Explain to me this probability puzzle

Jericho, your mistake is assuming that the first choice is irrelevent to the second. It's only irrelevent if you make your second choice randomly, but there's no need to make a random guess the second time round. There is a strategy that can be applied. (Just a note, I teach math, and I've used this exact problem before when teaching probability.)

The first guess has a 2/3 chance of being wrong. One wrong possibility (never the one you chose, and never the correct one) is then removed. Given that you probably chose wrong the first time, you can use that knowledge to influence your second choice. The door you didn't choose and Monte doesn't eliminate has a 2/3 chance of being the right door. I posted a full breakdown of how this works earlier in the thread, and Pielorinho's 100 choices example illustrates the problem perfectly (it really is exactly the same problem, just with more choices for the first guess).
 

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jerichothebard said:
Ahh... I see.
And you are correct - you have a 2/3 chance of having the full egg in your hand. But in this scenario - YOU'RE MONTY!


THAT'S RIGHT!

Stop right there.

Now, we have one more step. I'm Monty, and you're the guesser. You know, since you just told me this, that I have a 2/3 chance of having the full egg in my hand. Now I give you a chance to change your mind, to choose any of the three eggs.

You can choose the egg on the table. What's the chance that that's the right egg?

You can choose BOTH eggs in my hand, and if EITHER of them is right, then we'll proceed to the endgame. You just told me what the chances were that I was still holding the correct egg--what was it you said? 2/3, right?

Okay, so let's say you choose "the eggs in your hands" over "the eggs on the table." 2/3 of the time, that'll be the right choice, right?

If you're right, then I'll ask you the endgame question: of the two eggs in my hand, which one is the right one: the smashed one, or the whole one?

I'm going to assume that you'll choose the whole one 100% of the time, since obviously the smashed on isn't right. That means 2/3 of the time, you'll win!

As for your playing the game and being right 50% of the time, that baffles me. Do me a favor, wouldja? Play the game 30 times. Each time, choose 1 as your first guess, and then switch to the other one as your second guess. Tell me what you end up with. (I just went back to the game to verify my results and got an error).

Daniel
 

Incidentally, you can try something similar to the egg trick using playing cards.

Three Card Monty Haul
"Monty" holds two jokers and a king, spread out, face down. You choose one card, which the "Monty" person puts face-down on the table, and then the "Monty" person looks at the remaining cards. If either of them is the king, "Monty" turns the remaining joker face-up, showing it to you. If neither of them is the king, "Monty" turns one of the remaining cards at random face-up (Monty should decide before looking at the cards which to turn face-up if neither is the king). Then you choose whether to stay or to switch. You win if, after deciding, you've chosen the king.

I made my tired and cranky wife do this experiment with me, and it worked out exactly as I'd expect: switch every time, and your chances of getting the king increase dramatically. Try it with a friend and see!

Daniel
 
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It doesn't matter whether you keep the door you've choosen or choose the other door. They have the same probablity of being the winning door.

It's all a matter of probablity spaces. Your first choice comes from a probablity space that involves three choices, with one winning choice. The second choice involves an updated probablity space with one winning choice and two choices.

Let A be the probability that you have the winning door before the losing door is shown.
Let B be the event that you have been shown a losing choice.

Before the first door is opened you have a A= 1/3 chance of having the correct door. However, after the door is opened you have a new piece of information B. Thus your sample space is actually A|B (the sample space of A given that event B has occurred.) Event B reduces you to two choices with one correct choice. Thus you have a 1/2 chance of being correct.

Consider it this way. Suppose that you have a second individual walk in after the first wrong door is shown. You ask them to choose a door without them knowing which door you have chosen. They see two doors (C and D). One is the door you have chosen(C) and the other door (D). Each has a probability of 1/2 or being the correct door. If they were to independently choose door C, then they would have a 1/2 probablity of having chosen the correct door. Thus, so should you have a 1/2 proablity of having the correct door.
 

milotha said:
It doesn't matter whether you keep the door you've choosen or choose the other door. They have the same probablity of being the winning door.

Unfortunately, this is incorrect. Look at the "choose a number between 1 and 100" version given earlier. It is the same as the door problem, but with more doors.

The second choice involves an updated probablity space with one winning choice and two choices.

Correction - it is an updated probability space, and information. In effect, Monty gives you information when he tosses one door away. If you use that information, you improve your chances. If you throw away that information (or if Monty can move the car), then the odds come down to 1/2. But if you keep the information, your chances improve.
 

Pielorinho said:
[/i]

THAT'S RIGHT!

Stop right there.

Now, we have one more step. I'm Monty, and you're the guesser. You know, since you just told me this, that I have a 2/3 chance of having the full egg in my hand. Now I give you a chance to change your mind, to choose any of the three eggs.

See, right there is the flaw in your reasoning. There aren't three eggs to choose from. You didn't get what I'm trying to say a bit later - that there are only two real choices now.

You can choose the egg on the table. What's the chance that that's the right egg?

well, there are two viable eggs, one of which is a winner. That's 1/2


You can choose BOTH eggs in my hand, and if EITHER of them is right, then we'll proceed to the endgame. You just told me what the chances were that I was still holding the correct egg--what was it you said? 2/3, right?

Okay, so let's say you choose "the eggs in your hands" over "the eggs on the table." 2/3 of the time, that'll be the right choice, right?

Wrong. There aren't three choices, and you don't have 2 of them anymore. You have one, out of two.

If you're right, then I'll ask you the endgame question: of the two eggs in my hand, which one is the right one: the smashed one, or the whole one?

I'm going to assume that you'll choose the whole one 100% of the time, since obviously the smashed on isn't right. That means 2/3 of the time, you'll win!

As for your playing the game and being right 50% of the time, that baffles me. Do me a favor, wouldja? Play the game 30 times. Each time, choose 1 as your first guess, and then switch to the other one as your second guess. Tell me what you end up with. (I just went back to the game to verify my results and got an error).

Daniel
Results attached.



You're right that there is information gained when the egg is smashed. That information is:

"This egg is not a winner".


You do NOT tell me that the odds of your egg being the winner are 2/3 - why? Because there aren't 3 eggs in the game anymore, and you don't have 2 eggs anymore. You have 1. Out of 2.

I don't know how to make it any more simple.

2 eggs. 1 winner. 1 choice. 50% chance.



When we started the game, each egg has an equal chance of being the winner. That doesn't change because one egg is removed.




The 1 in 100 or 1000000 game as described is a fundamentally different game.

Here's why.

When you smash the egg, you are removing one egg from the game. You are not reducing the choices to two, you are removing one egg. The fact that this reduces the choices to 2 only happens to be a coincidence of the starting number.

The only way these are the same game is if you reduce the number by one in each game. In which case, the chances of being right = 1/2 or 1/99.

In the game Monty Haul played on TV, each choice has an equal chance of being right. No matter how many choices, as long as there is only a reduction of 1 choice at a time, the odds are equal to the number of choices.

You are right that there is information gained if more than one choice is removed at once - because that is a fundamentally different game than the game we are talking about, where you remove one at a time.

basically, he says, "here are 99 more guesses, and all of them are right." He packages all the losing guesses together with the win, which multiplies your odds of winning by switching by 99 times."

It's now a game with 99 right choices and one wrong choice.

You'd be stupid NOT to switch, because Monty gave away the game.

Imagine you start with three choices, and Monty removes two! That's a lot of information!






However, if you were to play the game with 100 doors, and Monty just keeps removing one door at a time and letting you guess all the way down to two doors, the answer still won't change! Your chance of winning = 1/2.



Back to the original question:
Imagine it with four choices, for a more clear example.

Four Doors. I choose #2.

Monty removes #1.

What's left?

#2, #3, #4.


What should I do? switch? stay?

It doesn't matter. Each door has a 1/3 chance of winning! Three doors. One winner. One choice.




See how the game doesn't matter until the second round? No matter what, the first bit of information isn't information at all - it's meaningless! It's just for excitement and drama.

Imagine if they only started with two doors. No drama, just a simple coin toss. BOOR-RING!

But by performing the sleight of hand with the two choices, they turn a simple coin toss into good *ahem* TV.



jtb
 

Attachments


jerichothebard said:
Results attached.

jtb

Honestly, all I can say about your 16/30 results is that they are a statistical anomaly. They do happen sometimes, especially with such small samples as only 30 tries. And 30 really is a small sample.
I've seen this problem demonstrated more times than I can count. I've watched students try it out. I've seen it written up in math texts. Every trial with enough samples bears out the 2/3 probability.
I also just played your game 100 times, in which I switched every single time. I won 66 times, which perfectly holds to the odds of 2/3.
The thing to remember is that the second choice does not have to be random.
 

Jericho, here is the right way to look at it in terms of probability spaces.

You make the initial choice, which by hypothesis has a one in three chance of being correct. Therefore you can divide the relevant probability space into two parts - the door you chose, which has an one in three chance of being the winner, and the doors you did not choose, which have a two in three chance of containing the winner.

Now Monte eliminates one of the two doors in the 2/3 space.

Do you get it now?

(Monte does NOT get to move the car around after you make your initial choice! The only way you could be right is if he did.)
 

http://www.cut-the-knot.org/hall.shtml

http://math.ucsd.edu/~crypto/Monty/montybg.html

Google for "Monty Hall". If you understand german, google for "Ziegenproblem". Try any of the games at least 50 times. The web is full of explanations - and since we are not talking about some obscure theory of black holes or something, but a simple propability application, I see no reason why you should doubt these sites.

Here is what you seem to miss: The situation at the end is dependant on the choice you make at the beginning, since it forces the host into a decision that is different depending on the choice. If you choose the car door, the host can randomly eliminate two doors. But if you choose a goat door, he can only eliminate one door - the other goat door.

20 or 30 tries are nothing for propability purposes. In the first few choices in the game linked below, I got lucky andhad a 50% win quote without switching, but that quickly moved towards 1/3rd.

I have attached an image that comes from here, the site of the university in Osnabrück, that shows the propability tree for the game.
 


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