[Guest]

"Welcome all, welcome all. Thanks for being here today. This is Doors of Wonder, the fantastic new gameshow where everything's made up, and the points don't matter!"

A man in a black jacket appears on screen, handing the flamboyant gameshow host an inconspicuous index card.

"That's right, that's what I said, the fantastic new game show where doors are dollars and choosing wrong is catastrophic!

"So who do we have here tonight? A young girl by the name of Bobby! Mind if I call you Bob? Bob it is!

"So Bob, here's the deal. You have four doors to choose from, and behind one of them is a solid gold Humvee! (Or if you prefer, a diamond studded swimming pool). Which door to you choose?"

"I choose four!" Bob declared immediately, having already made up her mind before the show had begun, and feeling clever for having done so.

"Four it is folks! And behind door number four is... wait, first let me show you what's behind door number one. Door number one... contains nothing! Now Bob, would you like to change the door that you chose?"

Bob was about to shout out "No way!" but then one of her friends in the audience yelled out faster "Yes, do it Bobby!"

So Bobby did. She chose door number two.

Then the gameshow host congratulated her, telling her that door number four was in fact NOT the gold Humvee door. He asks her whether she wants to change her door again, and with the help of her friend, Bob chooses door number three.

So far, this situation should seem very familiar, with the only difference being four doors instead of three.

However, there's two strange rules this particular gameshow has:

Any statement that the host makes has a 1/3 chance of being a lie.

Any time during the game, Bob can press a button, and this button will tell her whether the door she is currently choosing is the Humvee door or not. However, she can only use this button once.

Assuming that Bob always makes the best choice (I guess her friend is a "perfect logician," or perhaps a perfect statistician), what is the probability of her winning the Humvee?

Also, you can assume that the host will always say "Door number X... contains nothing!" and he will always say this twice. Also, this is how the host works: first he learns whether or not he needs to lie, then if he finds he has to tell the truth, he chooses randomly between the remaining doors. If he finds he has to lie, he will of course choose the Humvee door. However, if Bob is currently choosing the correct door, then he has to tell the truth.

I don't know the solution (I've only calculated her winchance without either of the conditions, and her winchance with the first condition, and I don't feel like doing any more right now)

GL HF!

[Guest]

it would be best to stick with you origanal answer

if you pick 4 then you have a 1/4 chance of winning if he shows u a door then u have a 1/2 chance of winng or 2/4

if u pick from the 3 remianing doors you only have a 1/3 chance of winning

If he shows yu another door then u have a 3/4 chance ove wining instead of a 1/2

[Guest]

it would be best to stick with you origanal answer

if you pick 4 then you have a 1/4 chance of winning if he shows u a door then u have a 1/2 chance of winng or 2/4

if u pick from the 3 remianing doors you only have a 1/3 chance of winning

If he shows yu another door then u have a 3/4 chance ove wining instead of a 1/2

First off, you should use spoilers.

Secondly, I don't follow that all, and I doubt that it's correct, but I don't know.

In my original post I said I'd figured it out, with the first condition (1/3 chance of lying without the button), but I was wrong.

I'm getting that if you change twice, you have a 1/3 chance of winning. If you just stick with your answer, I don't see it changing from 1/4. Neither of the other two possibilities (switch/stay, stay/switch) are likely to be greater than 1/3, considering there's only 5/12 left.

Btw, I figure that if the host selects the same door twice in a row, like "Door 1 is not it!" then after the first decision, "Door 1 is not it!", then Door 1 must be it, and you select Door 1 to win (so if you're planning to switch twice, but then this happens, you modify your plan). This makes the scenarios a bit more complicated (but made it a bit simpler in one case when I got my 1/3 answer).

My 1/3 answer is basically

1/4(2/3) +

3/4(1/3(1/3)) +

3/4(2/3(1/2(2/3(1/2))))

And I realize that makes no sense, but if you go through possibilities maybe you'll end up with the same equation.

I haven't even bothered to try throwing in the button yet...

[Guest]

Just to clarify DN, the host SAYS it is not in door number X but does NOT actually open the door and show what is behind it.

If I understand correctly then the host picking a door has no effect on the outcome. Let's say you pick number 1 then 1/4 of the time you are right and regardless of whether the host has to tell the truth or lie he will pick 1 of the other 3 randomly and 3/4 of the time you will pick wrong leaving 3 doors, but 1/3 of the time he will lie and point to the correct door and 2/3 of the time he will randomly pick from the remaining 2 doors. So the host will equally likely pick any of the 3 remaining doors. Since the answer from the host make no change in the probability for this round, the next guess (stay or switch) is just like starting over and hence same analysis applies and therefore in the end you have only a 1/4 chance of being right.

[Guest]

@Moshe, yes, he SAYS but does not SHOW

Just to be clear, everyone understands me when I say that if you ignore the two conditions (lying and button), and just do the 4 doors problem, Bob has a 5/8 chance of winning if she changes twice?

I can explain this if necessary

There is a 1/4 chance the first door you choose happens to be correct

If this happens, the host must tell the truth. Bob then switches. If the host gets the 1/3 chance of lying, he will eliminate the true answer, and Bob will not choose it. If the host gets the 2/3 chance, he will eliminate the one remaining, and Bob will switch back to her first answer, and get it right.

The chance of this is 1/6, and I also got two other win possibilities, both 1/12, which all add up to 1/3.

But I just realized!

If this happens, the host must tell the truth. Bob then switches. If the host gets the 1/3 chance of lying, he will eliminate the true answer, and Bob will not choose it. If the host gets the 2/3 chance, will he really eliminate the one remaining? Or is it a 50/50 chance, with the other possibility being the elimination of the door Bob is currently choosing?

Dangit, I'm confusing myself with my own problem

[Guest]

OK, I just re-evaluated and realized I overlooked the obvious. I still contend the 1st round was correct but after that things do change.

If you stay then switch:

1/4 (picked correct) host picks random, host picks random: Switch:=0%

3/4 (picked wrong) 2/3 host picks wrong, 2/3 host picks wrong: Switch:=100%

2/3 host picks wrong, 1/3 host picks right: Switch:0%

1/3 host picks right, 2/3 host picks wrong: Switch:0%

1/3 host picks right, 1/3 host picks right again: Switch to host's pick: 100%

net=5/12 right (fyi if you stay, stay except if the host picks the same 2xs then you have 4/12)

If you switch, switch

1/4 (picked correct) host picks random, switch to a wrong answer, 1/3 host picks right, switch:0%

2/3 host picks wrong, switch (1st pick) = 100%

3/4 (picked wrong) 2/3 host picks wrong, 1/2 switch wrong, 2/3 host picks wrong (1st pick): Switch:=100%

1/3 host picks right: Switch (1st pick)=0%

2/3 host picks wrong, 1/2 switch right, 2/3 host picks wrong (1/2 1st pick, 1/2 last pick): Switch:0%

1/3 host picks right: switch =0%

1/3 host picks right, switch wrong, 2/3 host picks wrong: Switch:0%

1/3 host picks right, switch wrong, 1/3 host picks right again: Switch to host's pick: 100%

net=5/12 right (fyi switch,stay would be 3/12)

I will consider the button later.

[Guest]

Dr Moshe: I agree with your first assessment that the hosts pick makes absolutely no difference whatsoever on the chances (if Bob hasn't used the button). I think I desagree with your second assessment that the second pick does make a difference. Did you take into account to possibility that the host might pick the same door two times in a row?

Anyhoo, if the hosts pick really has no bearing on anything at first then Bob should push the button immediately after picking the original door before the host has said anything.

Say Bob picks door 1. In 3/4 of the games the car isn't in Bob's door so we get:

1: 0

2: 1/3

3: 1/3

4: 1/3

Say the host picks Door number 2, then using Bayes Theorem:

Probability of 2 chosen if 2 is correct: 1/3

Probability of 2 being correct: 1/3

Probability of 2 being opened: 1/3

So 2 remains at 1/3.

Probability of 2 chosen if 3 is correct: 2/3 * 0.5

Probability of 3 being correct: 1/3

Probability of 2 being opened: 1/3

So 3 (and 4) remain at 1/3 also. The only logical thing for Bob to do is switch to 23 or 4 (Doesn't matter which one.)

Say she picks 2, and the host picks 3:

Probability of 3 chosen if 2 is correct: 0.5

Probability of 2 being correct: 1/3

Probability of 3 being opened: 0.5

So 2 remains at 1/3.

Probability of 3 chosen if 3 is correct: 1/3

Probability of 3 being correct: 1/3

Probability of 3 being opened: 0.5

So 3 becomes 2/9.

Probability of 3 chosen if 4 is correct: 2/3

Probability of 4 being correct: 1/3

Probability of 3 being opened: 0.5

So 4 becomes 4/9.

In 3/4 games Bob has a 4/9 chance of winning.

But in 1/4 of the games Bob finds out the Prize is in her door before we even begin and has a 100% chance of winning.

So, on average, Bob wins 7 times out of 12.

[Guest]

Based on reading Darth's second post, I believe the host will treat any door HE points to as being "opened" and out of play except with the correct door he has no choice but to select it (as long as you haven't selected it) when he is forced to lie.

As for your analysis, maybe I'm missing something but you don't factor in the chance that the HOST will choose door 1 even though Bob knows it isn't there, nothing says the HOST knows or honors the result of the button. I guess the game would be frustrating if the HOST could/did pick it even though you know it isn't in there.

HOWEVER, based on my 1st statement (host only pick wrong doors not picked before), if the second statement is true (host can't pick door after the button says it isn't there), then we have a problem:

Bob picks 1

Host picks 2

Bob uses button and 1 is wrong

Bob switches to 3

If 4 is the Humvee and Host must tell truth, then Host has NO DOORS to pick (door 1 out from button, door 2 was alread out from host, door 3 you have selected and door 4 is the prize).

If I don't use the button right away, I believe I can get the odds up to ... (see spoiler)

I am using a switch switch strategy:

pick correct on 1st guess = 1/4 Host picks wrong door 1, You pick wrong door, now host picks other/last wrong door 2/3 and picks right door 1/3 so

right 2/12 and wrong 1/12

pick wrong on 1st guess = 3/4 Host picks wrong 2/3, you pick right 1/2, Host must pick wrong 1 = you're wrong 1/4 ******see later******

you pick wrong 1/2, host pick wrong 2/3 you're right 2/12

host pick right 1/3 you're wrong 1/12

Host picks right 1/3, you pick wrong 1 , host picks wrong 2/3 you're wrong 2/12

Host picks right 1/3 but then you know he picked the right one so this is the only case you don't switch to the unselected door, but rather select the same as the host and you're right 1/12

SO: you are right 5/12 and wrong 7/12, BUT if you use the button on your second choice, then the line marked above (*****see later*****) becomes a win when you realize you have selected the correct door and the 1/4 wrong becomes 1/4 right and changes your 5/12 win to 8/12 or 2/3 win !!!

[Guest]

As for your analysis, maybe I'm missing something but you don't factor in the chance that the HOST will choose door 1 even though Bob knows it isn't there, nothing says the HOST knows or honors the result of the button. I guess the game would be frustrating if the HOST could/did pick it even though you know it isn't in there.

Ah! Of course you're right. Given that, then once again the choice of the host makes no difference on the probabilities whatsoever. Poor Bob is stuck at a 50/50 chance on average.

Which means that the riddle is a bit trivial unless you are correct in assuming the host will not pick the same door twice. I'll wait for the OP to clear up what happens in the situation you described before working on this further.