Delicate Genius Blog

Very cool dg – let me know when the cone of silence lifts!

Very cool simulator! ;-D What amuses me even more, is why people are getting confused over this little dilemma, which, in my opinion, is hardly difficult. :-p. I answered it and provided the correct reason for it (a complete ego-inflater, i know) last year when I was merely 15.

What intrigues me is that the % of choosers is 2/3 swap and 1/3 stick.
How come?
Is ti that the players are all non-naive
Is this the average within each player’s trials???

I would say more people are swapping because the problem says in the first instance you have a greater chance of winning if you swap. It’s almost telling you to swap, sub-conciously…I think it’s more of a pyschological effect than anything else.

We naturally want to win and we trust the problems claim with only slight apprehension because there must be some tuth to it otherwise it wouldn’t be such a well known problem. Perhaps were just dogs in some guy name Pavlovs’ backyard.

As far as a simple explanation of the actual problem itself goes, perhaps Syazana can help us out…?

Vincenze.

to explain this very, very simply, one should switch because the chances of not choosing the door with the money behind it in the first place is 2/3. the fact that the host reveals an empty, rewardless door changes nothing. i doubt if this is mathematically accurate, but it works for me…

Well – it was momentary — now the totals are 50 -50 for swappers and stickers; o r maybe I was confused by the winners row; now I’ll never know …

anyway to Syazana: your explanation is simple – a little too simple.
The key is that the host MUST choose a door with nothing behind it. So he does tell you something on the 2 out of the 3 possible choices, when you were wrong. You just don’t know when they were so you should always switch.

Imersion, there is a Perl program that simulates the probabilities, offering insight into the paradox. From their website:

“Two strategies are simulated: stick with initial choice every time, or switch from initial choice every time. The program performs a number of games and counts how many times each strategy wins the game.

The result of the simulation almost always shows that the switch strategy wins about twice as many times as the stick strategy. This is empirical evidence that switching is a better strategy in this game.”

If the totals were 50/50, it was only a matter of time before they leveled out to 66.66/33.33, or closely thereabouts. It has been mathematically proven that switching doors results in a significantly improved chance of winning. To see a visual representation of the use of Bayes’ theorem to confirm the findings, search Wikipedia for “Monty Hall problem.”

Yoni,

I’ve heard many people tell me that it checks out mathematically, that it’s a simple problem (even though it caused a storm when first proposed, especially in math circles)… using bayes theorem, and you’re right there’s plenty of simulators about…but over the years I’m yet to receive a complete explanation that made me say “oh, it all makes sense now”.

Nothing personal, I’d just love for someone to clear it up for me…

maybe I’m just a bit statistically impaired. :)

Vincenze.

Vincenze, here’s my attempt. Imagine you are given a choice of two games to play.

Game A rules:
There are three doors, and the prize is behind one of them. You choose ONE door, and if the prize is behind that door, you win.

Game B rules:
There are three doors, and the prize is behind one of them. You choose TWO doors, and if the prize is behind ANY of the two, you win.

Obviously you would always choose to play Game B. Now I will show that Game A translates to sticking, and game B translates to switching in the original game.

If you decide to stick, your initial choice means “I choose this door”. Then you ignore the host opening one of the other doors, and wait for him to open the door you’ve chosen initially. This is exactly Game A.

If you decide to switch, your initial choice actually means “I choose the OTHER TWO doors, NOT this one”. Then the host opens BOTH these doors for you, in two stages: first to show a goat, and then to show what’s behind the door you “switched” to. If the prize is behind EITHER of them, it is yours to take. This is exactly Game B.

Fun… Intersting… Makes one think…

nice!!

Vincenze-
I’ll attempt the simple explanation, so here’s the short version. I’ll give you the long version if this doesn’t do it for you.

The problem with the “Monty Hall problem” is that there is no way to solve it without making an assumption. And I think most people that look at this using logic are unwilling to make the assumptions that Marilyn makes. As the problem is written, she knows that Monty knows where the prize is, but she ASSUMES he always chooses a door, and the door he chooses always has no prize.

Now when I look at the problem logically, I am not willing to assume anything, because assumptions are illogical! So even if he does know what is behind the doors, how do I know he picks a door every time? How do I know he isn’t trying to trick me? Maybe he just picks a random door? Without assuming anything, I see a 50/50 chance that I have the correct door, which makes sense, and is correct.

But using her assumptions, there are three possibilities-

Number one- you chose loser door A, he shows you loser door B. He has no other choice under the assumptions. If you switch, you win. (1/3 wins)

Number two- you chose loser door B, he shows you loser door A. He has no other choice under the assumptions. If you switch, you win. (1/3 wins)

Number three- you choose winner door, he shows you either loser door. If you switch you lose. {So each of his choices counts for only ½ of the possibilities for that door (1/3 of your choices), so each is 1/6 loses, or added together, 1/3 loses.}

Therefore, two out of three possibilities, you win by switching. This is counterintuitive, but if you diagram it out, it makes sense.

So, her math is correct- but only if you follow her assumptions (which is what the simulation above does also).

The reason there was such a huge outcry about her answer, especially from the “learned” community, is because a true scholar or true logical person does not make these assumptions when solving a math problem.

Dg

I have been working on a variation of the the problem.

As many are aware, the fact that Monty knows where the prize is and will never eliminate it gives you the 2/3 edge by swapping. Its been said many times times that if the host’s eliminations are random (i.e based on no knowledge), but nevertheless an empty door is succesfully eliminated, then your chances of winning are now 50/50 either by swapping or sticking. And there are some simulators which allow you to specify whether the host knows or not. If you choose ignorance, you are given the results of the games in which the host didn’t prematurely reveal the prize. They do indeed level out as half won by swapping, half by sticking. What you’re not told, however, is how many games were voided.

On average we know that if the host’s eliminations are random, the game will be voided 1 out of every 3 games. And, on average, the two other games will see a win by swapping and a win by sticking. For this reason, the 50/50 probability is holds up.

Now consider: the game is played as above six times. To make it simpler, there are six rows with three doors each. And of course, behind one of the three doors in each row is a prize. So there are six prizes to be won.

The player chooses a door from each row. Question: at this point, how many prizes is the player most likely to have found? The answer here is not contraversial – it is of course 2 prizes. That is the most likely outcome if you have no other information.

Ok, so to complete the game the host opens a remaining door in each of the three rows. If the host discovers a prize in any row, then that particular row is disqualified and you have lost that prize. Assume that the host DOES NOT know where the six prizes are. Nevertheless he successfully aviods eliminating any of them.

So you are now left with a choice of keeping your originally chosen six doors or swapping to the remaining six doors.

What is your best option to increase the number of prizes? Does it make no difference?

One thing we can agree on: if, as expected, two of the six games had been voided by the host eliminating the prize – then for the remaining four it would have been 50/50 odds of getting a prize. In that case, the likely number of prizes you would end up with is still two.

However, even though the host’s eliminations were random, none of the six games were voided. IF the odds are still 50/50, the it should make no difference whether you stick or swap. But your expected number of prizes is likely now to be 3 not 2.

It has been claimed that so long as the all host’s eliminations were random, the probability is 50/50 for the non-voided games. Is this true, regardless of how many games were voided?

In the case of no games being disqualified, has the probability moved away from 50/50, and if so in which direction – swap or stick?

Bear in mind that when you first made your six selections the odds were 1/3 in each time, and you estimated you would probably have 2 out of the six prizes. After the host has randomly eliminated an empty door from each of six rows, does that change your first estimate? If your first estimate still holds, then the odds – as with Monty Hall – have become 2/3 in favour of swapping. If by sticking you probably end up with the 2 prizes, then by swapping you would end up with 4.

On the other hand, if no games were voided this might be a clue that you originally chose the correct door more times than not. After all, a game can only be voided if you originally chose an empty door. Normally you would expect two out six games to have been voided. Instead none were.
This suggests you are more likely than before to have selected above 2 prizes. Is the likely figure now 3 prizes? That brings us back to 50/50. Could be it be 4 prizes? If then the odds are 2/3 in favour of sticking.

So there we have it – conflicting answers. Which is corect?

Simon- you lost me a little…

Let’s say the host shows a door every time randomly (he doesn’t know where the prize is).

You have three possible picks, he has two possible picks each time, yielding 6 possibilities. Now for two of the six, you can’t switch (he showed you the winner, or the “void” scenario). These each count as 1/6 of the possible final outcomes. Four 4 of the six times he shows you a loser dooor, so you have two choices, switch or not. Each of these choices is worth 8/12, or 2/3 of the total possible outcomes.

So here are the six possibilities for the first two choices (your pick, his door).

Number one- You pick loser door A, he shows you loser door B. This represents 1/6 of the total possible outcomes, but since you have a choice of switch or not, each of those is half of this number, or 1/12. If you switch, you win. (1/12) If you don’t switch, you lose (another 1/12).

Number two- You pick loser door A, he shows you winner. You lose. (1/6)

Number three- You pick loser door B, he shows you loser door A. If you switch, you win. (1/12) If you don’t switch, you lose (another 1/12).

Number four- You pick loser door B, he shows you winner. You lose. (1/6)

Number five- You pick winner, he shows you loser door A. You switch, you lose. (1/12) You don’t switch, you win. (1/12)

Number six- You pick winner, he shows you loser door B. You switch, you lose. (1/12) You don’t switch, you win. (1/12)

So, if he always picks a door randomly, 4/12 (1/3) of the time you lose without even having a choice (void). 2/12 (1/6), you lose if you switch. 2/12 (1/6), you win if you switch. 2/12 (1/6), you lose if you don’t switch. 2/12 (1/6), you win if you don’t switch.

Now in the above scenario, we are eliminating 4 of the 12 possibilities, because we are told that Monty shows us a door with no prize behind it (even though it was random in this solution). Therefore, you have a 50-50 chance of winning if you switch. In this scenario, Marilyn is wrong. Switching does not benefit.

So, the big problem is still that the answer is impossible to come to without assumptions. What if he only sometimes gives you the option of another door? What if he tries to trick you into switching? What if he tries to trick you into not switching? It is impossible to figure the odds for all of these things.

Marilyn’s answer is wrong in my mind because she can not logically come to that answer without making certain assumptions. That’s why there was such an outcry. If she would have fully stated the conditions of the problem as she saw them, people would have initially questioned the answer, but quickly figured out she was right.

Hi Honolulu

I think we’re already agreed that: if you know the eliminations are random, then on average you have a 1/2 chance for all non-voided games. Non-voided games are those in which the host does not reveal the prize. Voided games are those in which he does.

Why are the odds 50/50? Again, I think we’re agreed. Its because on average, the host can be expected to void one third of the games. These voided games are ones you would have won by swapping, if the host had known to avoid revealing the prize.

Now we get to the contentious bit. Compare two scenarios.

Scenario 1: you have the following information

I played the game six times.
In each game, the host definitely knew where the prize was.
In each game I picked a door and the host opened one of the other two doors.
In each game, the door the host opened was empty.
In each game, I swapped to the remaining door.

Question: how many of those six games did I most likely win by swapping?

Answer: 4. We’ve already estabilished that when the host knows to avoid eliminating the prize, I was likely to win 2 in every 3 games by swapping. If I had stuck to my original selection in all six games, I would most likely have found the prize twice – the odds were only 1/3 in each game that my first selection was right.

More contraversial:

Scenario 2: you have the following information

I played the game six times.
In each game, the host definitely did not know where the prize was.
In each game I picked a door and the host opened one of the other two doors.
In each game, the door the host opened was empty.
In each game, I swapped to the remaining door.

Question: how many of those six games did I most likely win by swapping?

Answer?

Bear in mind, you know the host’s eliminations were random. But you also know that no games were in fact voided.

Are the odds still that I won 4 games if I swapped and 2 games if I stuck?

Best
Simon

Your scenario 1- exactly right. Under those condiutions (the ones Marilyn assumes), you win 2/3 of the time if you switch.

Your scenario 2- Despite his picking the empty door 6 straight times, odds are still 50/50 when he picks randomly. His random pick has no effect on your odds, so there is no benefit of switching. Odds are you’d win 3 and lose 3 (of course that may not be true in the small sample size of 6 games…).

Another way of saying it- even though no games were voided, you eliminated that possibility of a void game by setting the condition that the door he picked was empty- so your odds are 50/50 still.

Ok, here is the case for saying that the odds are still 2/3 in favour of swapping:

In both scenarios, before the host opens doors, the odds are definitely 1/3 that you will find the prize in your selection

In Scenario 2, as with Scenario 1, you are likely to have originally picked the prize twice out of six games. Whatever the host does subsequently, those were the original odds.

What’s most likely to happen in Scenario 2 is that the host will void two games by revealing the prize. But this does not alter the original odds for your selection – two prizes out of six. It is simply that there are only four games left and in the two that got cancelled you picked an empty door. So out of the remaining four games, you still say you probably picked the prize twice.

This most likely sequence of events is what confirms the 50/50 odds, when
the host doesn’t know where the prize is.

Now what happens if the host is ignorant, but still never revealed the prize in any of the six games? Does the host’s state of knowledge or lack of it have any bearing on what you probalby did when you made your selection?

I would use the following argument:

In any six games, on average you are likely to have originally selected the prize twice, regardless of what the host did afterwards.

On average, the host’s random eliminations will mean that most of the time you probably select the prize twice out of four games. This is the most likely outcome.

But if you are left with six games because the host never revealed the prize, this does not alter the original odds for your selection, which took place before the host did anything. Therefore you can still say you probably found the prize twice out of six games. If so, you would of course have won four times out of six by swapping.

Naturally, the host’s ignorance about where the prize is affects the likelihood of how many games will be voided. But it is the actual games voided that take away your advantage and reduce the odds to 50/50, not the host’s ignorance.

The average odds for your first independent selection is two prizes out of six. Do you ever have grounds to say that you originally selected three prizes out of six instead of two?

Now having made the above argument, I actually believe there is a flaw in it. Feel free to identify the flaw if you can.

But apart from the flaw, you see what I’m getting at :)

Here’s the flaw- the host’s original state of knowledge has no bearing on your original odds, in this you are correct. However, in your scenario 2, you stated that the host randomly showed empty doors 6 times- that doesn’t change your original odds (1 in 3 chances of winning any single game, or 2/6 of the total games won). But, it does alter your odds after his selection- the fact that he showed an empty door (randomly) does change your subsequent odds- he has in effect eliminated 4 of your twelve possible outcomes. Therefore, you now have 8 possible outcomes, 4 you win by switching, 4 you lose. It doesn’t matter if you play one game or 6 games or a million games- if he picks randomly and shows an empty door, your odds are 50/50 on the remaining two, doesn’t matter if you switch.

If he nonrandomly picks an empty door every time, it does change things as described above… good to switch (2/3 wins).

So, the flaw is that a random pick doesn’t change your possible outcomes (12 of equal weight), but a nonrandom pick does (8 possible outcomes, but the two where you switch and win only have half the weight of the two where you switch and lose).

Sorry, this time you lost me.

You said “the fact that he showed an empty door (randomly) does change your subsequent odds- he has in effect eliminated 4 of your twelve possible outcomes.”

Twelve possible outcomes? I’m not aware which twelve you’re referring to and which four have been eliminated. I never did mention twelve outcomes, so you have me guessing slightly.

One interpretation of what you mean is that the player makes six selections and the host makes six eliminations. If that’s what you mean by twelve outcomes, then I understand you. In that case, on average, four outcomes will be eliminated if the host is randomly opening doors and reveals two of the prizes. But no outcomes will be eliminated if the host never reveals a prize.

If the host doesn’t know where the prizes are, the most likely scenario for those twelve outcomes is: you discovered 4 empty doors and 2 prizes in your original selection. The host then reveals 4 empty doors and 2 prizes in his elimination. By eliminating 2 prizes, the host eliminates 2 of your empty doors. (In that sense, four outcomes have been eliminated.) You are left with with 4 games out of which you orignally chose 2 prizes and 2 empty doors. This most likely scenario is what validates the 50/50 odds. If you stick or swap in all four remaining games, you will end up with 2 prizes and 2 empty doors.

However this is only possible if the host voids two of the games.

Now you did say that the host’s subsequent opening of doors “doesn’t change your original odds (1 in 3 chances of winning any single game, or 2/6 of the total games won).

I come back to that point: the odds are that you found the prize twice out of six games in your original selection, regardless of what the host did or didn’t do afterwards.

So what happens if you found the prize twice out of six games originally, but the host randomly opens six doors without revealing the prize? Obviously by sticking you will have 2 prizes and 4 empty doors. By swapping you will have 4 prizes and 2 empty doors – the same as Scenario 1.

So we’re agreed that when you first make your first selection, you estimate that you probably have selected 2 prizes and 4 empty doors.

Now the question remains: if subsequently the host randomly opens six remaining empty doors, do you have grounds to revise the first estimate of what you did before the host started opening doors. In other words, do you have reason to say that you originally selected 3 prizes and 3 empty doors, because of what the host happened to do afterwards?

If the answer to the question is no, your original odds still stand. Then you are in the same position as if the host had known where the prize was – and the odds are 2/3 in favour of swapping.

If the answer to the question is yes, you are obliged to set aside the original odds you gave for your first selection.

Therein lies the challenge. If the host knows where the prize is, we’ve established that you have no reason to revise the odds for your first selection. So swap.

Do you have grounds to revise the odds for your first selection because the host is ignorant? Does an ignorant host give you any more information than a knowledgable one?

Aloha Simon-

Forget about 6 games for now… the odds are no different for one or for six. The 12 possibilities are for one game in which the host randomly picks a door.

1. You pick loser A, he picks loser B, you switch- you win
2. You pick loser A, he picks loser B, you don’t switch- you lose

3 and 4. You pick loser A, he picks winner- you lose, or void (this carries the same weight as you pick loser A, him loser B- you just don’t get a choice to switch)

5. You pick loser B, he picks loser A, you switch- you win
6. You pick loser B, he picks loser A, you don’t switch- you lose

7 and 8. You pick loser B, he picks winner- you lose, or void (this counts the same as you pick loser B, him loser A- you just don’t get a choice to switch)

9. You pick winner, he picks loser A- you switch, you lose
10. You pick winner, he picks loser A- you don’t switch, you win

11. You pick winner, he picks loser B- you switch, you lose
12. You pick winner, he picks loser B- you don’t switch, you win

So for any one game, there are 12 possible outcomes. If he randomly picks, there is a 4/12 chance of no option to switch- because he showed the winning door, so you lose, or the game is voided. So 8/12 chances give you the option to switch. Of the 8 possible outcomes of swtiching, 4 win, 4 lose. So you have a 50/50 chance of winning if you switch. Now play the game 6 times if you’d like. Each time, you have a 50/50 chance if you switch, so odds are, you win 3 by switching, lose 3 by switching.

Now, let’s look at one game under your stipulated rules- the host randomly picks and shows you an empty door. That eliminates the possibility of 3,4,7, and 8 above. Under your conditions (random, empty door) there are 8 possible outcomes, 4 in, 4 lose by switching. Your original odds in this game were 4/12 of picking the right door, your conditions you stipulated jump to the next step and eliminate 4/12 of the possible outcomes.

OK, now let’s look at 6 games. If the ignorant host shows you an empty door six games in a row, it doesn’t matter! For each game you have a 50/50 chance of winning if you switch, same if you don’t switch. If the host isn’t ignorant, and always shows one of the loser doors, then you benefit by switching. Look at my response above for the possibilities, then diagram these out on paper.

So let me answer your questions-

“Now the question remains: if subsequently the host randomly opens six remaining empty doors, do you have grounds to revise the first estimate of what you did before the host started opening doors.”

I see what you are getting at- If you probably picked the right door 2 out of 6 times, and the host showed you the other empty door in each game, you still probably picked the right door 2 out of 6 times so you are better off switching! OK… makes sense, but I think I see the flaw. Each game is independent of each other, so the host’s random picks which game up empty do allow you to revise the first estimate. Now you look at it as 6 games of two doors each, so you probably picked 3 out of 6. Or, a better way is to look at it like above- each game is independent, each is 50/50… so whetehr it is one game, 6 games, or a million games…

“In other words, do you have reason to say that you originally selected 3 prizes and 3 empty doors, because of what the host happened to do afterwards?”

Yes, you do. But remember, when he picks nonrandomly, you don’t Confusing!

Very well, I see your twelve possible outcomes refer to one game. That’s now clear.

Curiously what you’ve described is actually six possible combinations between your selection and the host’s elimination. Now each of those six combinations has two ‘sub-combinations’ which depend on whether you stick or swap. Fair enough.

We can also say that where the host has picked a winner, you have no decision to make because you know the outcome – that is why I call it the void scenario.

Out of your twelve outcomes, there are indeed four possible void scenarios. And one thing we’re already agreed on is that when you play each game individually, a random elimination will leave you with 50/50 odds.

What is the justification for those 50/50 odds? It is, in your own words: “the 4/12 chance of no option to switch- because he showed the winning door.”

The odds were still 1/3 that you originally picked the right door. These odds hold true with one game and become the average with repeated playing.

The only factor that can increase your 1/3 chance to 1/2 is the fact that you can expect a third of the games to be voided. This too will be the average. And in order for the 50/50 to work, it is imperative that all games where the host picks a winning door do indeed get voided. Why? Because if they count as games that you lost, you will, on average, lose 1/3 times – whether you stick or swap. If such games are disqualified, you will on average lose 1/2 times with the non-voided games – whether you sick or swap.

So the 4/12 chance of a void scenario is what allows you to say for each game that it makes no difference whether you stick or swap. The odds are 1/2. However if the void scenario were to never happen – as with Monty Hall – then this maintains the integrity of your original odds for your selection – 1/3. Thus 2/3 if you swap.

And so we’re back to my original proposition. You see Honolulu, I didn’t describe a single game in which the 4/12 option could obviously happen.

I deliberately described six games in which the 4/12 option did not happen. Here, your decision whether to stick or swap does not apply to one game at a time in isolation, but to all six games. – and only after the host has made all his random eliminations.

So if you happen, albeit by chance, to get a run of games in which the 4/12 option, so crucial to your argument, never occurs – you are once again left with six games in which you are more likely to have originally picked the prize twice rather than three times. Thus, the host’s ignorance as to where the prizes by itself gives you no reason to abandon your swapping strategy – unless actually accompanied by the expected 4/12 void cases.

Now if there is a flaw in this reasoning, I don’t beleive you have identified it so far. :)

Simon

Honolulu

As I mentioned before, I beleive there is a hidden flaw in the argument, but it has yet to be identified.

That said, I think my proposition may be even clearer, and the mystery even greater, if the puzzle is expressed like this.

There are six rows of three doors. Behind one door in each row is a prize. You select one door from each row.

Enter Host 1, who definitely knows where the prizes are. Out of the doors you didn’t pick, he opens an empty door from each row. He then closes them and leaves. You make a mental note which six doors he opened.

So far, this is classic Monty Hall territory. At this point we’re agreed that if you stick to your selection, you will probably have found two prixes. If you swap to the doors Host 1 didn’t open, you will probably find four prizes.

However there is one more ritual to be performed before you decide.

Enter Host 2, who definitely does not know where the prizes are. Randomly he reveals exactly the same six doors as Host 1. You have been shown nothing new.

Do you now estimate that you originally chose three prizes instead of two and conclude that will make no difference whether you stick or swap? Or do you go with your first assessment – 2 prizes if you stick, 4 prizes if you swap?

Best
Simon

Very good game, although im confused as to why the percentages of wins for swappers / non-swappers adds to 110%?

They represent 65% of swappers winning and 45% of non-swappers winning. 2 different groups.

One point that’s interesting is that while the per centage of swappers who win is near as dammit bang on right, the percentage of stickers who win is nearly 12% higher than it should be. Why is that?

@nick…

you’ll notice the total for swappers is about twice that of non-swappers, in general as the number of trials increases, the accuracy will also increase.

So by the time the number of non-swappers gets to 4000 it should be closer to the expected result.

:)

Ah yes. I see that. thank you Vince.

Ive not followed all the earlier posts in fine detail, but the simplest explanation I have come across follows the same principles but changes the details slightly. Suppose that, instead of there being just 3 doors, there are one million! As before, just one of them conceals the prize. The odds of selecting the right door are one in a million. Crucial (as noted above) is that the gameshow host knows which that door is. He opens all the other doors, leaving one shut which will also be wrong only if the initial selection (one in a millon chance!) is right. Do you stick or switch?

The priniciple is exactly the same when there are only three doors. Only the odds are different.

I only heard this problem for the first time today, so I don’t really know what I’k talking about, but I have a scenario for you.

Suppose we are told that Monty opens Door B to reveal a goat. We are not told which door the contestant has previously chosen.

So based on this information alone, we know that there is a 50/50 chance of the money being in either A or C.

Now lets assume that the contestent has chosen Door A. The conditions have not changed from above and we have established the 50/50 chance, so why does he suddenly have a reason to change just because we know he chose Door A initially?

This is really bugging me.

Cheers

Hi Chris. Where your scenario falls down is that while we don’t know what the contestant chose, the contestant does.

What’s key is the power and knowledge of the gameshow host. He both knows which is the prize window and, *if the prize window comes under his domain,* will offer it to the contestant to switch to. That is more likely to happen than not, because the contestant is more likely to randomly choose either of the two empty windows than the sole prize window. Then the remaining empty window and prize window will both pass to the gameshow host, who will obligingly eliminate the empty window, leaving the contestant only to switch to land the prize.

But if the contestant chooses the prize window with his first guess, the gameshow host is powerless to do other than offer an empty window for the contestant to switch to from his original, correct, choice.

The first scenario is two-in-three likely to happen. The second scenario is one-in-three likely to happen.

So it makes sense to switch. QED?

There must be something wrong. There are now nearly 4000 non-swappers and still a win-rate of approx. 42%. It should be nearer to the 33% by now…

I found an error in the code.
When the door with the prize is the first one selected, the code will always select a certain one of the latter two. This means that if this one is _not_ selected, you will _know_ that the prize is there.
Confusing?
To illustrate:
Select door number 3. If the prize is behind door 2 or door 3, door 1 will be opened. This means that IF door 2 is opened you will know for a fact, that the prize is behind door 1.
The code should select one of the remaining doors randomly if the game should be ‘fair’.
Does this matter?
Yep!
If you are a non-swapper you would of course still swap when you know for a fact where the prize is. This would not count in the non-swap statistics since you swapped. This happens for all games where the prize is under door 1 and you choose door three (lets just say that the user always chooses door three – the user may do that).
We are left with the games where the prize is behind door 2 or 3 – 50% chance that you chose correctly the first time.
This means that using this ‘cheat’ it should be possible to pull the non-swap statistics toward 50% and certainly explains why it is greater than 33% which is the theoretically correct percentage…
Just my 42 cents…

…similarly – if you only swap when you _know_ where the prize is, this will be pulled towards a 100%…
So you can improve the 66.6% vs. 33.3% towards 100% vd. 50% – that’s quite a cheat!

why I can nerver get prize when I choose a door among 3 doors?

if that simulator assign the prize to ‘choosen door’ or ‘not removed door’,

is it meaningless?

sorry for little english. I’m not native.

If you are a swapper, you will win whenever your inital guess is wrong. So you will likely win 2/3 of the time. Draw a picture. Case closed.

Do what John Losse suggests and draw a picture – it helps!
A=Loser; B=Loser; C=Winner

Choose A:-
-and stick = Lose
-and swap = Win

Choose B:-
-and stick = Lose
-and swap = Win

Choose C:-
-and stick = Win
-and swap = Lose

Stick = Win 1 out of 3 times
Swap = Win 2 out of 3 times

For all those to visit here in the future, visit the website http://www.montyhallproblem.com/ for even more explanation(s) The way I see it, on your first pick, you only have 33% of being right, so when the other remaining losing door is revealed, of course you will then pick the winner more often, got it now?

This problem reveals the limitations of statistics:

From the point of view of the programme’s accountants, the statistics matter – that people are twice as likely to pick the wrong door affects how many cars they have to budget for. (This figure is of course affected also by how many people will swap and how that affects the final number of ‘rights’).

So, over multiple games, the statistics matter and their implications matter – TO THE ACCOUNTANTS.

BUT to a contestant, the statistics are irrelevant: s/he has one single go: the fact that most contestants will gain by swopping is of no interest at all to ONE contestant. The single contestant is given the information that a particular door is non-car, thus effectively taking it out of the choice range for all but the loony. This, in a single-shot game, with no invitation to return for the next statistically-valid number of weeks, leaves the contestant 2 doors to choose from, a 50:50 chance. If s/he is statistically clued-up and knows that swopping statistically improves the chance of picking a car to 67:33, and then gets it wrong, the fact of having chucked away a car for the extra ‘third’(67 minus 50 = 17 = about a third of 50) of a chance of being right is not going to be a consolation, so a contestant who wants to remain sane SHOULD NOT SWOP.

So now I understand why it’s 2/3 winning chance.

By switching, instead of getting only 1 prize, you will now get 2 prizes. It follows that the chance of a car in those 2 prizes is 2/3.

The Ace of Spade simulation explained in the Wikipedia shows this clearly.

Tom,

There is only one prize (per contestant that is). You never get two

prizes and even if you did the chance of getting one prize and not the

other would be 1:2 not 2:3.

I think what you may have meant was that you effectively get 2 chances at

the prize but this is not the case because you can open one door only in

the end.

The essential thing (as everyone has said) is that the host HAS to open

one of the two empty doors and this is what causes the phenomenon.

Interesting to also ponder the question: “There is at least one boy in a two-child family. What is the probability that the other is a girl?” The answer is 2/3!!

(B,B) (G,B) (B,G)

The facts you have provided can not be used to back up your theory.

if there are two door’s one with a priz and one with out the proberbilty of picking the one with the prize will always be 50/50. you can not prove this over wise by the number of times you do this!

there is nothing to say that if i do this 100 times the prize is going to be under each door the same number of times! there for making your test Inaccurate, incorrect and usless!

Doug, you didn’t really show much capacity for rationale thinking.

Basically the most illustrative way to do this is rather than think of 3 doors think of a million. Imagine I tell you I’m thinking of a number between one and a million. You guess 77. I then tell you the correct number is either 77 or 1048 and I will give you the option of staying with 77 or switching to 1048. If you think your not better of switching your guess, you’re a fool. (ie the odds are not even close to 50/50)

Your phrase, simply charm
There is nothing to tell – keep silent not to litter a theme.
I think, that you are not right. Let’s discuss it. Write to me in PM, we will talk.
.. Seldom.. It is possible to tell, this :) exception to the rules
I like this phrase :)

It doesn’t work if the game show host doesn’t give you the choice to switch after you’ve already chosen the wrong door.