Horse …
water …
… cannot …
Right, one of the 1/3 chances is now gone. The remaining two go up to 1/2. At least *someone *is on my side. 
Great username / post combo ! :rolleyes:
A little rude… when come back, bring water.
It is amusing that the debate is between the two options that are not the way far ahead winner.
And I misspoke here, it actually is 33% and 33%. Because in the first scenario two things must be true.
(1)You and the other contestant had won the car between the two of you
(2)you must be the one to remain standing
Sorry.
Here’s a way to look at the problem which should lead very methodically to the correct answer.
Monty rolls a six-sided die. If 1 or 2, he places the car behind curtain #1; 3 or 4, #2, 5 or 6, #3. He asks the two players to pick different curtains. If one picks the car he opens the other’s. If the car is behind neither selected curtain, he opens yours if the die was even, his if odd.
(1) Do you agree this is a fair restatement of the problem?
(2) The chance that the die was 4 is 1/6. If you ask ‘Was the die 3?’ and receive the truthful answer ‘No’, do you agree that the chance has increased to 1/5 ?
(3) List exactly what happens in each of the six cases defined by (1).
A variant problem:
I’ve chosen a card from a deck. You think of a card.
Now fifty other people also choose a card. 51 of the 52 cards have been chosen.
What are the chances that y’all HAVEN’T chosen the right card among you?
1/52, right?
Now I’m going to reveal that 50 of you are losers, showing you that your card is wrong. One player is left over, and that player may switch to the unchosen card, or stick with theirs.
IF you’re that left-over player, do you stay or switch?
There’s a 51/52 chance that the winning card was among the chosen cards, but I just proved that if it was in that group, it had to be yours, since all the others were wrong. There’s only a 1/52 chance that the unchosen card was correct.
Stick with your choice, given the fact that you’re still standing.
For example, let’s say you were player number 18, and among the remaining cards, you chose the five of diamonds. The remaining players choose all the cards except for the two of hearts. I reveal that it’s definitely not any of the other chosen cards, but it might be yours, or it might be the two of hearts.
It’s probably yours.
If anyone wants to try a game like this, let’s model it: I’ll think of a number between one and ten inclusive for each “game”. Pretend you’re nine people and make nine different guesses as to what the number is. I’ll definitely eliminate eight of the guesses. The remaining “person” can stay or switch. It’ll look like this:
I predict that in 9/10 of these games, the non-eliminated player will win by sticking with their number, and that similarly, in the OP’s example, in 2/3 of the games, the non-eliminated player will win by sticking with their number.
FWIW, I got it wrong when I first read the problem, and appreciate the clear explanations in the thread.
So this is one of those problems where several different solutions seem to make sense on the surface. I admit I’m not usually that smart when it comes to probability math, but I can’t find a logic flaw in my previous reasoning. Which one of these statements is incorrect, and what is the explanation of the error in logic?
- If 2 contestants pick 2 of the 3 doors, then 66.7% of the time one of them will win
- If statement #1 is correct, then in a scenario where no switching is allowed, Monty removes a contestant, leaving 1 contestant, and the unchosen door, and 66.7 % of the time it will be revealed that the remaining contestant is a winner.
- If statement #2 is correct, then 33.3% of the time, the prize is behind the unchosen door, and thus if the remaining contestant had switched every time, he would be a winner 33.3% of the time.
- Not switching ever, it should be expected that the remaining contestant wins 66.7% of the time. Switching always, he will win 33% of the time, and switching sometimes will result in an expected win % between 33.3% and 66.7%. Therefore the best chance to win is to never switch
Renaming curtains A, B, C to avoid number confusion.
I choose curtain A, Bob chooses curtain B.
According to what Monty rolled, these things happen:
- Monty opens BOb’s curtain; I chose the car.
- Monty opens BOb’s curtain; I chose the car.
- Monty opens my curtain; Bob chose the car.
- Monty opens my curtain; Bob chose the car.
- Monty opens Bob’s curtain; I didn’t choose the car.
- Monty opens my curtain; Bob didn’t choose the car.
Do you agree?
Now, the problem further states that Monty just opened Bob’s curtain. That gives me three possibilities as to what Monty rolled:
- Monty opens BOb’s curtain; I chose the car.
- Monty opens BOb’s curtain; I chose the car.
- Monty opens Bob’s curtain; I didn’t choose the car.
Note that in two of those three possibilities, I was the one that chose the car. I win. Only if Monty rolled 5 will I win by switching.
Correct. The simplest way to see this is in the Spoiler I posted earlier.
The chance for curtain #3 was 33.3% to start with and hasn’t changed. The combined chance for curtains #1 and #2 was 66.7% to start with, and is still 66.7%. Monty’s action has changed that 66.7% from a (33,33) split to a (67,0) split.
The card proof works with the original Monty Hall, not with this one. You just revealed that 50 of us are losers. But you were not allowed to touch that 52nd card. It has a 1/52 chance. My card has a 1/52 chance. The fact that I remain standing tells me nothing! Any one of us could have remained standing. One of us HAD to remain standing.
This is where I disagree with you.
The phrasing of the question implies you *are *the one left standing. (“Should this contestant switch?” clearly refers to the person left).
Condition 2 has already happened. Monty’s asking you to stay or switch. The eliminated player never gets that choice.
The probability of a flipped coin coming up heads is 50%, but the probability of a coin that was already flipped and came up heads showing heads is 100%.
The problem is that “I” am not the remaining contestant. I am one of the original contestants, unchangeable. Maybe I remain, maybe I don’t.
This is completely different from the classic Monty Hall Problem.
In this case, 2/3 of the time, somebody picked the door with the Big Deal, and since the probability of showing a contestant-chosen door that was not the Big Deal is 100%, this does not affect the probability that somebody chose the Big Deal - it only reduces the possible contestants that did it from two to one.
Let B be the Big Deal, X be Consolation Prize 1, and Y be Consolation Prize 2.
There are six possible selections for (First Player, Second Player, Neither):
X, A, B - A is shown
X, B, A - B is shown
A, X, B - A is shown
B, X, A - B is shown
A, B, X - either A or B can be shown
B, A, X - either B or A can be shown
In the first four, switching is bad; in the last two, switching is good.
The main difference between this and the Monty Hall problem is, the probability of the Big Deal being chosen is 1/3 in the original and 2/3 in this version.
BTW, in reality, (a) the contestants were never given the option to switch, and (b) the door with the cheapest prize was always revealed first, even if nobody chose it (“You took 1, and you took 2, and nobody took 3; let’s see what’s behind Door Number 3”).
OK. but somebody has to be left. The question is, does being the one left give you any useful hint on where the car is. I continue to say no.
OK let’s see:
A. Scenarios at start:
- I chose car, Bob has nothing, unchosen door has nothing
- I have nothing, Bob has nothing, unchosen door has car
- I chose nothing, Bob has car, unchosen door has nothing
All three are equally likely. Correct?
B. When Monty opens Bob’s door that means that (3) above is eliminated. It cannot have been true, because otherwise Monty would have opened my door and I’d be out of the contest. So now we have only two choices left. Since each one is equally likely (see (A)), my chances of winning the car are now 1/2. If I switch, they stay 1/2.
QED
Not really sure why people are having a problem with this.
I may as well mention the loosely related problem from the game of contract bridge.
You and your partner (the dummy) have nine hearts missing onlly the Queen, Jack, Trey and Deuce. You cash the Ace; LHO plays a small card, RHO one of the honors (say the Jack for definiteness). You lead toward dummy’s King, LHO following with the other small card. Should you play for the Queen to be with LHO or RHO?
There are only two possibilities that remain for the original holdings:
[indent]LHO: 3 2 RHO: Q J
LHO: Q 3 2 RHO: J[/INDENT]
The initial chances for these two cases were about the same. (1st case was slightly more likely than 2nd.)
What is the percentage play now?
Folly, try it out. I’ve got a deck of cards in front of me, for real, and I just chose one at random. I want you to make 51 different guesses as to what it might be. Of your 51 guesses, I’ll eliminate 50.
(To make it easier, instead of writing 51 cards, you can just tell me which one you’re declining to guess, e.g., say, “I’m guessing every card except the 6 of diamonds”)
Seriously, dude, LET’S DO THIS!
Hey, Folly! Good news: The two of us are on the show together. Let’s make a deal: If either of us wins, we’ll split the car. You get the front half, I get the back half. Sounds good? Cool, let’s work together on this.
OK, there are three doors. We get to pick two of them. Awesome, that gives us a 2/3 chance of getting the car.
Now, Monty opens my door… bummer, there’s a goat behind it. Ah, well, no matter. Still your door to go. Sounds to me like we still have out 2/3 chance. Agreed? Tell you what, though, I’ve been thinking about this. If you win the car, just keep it. I don’t really want half a car anyway.
Now tell me how this is different from the scenario in the OP.