1/3 + 1/6 is 1/2, not 2/3.
I’m getting really confused by some people explaining the original problem, some people explaining the new problem, and some people explaining what they think the new problem is.
PLAY THE FUCKING GAME! Do you tell him:
- Switch! He’ll switch.
- It doesn’t matter! He’ll switch.
- Stay! He’ll stay.
Argue afterward, but first, PLAY THE FUCKING GAME!
Fine, me and 50 others choose all cards but the Ace of Spades. We are holding our cards. You are Monty Hall. You have a sticky note in your hand. The sticky note says The winning card is “…” We are all looking down at our cards. *
- We move north. We are eaten by a grue.
Oh, I feel dumb.
I wrote a simulation (C#) . Inspect it and see if I wrote anything wrong in the model:
class Program
{
static void Main(string[] args)
{
int[] totalCurtainsHoldingGoodPrize = new int[3];
int[] myGuesses = new int[3];
int wins = 0;
int losses = 0;
int seed = (int) DateTime.Now.Ticks % int.MaxValue;
Random randomInts = new Random(seed);
for (int simulation = 0; simulation < 1000000; simulation++)
{
int goodPrize = randomInts.Next((int)3);
totalCurtainsHoldingGoodPrize[goodPrize]++; // sanity check - make sure each curtain gets the prize a fair amount
int theirGuess = randomInts.Next((int)3);
int myGuess = randomInts.Next((int)3);
while (theirGuess == myGuess)
{
myGuess = randomInts.Next((int)3);
}
myGuesses[myGuess]++; // another sanity check
if (theirGuess == goodPrize)
{
// Monty shows my curtain. I don't get to guess
continue;
}
else if (myGuess == goodPrize)
{
// Monty shows me their guess. I'm going to stay pat and WIN!
wins++;
}
else
{
// we both picked wrong.
// Monty shows one of us the other person's, at random.
int montyEliminatesThem = randomInts.Next((int)2);
if (montyEliminatesThem == 1)
{
// I'm going to stay pat on mine and lose.
losses++;
}
else
{
// Monty eliminated me. Another push.
continue;
}
}
}
System.Diagnostics.Debug.WriteLine("The prizes were distributed like this: {0}, {1}, {2}", totalCurtainsHoldingGoodPrize[0], totalCurtainsHoldingGoodPrize[1], totalCurtainsHoldingGoodPrize[2]);
System.Diagnostics.Debug.WriteLine("My guesses were distributed like this: {0}, {1}, {2}", myGuesses[0], myGuesses[1], myGuesses[2]);
System.Diagnostics.Debug.WriteLine("I won {0}, and I lost {1}", wins, losses);
float percent = (float) wins / (float) (losses+wins);
System.Diagnostics.Debug.WriteLine("That's {0} percent", percent*100);
}
}
Here are the results (one million runs):
The prizes were distributed like this: 334035, 333505, 332460
My guesses were distributed like this: 334049, 332764, 333187
I won 332990, and I lost 166330
That’s 66.6887 percent.
So, yes, I’m back to ‘it’s better to stay put.’
Well, by pick, I meant pick you for “the deal”, not pick your curtain to show you already lost. It seems like you understand it now anyway, thanks to zut’s excellent explanation.
In the OP’s scenario:
If the unpicked door has the car, Monty opens 1 door (which has no car).
If your door has the car, Monty opens 1 door (which has no car).
For the scale-up:
If the unpicked door has the car, Monty opens 98 doors (with no car).
If your door has the car, Monty opens 98 doors (with no car).
Seems to work for me.
THANK YOU!
Okay. All of y’all, stand up. Hold on, lady up front holding the King of Clubs? You stay there. Everyone else, move north.
CRUNCH CRUNCH CRUNCH
BURP
Yikes.
Thanks for your patience, lady holding the King of Clubs. You may choose to stick with your King of Clubs, or you may choose to switch to the Ace of Spades. If you end up with the right card, YOU WIN! If you end up with the wrong one, you’re moving north, too.
Folly, that lady grabs your hand. Do you tell her to switch or to stay put? Keep in mind that if you tell her it doesn’t matter, she’s gonna switch. If she loses, you’re going north with her.
[edit: I just noticed that you included yourself in that group of 51, which would mean you’re getting eaten anyway. Who cares, that lady is really nice, what advice are you gonna give her before you go North?]
Except in the scale-up you said:
If the unpicked door has the car, that’s gonna result in 99 doors opening.
Sorry, you’re right: I meant to say all picked doors with goats, except your door. You get to keep standing either way, or there’s no game.
Right.
Let’s say that I pick door 1, and the douchebags pick doors 2-99, and the car is behind door 100. How many doors does Monty open?
I tell the lady, you were either chosen at random (good thing, what with the grues) because the sticky note has Ace of Spades written on it. Or the Stick note has King of Clubs written and on it and therefore you were not chosen at random.
Then I ask the lady. Do you think you were chosen at random or not chosen at random. And what do you think are the probabilities of each scenario?
She panics at your babble and switches, because you didn’t tell her anything useful.
OH NO! KING OF CLUBS WAS RIGHT! Seriously, when I drew a card at random, that’s the card that came up. Your poor understanding of the odds led to her gruesome demise.
Want to try again? If your understanding is correct, and we try this game ten times, and each time you try to Socratic method the poor schmuck instead of giving them advice, half the time their decision to switch will lead to salvation.
Let’s go: this time, what card doesn’t get picked?
Folly, that guy with the nine of spades is watching. What do you tell him: switch to the six of spades, stay, or “how do you feel”? I already know which card is right.
Damn it. You had time to quote me before my edit. I do hate it when that happens. Yes, I made a slight boo-boo at first. But the revised scale-up still works.
LHOD – are you saying you should stick or switch? Also, are you saying that it doesn’t matter that sometimes a given contestant will be eliminated before being asked?
I don’t know the answer, but it seems that since there are times when a given contestant will be eliminated that the odds should change from the original Monty Hall problem.
If 2 doors are chosen, the odds that one will be a winner is 2 in 3
ok.
Now add a stipulation- 2 doors will be chosen. After they are chosen, one door that is a) one of the chosen doors, and b) not the winning door, will be eliminated.
knowing this will happen, does it change the odds that before it happens, there is a 2 in 3 chance that a winning door is selected with the initial 2 choices? No.
Ok, so play it out. The doors are chosen. One of the choices that is incorrect is eliminated, as stipulated. The odds that the remaining choice is the winning door is still 2 in 3. The odds therefore that the other door is the winner is 1 in 3. It doesn’t matter if it’s you or me or some random guy, whoever is in that position of being the remaining contestant has a 2/3 chance of winning if he doesn’t switch, and a 1/3 chance if he does switch.
Yes. Half the time, she will lose by switching. So? Half the time she’ll win by switching.
Unless of course Bup’s code up there proves me completely wrong. I’m not a coder, can’t follow it.
Yeah, that’s the correction I was saying was necessary. Sorry about the premature quoting! 
Absolutely on the “stick” side.
No, it absolutely matters. The fact that Monty eliminated someone (or, with the cards, many someones) besides you is giving you additional information.
Try the card game if you’d like. You’ve got 51 minions. Which card don’t you have them pick?
No, that’s the *hypothesis *(of yours). It remains to be tested if she wins half the time if you do this ten times.
LET’S GO, I SAID!
Try it again. First time, maybe you’re right: fifty fifty chance switching was gonna kill her. Maybe my hypothesis, that you doomed her to a 98% chance of death, is wrong.
But if you try it ten times, I’ll be a tiny bit surprised if one person lives by switching, significantly surprised if two live, and completely flabbergasted if five do.
So you’ve done it once. The nine of spades guy is asking for advice. Go ahead, tell him to switch, see what happens if you think it doesn’t matter! Then try again! Again! Again!