Let us try this again with a slight change to the rules. This is another riddle that arose during our family weekend gathering where conundrums are always part of the fun. If you wish to post hints or spoilers that is OK, just please use the spoiler box. Read the riddle carefully and then vote before perusing any of the responses.
On the old game show Let’s Make a Deal, the host Monty Hall would select two of the show’s winners to give up their prizes to try and win “the big deal of the day.” Once these two players were chosen, Monty would show these contestants 3 doors. Behind two of the doors were average so -so prizes. But behind one door was “the big deal” which was often a car, a boat, or some such expensive fabulous prize.
One contestant gets to select a door first and the second chooses from the remaining two doors. At this point of the game Monty always opens a selected door and shows one of the contestants that, alas, he/she won only a so -so prize. There is polite applause.
Monty then turns to the other contestant and before revealing the prize behind their chosen door, he always gives the remaining player an option. That player can switch his/her door for the other unchosen one and instead take that hidden prize.
All I ask is that you carefully read the riddle first and then vote before piling on. This is a public poll btw. And vote before reading responses after this one.
The way you have changed the problem does change the outcome, yes.
You made it doors when Let’s Make a Deal used CURTAINS.
[Spoiler]The crux of the original problem was that 2/3 of the time, Monty was showing you the only prize he could - 2/3 of the time you had picked the wrong curtain, and he’ll never show you the big prize, so he has only one curtain he can show you. 2/3 of the time, by process of elimination, he’s “shown” you the good prize.
Now, though, he always has a contestant who’s wrong. The other contestant has a 50/50 shot of having picked the correct curtain in the first place.
So switch or don’t, it doesn’t make a difference.[/spoiler]
I think this is no longer the “classic monty hall problem” and that Bup has the right of it.
The key being that now Monty has to choose to reveal one of the two doors chosen by the contestants. Since this is so, you gain no new information about the third door like you do in the “classic monty hall problem” [/scarequotes]
Superficially, it seems like a difference. But when you think about it, it really doesn’t affect the odds.
If there’s one prize and two contestants then it’s certain that at least one of them will pick a losing door. Monty knows where the prize is so he can always pick a contestant who chose a losing door at his first reveal. As far as the remaining contestant is concerned, the game is no different than if he was playing alone and Monty had opened an unchosen door.
Well, no, I think (and I answered the poll wrong).
At the start the chance the contestant picked the right door was 1/3. Because Monty did not pick HIS door to open, his chances immediately improve to 1/2. Then, whether he switches or not, they stay at 1/2.
The difference between this and the original Monty Hall thing is that in the original Monty could never open the contestant’s door after the first choosing.
I am interested to see how the poll turns out. Should be fun.
[Spoiler]I suppose I should have warned everyone to not assume anything…though in my defense I did ask everyone to read the riddle before replying. And my sister did not give us any warning…so fair is fair.
Please use the spoiler box and…assume nothing… Heh-heh:D[/spoiler]
So far, I’m the only person advocating standing pat. Here’s why:
The priors are: you and the other contestant have each selected a door of the three doors, and they’re different doors. Monty is going to select a door that one of you picked, that isn’t the big prize.
[spoiler]Chances are 2/3 that one of you picked the right door, and only 1/3 that neither of you did.
If one of you picked the right door, and Monty’s shown what’s behind the other guy’s door, then you’ve won. Stay put.
If neither of you picked the right door, then Monty could have picked either of your doors to show. So starting from the beginning, there’s 1/6 chance that neither of you picked the right door, and Monty showed your door, which didn’t happen. And 1/6 chance that neither of you picked the right door, and he showed the other guy’s door, which did.
So your chance of already having the right door is (2/3)/(2/3 + 1/6) = 0.8. Stand pat!
[/spoiler]
And yes, this is different from the Monty Hall problem as described in Cecil’s column way back.
I am getting ready to head home from the weekend vacation in Montreat, North Carolina. It is a 7 hour drive so I will not be able to follow this thread for a while. Tomorrow after work I will post the correct answer to this riddle… or what I believe is the correct answer anyway. Feel free to contradict. I suspect there will be consensus by that time, though. Maybe.
I only wish I had put a warning at the start that the riddle was a variation so that those who posted quickly would have had time to examine their answer and either stick by it or change based on their further thoughts. I apologize. I certainly have enjoyed the discussion here today. There is still fresh juice to be squeezed out of what seems a tired old puzzle.
I think the problem is that you are assuming a 50/50 chance that Monty picks your door or the other door, but it’s not. Monty knows where the big prize is. I’m sticking with it doesn’t matter if you switch or not.
