Then what is incorrect with post 49?
No, actually I’m just making a typo. It’s better not to switch 2/3 of the time. Sorry about that.
This is a subtle error that is common to many of these parlor-game type probability questions. It basically comes down to this: in scenario 1, Monty Hall has to pick you. In scenario 2, there’s only a 50/50 chance that he picks you. So the odds of scenario 1 given that he picked you, are 2/3. Stay with your original pick. It matters.
Simple. Post 49 discusses the probability of any one of the contestants winning - either you or the other one. The question in the OP is about the probability of YOU winning AFTER the other one was eliminated.
Just for fun I haven’t read the solutions offered by others in this thread … so forgive me if my answer is obviously shown to be false.
I reason that it is better not to switch. Here’s why.
There are two contestants picking from three doors. Between the two of them, they have a 2/3 change at the big prize.
Monty then shows them that one of those two did not win. That means that there is a 2/3 chance that the other fellow already picked the winner.
The additional information provided by Monty was which one of those two was not the winner.
The remaining guy has a 2/3 chance if he stays, and only a 1/3 chance if he switches. He (or she) has effectively been given two chances to pick - his or her own, and the other contestant’s.
I’ll play your way.
51 people choose everything but the 6 of spades (I chose one of those cards, but I’m not saying which).
I don’t care which one you are. But I have a question. You there–in the back–dude that chose the nine of spades?
Hi there, dude. Stand up.
Everybody else, SIDDOWN. You lose! Your cards are not right! I spit on you!
Okay, guy with the nine of spades. I’m going to offer you a choice, You saw that all the other choices were wrong and those people are pitiful losers. Either yours is right, or the lone remaining card, the six of spades, is right. What do you want to do?
So are you saying that in this scenario that after eliminating 50 players the lone player left has a 50:50 chance against Monty? And that it doesn’t matter if he switches cards or not?
Run that scenario 100 times and see what you come up with. I say the lone player wins better than 95% of the time by keeping his card.
Do explain again. In scenario 1, Monty has to pick Bob, 100%. In scenario 2, Monty has to pick Bob, 100%. In scenario 3. Monty has to pick me, 100% and I am eliminated.
No, post 49 discusses the probability of the remaining contestant winning when he switches vs. when he doesn’t switch, which is precisely what the OP is asking. So which statement in post 49 is incorrect?
When Monty opens Bob’s door, that tells you that scenario (3) is eliminated. That’s an extra piece of information you didn’t have before. But it also tells you something about the relative probabilities of scenarios (1) and (2).
In scenario (1), Monty will *always *choose Bob’s door. In scenario (2), he has the *option *of choosing either your door or Bob’s. So, let’s renumber:
- You chose car, Bob has nothing, unchosen door has nothing, Monty chooses Bob’s door
2a. You have nothing, Bob has nothing, unchosen door has car, Monty chooses Bob’s door
2b. You have nothing, Bob has nothing, unchosen door has car, Monty chooses your door - You chose nothing, Bob has car, unchosen door has nothing, Monty chooses your door
Scenarios 1, 2, and 3 all occur with 1/3 probability, but scenario 2 is further split by Monty’s random choice. So we’ve actually eliminated both scenario (3), and scenario (2b), leaving only (1) and (2a).
Staying is the right choice for the contestant giving the choice (2/3 vs. 1/3).
One assumption this is based on, that is not stated in the problem, is that the remaining contestant knows which door was chosen by the other contestant.
I think the possibility that neither contestant knows for sure which door the other contestant chose still fits within Biotop’s problem statement. For example, suppose the 1st contestant whispers their choice in Monty’s ear, 2nd contestant is told to pick a door, and is told to pick again only if they choose the same door as the 1st contestant. There is also the possibility that neither contestant knows if they are the 1st or the 2nd chooser (unless Monty has to tell them to choose again) which (I think) would only matter if neither contestant knows which door the other chose.
Note the phrasing: “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.” that could be more clearly written “At this point of the game Monty always opens a selected door and shows both of the contestants that one of them has won only a so -so prize.” (emphasis added)
While bup asks his dude holding the 9 of spades if he wants to switch, of course, I’ll play the same game with anyone else. Hampshire, wanna let your minions try their luck?
Thank you. This makes it clear. Yes, one should stay. 2/3 chance - 1/3+1/6 vs 1/3
Like with the original Monty Hall problem, I think this can be more easily visualized by scaling it up.
You’re on the show with 98 of your closest friends. Or 98 douchebags, doesn’t matter. You are shown 100 doors. 99 have a goat, 1 has a car.
You all pick a door. 99 doors picked, 1 unpicked.
Monty then opens all the doors for the contestants who picked doors with goats. As it happens, this eliminates the 98 douchebags. Two doors left. Yours, and the unpicked one.
Still want to switch? Still think it doesn’t matter?
Yes, the chance of being the last contestant standing wasn’t great. But now you are. And that is useful information indeed.
That doesn’t quite work as a scale-up, because on the off-chance that the unpicked door has the car, you’d have Monty open all the contestants’ doors.
Doesn’t matter! why is my card better than the lone remaining card!
What’s the difference between us? Neither of us was allowed to be eliminated.
The six of spades wasn’t eliminated because the rules didn’t allow it (Note: key difference from original monty hall problem). The 9 of spades wasn’t allowed to be eliminated because we are throwing out every result where we aren’t the last man standing.
no it’s not. It’s really not.
I notice that you’re still refusing to play, EVEN THOUGH THAT’S THE BEST WAY TO TRY IT OUT.
There are two possibilities:
- The six of spades–the card not chosen–was the correct card. If that’s what just happened, then I selected a second card at random and told the dude holding it to stand up and offered him the choice. (I could have offered it to anyone, but it just so happens that I randomly selected the poor loser holding the nine of spades to give a choice to).
- The six of spades–the card not chosen–was an incorrect card. If that’s what just happened, then I told the person holding the correct card to stand up and offered him a choice. (I couldn’t, under these circumstances, have offered the choice to anyone else, since the rest of them were losers).
That dude is turning to you, panic in his eyes. He’s planning on switching, because he’s vaguely aware of the Monty Hall problem, but if you tell him to stay, he’ll do it–and if he wins, he’s giving you half of the tropical island he stands to win. WHAT DO YOU TELL HIM?
If you tell him it doesn’t matter, he’ll switch.
PLAY THE GAME!
Either advise that 9 of spades dude, or corral your own minions and select 51 cards for a new game. Just trust me long enough to try it–you can argue after the game is over, to your heart’s content, but please just try it.
Now we’re getting somewhere. Why are the probabilities of your two possibilities different? Like you said, there is nothing special about the person selected.