ill posed; arguing whether the answer is 2/3 or 1/2 is pointless because both positions are wrong.

Ham Fisted
Friday, January 23, 2004

Here is a restatement of the problem which maintain the spirit of the original problem but alleviates doubts that the problem is ill posed.  Unfortunately, the explicit rendering of the problem in this way reveals much of the thought process required in the solution.

1. Three doors are presented to the contestant.  One door hides a prize.  The other two doors do not.  Monty knows which door hides the prize because he randomly selected which door would hide the prize before the game began. Also assume that the winning door does not change during the game. (For example, if the contestant knew that the winning prize was always behind door 1, that could change the solution.)

2. The contestant selects a door. After the contestant selects a door, Monty ALWAYS reveals a losing door.  If the contestant had chosen the winning door, Monty randomly reveals one of the two losing doors that the contestant did not select.  If the contestant chose a losing door, then Monty reveals the other losing door.
(For example, if Monty only let you switch when you had picked in the winning door, that would change the answer.)

3. You must also assume that the contestant has no knowledge about the doors, except that the revealed door is a losing door and that exactly one of the two unrevealed doors hides a prize.
(For example, if you have inside information about where the prize is hidden, that would change your answer.)

4.  Assume that your only goal is to maximize the probability that you will win the prize.  (For example, my wife objects to the solution because she has faith that her original guess would be "lucky".  She associates a high penalty with losing the game after switching, so that changes her answer. )

5.  The contestant is presented with the choice to keep the original door or to switch to the lone unrevealed door.

Should the contestant switch?  Yes, the contestant should ALWAYS switch as laid out in the solution.

If you doubt the answer, try it out experimentally.  Have a friend play "Monty" with three playing cards where one of the cards is designated the prize, and try both strategies (always stay, always switch).  Make sure your "Monty" follows all of the rules laid out above.  If you play enough, you will find that a "always switch" strategy will win at twice the rate of "always stay".  Based on my experience, some people find the solution so counter-intuitive that this experimental approach is the only way to convince them.

P Hyden
Monday, January 26, 2004

I would like to make the solution simpler.  Assume there are 100 doors and you select 1 of the doors.  You have a 1% chance of getting the correct answer.  Let's then assume you are shown 98 empty doors.  Do you then select the last door or keep your own?  As the previous person suggested -- you should switch.  The reason is that by having shown you 98 doors and letting you switch, you have in fact been "given" 99 choices (98 empty ones and 1 unknown).  Your original selection only gives you 1 choice.  Therefore, by switching your winning percentage is always
(n-1)/n.  In staying with your own choice, your winning percentage is 1/n.

If that is confusing, look at it from the reverse.  (a) You select 99 doors.  (b) You are shown 98 empty doors. (c) Do you switch to the door you did not select or do you keep the 99 doors you selected (98 empty + 1 unknown).  Sticking with your choice gives you a 99% chance of winning.  If you select the door you did not select, you will only be right 1% of the time.

Kishi A. Teixeira
Thursday, February 5, 2004

U can find a very good explanation for the Monty Hall Problem in the following link:

vijayakumar b
Tuesday, February 10, 2004

Think again Jeppesen; conditional probability is often very nonintuitive. Write out a table of possibilities.

But before we get to the goats and cars, consider this:
i flip a penny and a dime and hide the result from you. "at least one of the coins came up heads", i announce. what is the chance that both coins came up heads?

Now by your logic you would tell me that since the coins did not touch or interact that their respective states are arbitrary, telling you one means nothing about the other and so the probability of the other being heads is of coures .5

However, how many results are possible? lets investigate simply with a table:

Coin 1    |  Coin 2
    H                H          -Case1
    H                T          -Case2
    T                H          -Case3
    T                T          -Case4

now given our requirements, case 4 is out because there are no heads, BUT since it is not determined which coin comes up heads we have three cases left in which the requirements are met, and in 2/3 of the cases the other coin is a tail. Sneaky huh?

Now on to goats, cars and monty hall;
with our case structure here are the door possibilities:

Doors:    1  2  3
case1      C  G  G
case2      G  C  G
case3      G  G  C

Now pick any door and 'stand behind it' both literally and figuratively. Assume monty kows where the car is and will always show you a goat. pick a door and eveluate what would happen when monty opens another door.

now find out what happens if you stayed with that door OVER ALL THREE CASES. you will find that if you switch, you win the car twice as often, watch:

Suppose i select door 3. In case1, monty opens door 2 to reveal a goat - i switch and YAY, i win the car.
Case2 monty opens door one to show me the goat - i switch and YAY, i win the car.
Case3 monty opens either door depending on his fancy to show you a goat, if you switch, sigh a loss, but its the only one!

You see, if you stay, you will only win in the case where you selected the right car from the beginning (1/3 of the time). When monty opens a door he gives you  more information and you Must use it. Switching wins you the car the remaining 2/3 of the time

Jeremy Slickcode
Tuesday, February 10, 2004

How about this explanation: the probability of winning if you don't switch is simply the probability that you pick the right door on your first try. The probability of doing that is 1/3. Hence the probability of winning if you do switch is 2/3.

Lawrence Wang
Wednesday, February 25, 2004

Guess I should have read the post directly above mine before posting. :)

Lawrence Wang
Wednesday, February 25, 2004

The answer is "always Switch", and the probability of winning is 2/3 if you do.  I've never liked this answer. 

It always seemed to me that since Monty would open an 'Other' door, you REALLY only had two choices -- the door you picked, and the one he didn't open.  I could argue this point forever.  It wasn't until I did a game simulation, and ran it 1000 times, that I concluded son-of-a-gun, somehow my argument is wrong, the answer is always switch.

I still don't know WHY my argument is wrong, it still seems logical to me that you 'really' have a 1/2 chance if you stick, but the simulation results were quite compelling.

Monday, March 1, 2004

OK, I can explain why the 2/3 answer is correct.  It really is easier with the 100 door example.  You pick one door.  CLEARLY, you have a 1/100 chance of getting that one right.

Now, Monty opens 98 other doors, 'carefully' picking to make sure he didn't open the one with the prize.  Now the two choices you make are NOT independent.  You KNOW you had a 1/100 chance of picking the right door first.  Now, you KNOW that other door has a 98/100 chance of being the right door. 

It's not true that you always had a 1/2 chance -- there really is an increase of 'knowledge' about the system with every door Monty opens.  And it is taking advantage of that 'knowledge' to switch.

Monday, March 1, 2004

Best explanation I've seen.

Joe Bennett
Friday, March 19, 2004

I didn't try experimenting yet, but all the explanations I've read so far didn't really convince me in a better success probability of switching.

As I see it, there are 4 possible cases (not 3, as all explanations claimed). Let's assume that prize is behind door 1.

case A: selected door 1, monty revealed door 2
case B: selected door 1, monty revealed door 3
case C: selected door 2, monty revealed door 3
case D: selected door 3, monty revealed door 2

In two of the cases (A and B), if you switch you lose.
In two of the cases (C and D), if you switch you win.

It's a tie.

Wednesday, March 31, 2004

This thread has been coming up every couple of months for two years or more.

Let's see what is wrong with Dino's logic.

case A: selected door 1, monty revealed door 2
case B: selected door 1, monty revealed door 3
These two cases are correct and you lose in both cases.

case C: selected door 2, monty revealed door 3
case D: selected door 3, monty revealed door 2
No, in both these cases Monty will reveal door 1, which is where the prize is. And thus in both cases you win.

So is the poster right in saying that you have a 505 chance whether you stick or switch. No.

The point  is that the cases don't have equal weight.
In fact there are three cases
Case A: selected door 1 Monty reveals door 2 or door 3
Case B: selected door 2 Monty reveals door 1
Case C: selected door 3 Monty reveals door 1

It's clear that you have twice the chance of winning if you switch.

Incidentally this has used in the game of Bridge for well on thirty years. It is known as the principle of restricted choice.

Stephen Jones
Monday, April 5, 2004

Though I did not beleive it at first, I wrote a quick application, ran it through 2 million runs, and can see that as the number of repeated trials goes towards infinity, the values tend towards :

Dont Switch : 1/3 or exactly 33.2952 %

Switch Door : 2/3 or exactly 66.7049 %

Josh McGeehon
Monday, April 5, 2004

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