Hang on - i thought you should only switch when the host *knew* which one had the $100 bill. To quote wikipedia: "If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch"

Good call. So let's say I do know, and Carrie was not removed "randomly" but because I knew she was one of the two losers. The question for you and Bob remains.

As James pointed out, the question is not worded correctly. The host needs to purposefully open an empty from the set of the other two contestant's envelopes, otherwise there's no advantage to swapping envelopes or holding on to yours, as everything has equal probability.

If this is what you meant, then it is in your advantage to get your hands on Bob's envelope and it's in Bob's advantage to keep his envelope. So whoever gets to make the last swap will have the best odds.

I show both a pictoral example and empirical data in this blog entry: scottonwriting.net/.../13534.aspx

Thanks

The interesting thing about the Monty Hall problem (and this one as a consequence) is that it relies on the host's knowledge. What if the host had no idea where the car was, or perhaps even any idea of which door the contestant had chosen, but *just happened* to pick one the contestant hadn't chosen which *just happened* to not be the winner? In this case, the odds would not be increased, as others have mentioned, so if you're the contestant you're only gaining valuable knowledge if the host knows something you don't.

I too am somewhat skeptical of any impact a host's knowledge could have in your example.

Regardless of the host's knowledge, you are most likely to choose incorrectly the first time, and the car/money is most likely to be in one of the other two doors/envelopes.

I think the "knowledgeable host" caveat could refer to the stipulation that the host must always uncover an empty door/envelope. If, in cases where a "clueless" host sometimes uncovers the car and the host "wins" and you "lose" -- then your chances overall have been altered. But in a single case where a the host uncovers an empty door/envelope, even if by accident, I contend that all the standard monte hall enigma principles apply.

Yeah, ok, I keep going back and forth on some aspects of this problem. And I know from reading the wikipedia article that the host's knowledge or lack thereof is a point of contention. But let me reverse my prior stance and say that I don't think it matters. As Carl just said, it doesn't matter if the host *knew* the revealed door was a loser, only that he *happened* to reveal a losing door, *this time*. In that case, switching should yield you a 2/3 chance of winning and staying a 1/3.

The problem you've outlined above isn't a conditional probability problem, so the '2 in 3 chance if you switch' thing doesn't apply. You're just distributing three envelopes, and opening one at random, with no constraints. So, having found nothing in one envelope, the odds have simply been narrowed down to 50% for each remaining person, and there's no point in switching - the logic of the original problem doesn't apply here.

It's like having two contestants, both of whom choose a door, then having Monty open the final door to reveal a goat - that's the only door he could have opened, and so if it happens to be a goat then it's by chance. The odds of each remaining door having a car behind them remain the same.

That's why it does matter that Monty knows which doors have goats behind them, and is compelled to choose a goat-concealing door when making his decision. Without this element, the events are independent, and so if a goat is revealed by the host by chance (rather than a car, which has a 1/3 probability) then the subsequent decision to stick or twist makes no difference.

miss point,

If you write the code to simulate monty hall and figure that he doesn't know which door holds the car, or even which of the three doors to open (so he picks one of the three at random to show), but you subsequently throw away the cases in which he picks the contestant's door or the door with the car, then you are still left with results that favor switching 2/3 of the time (try it). By eliminating those cases from consideration, the odds are altered such that they mirror the scenario in which the host has knowledge of where the prize is located.

Likewise, if you had 2 contestants in 2 different rooms unaware of one another, and Monty Hall *happened* to pick the third door (the one neither of them had picked) revealing the goat, then from each's perspective they would be facing the Monty Hall problem, and would seem to have a 2/3 chance of winning by switching. This is the same scenario the envelopes problem describes.

Wow this is a tough one. It's even tougher than my year end exams. You must be a fan of Alfred Hitchcock.

This is a good mystery question. I shall use it on my colleagues and see what their answers are.

I have to agree with ssmith and others who are baffled by the answer that the cluelessnes of the host could affect the probability.

That's concerning the monty fall problem (host slips in a banana and grabs from a door and reveals it accidentally). However the description says (to not make the show be canceled) that it happens that he reveals a goat. It should give more data like what happens if a host reveals the car accidentally. Maybe it assumes that the lucky slip in a banana reveal a goat case is just one run and in other runs the host might reveal a car. But it's not defined what happens then.

I have found (and tried simulating in C code) two cases:

1) If the host accidentally reveals the car the show is canceled and we actually discard that run in the simulation. It goes back to the old problem and the simulation shows 2/3

2) If the host accidentally reveals the car then the game continues and the contestant looses (whether he switches or not he always gets a goat). With that rule in the simulation it gives 1/2 after many runs.

Why say that he randomly slips and grabs a door that happen to be a goat? Why not just state that there are new rules that give the freedom to the host to choose whatever he wants (but he doesn't know what's behind the doors now)?

p.s. I think one reason for this confusing description is that some people needed to give an excuse to mathematicians who wrongly assumed on the very first problem where Marilyn was right. They wanted to describe a slightly different problem (with the same data yet the host chooses randomly but he is lucky to not choose a car) to say that Marilyn had a misleading description of the problem and that they meaned the slight alternative. If they discribed the rule that the host can select the car then it would be a different problem. Although when Marilyn was asked about the random host, she said if the host is clueless yet already has revealed randomly a goat then it's 1/2 either way but if he knows then switch is to your advantage (2/3). Nobody specified a good rule but somehow they wanted to reform the old problem to give a 1/2 in a twisted and misleading kind of way imho (randomly he can chose either car or goat, but he never actually choses car while it's supposed he can chose a car but he doesn't. Ugh! Schroediger's cat anyone?)

p.p.s. About the three envelope problem, interesting. I think I sometimes have some insight of what might be the logical fallacy here but I need to think about it a little.

A little insight (not a proof) on the envelope problem.

Say we have Mike, Bob and Carrie.

Three starting posibilities:

0 = empty envelope

1 = 100$

M B C

1 0 0

0 1 0

0 0 1

Ok, in the case we reveal Carrie with see that only in two of the three rows we could do that. If we discarded the third row and had a challenge between M and B it's clear that it would be 1/2.

Someone might say that why not in the third row reveal someone else rather than Carrie to not reveal the price? Then in one of three cases (and on the third row) we'd have to say reveal Bob. Then in one of three cases we instantly open Bob's envelope which happen to be 0 and so we instantly make him loose and don't give him the chance to make a switch if he wants. This reduces his possibilities. But that would happen with Mike too in another run.

I think now that the important thing is. In Monty Hall, the contestant choses a door and then the host isn't allowed to reveal his own door. He only reveals a goat from the other two. And he always finds goats to reveal. He wouldn't if we restricted him to one door.

The envelope problem would make a 2/3 for Mike if there was a special rule that the person with an empty envelope that he opens must not be Mike. Then Mike gets the advantage to switch and the other of the two who was lucky to not have his envelope opened gets an advantage if he keeps his envelope.

I have some doubts about these thoughts but that would be nicely solved with a simulation. However we should think of the rules of the simulation. Although there is something bugging me when thinking about it..