Explain to me this probability puzzle

[Umbran]

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

No, it isn't. Because the full egg cannot move.

After your first choice, there's a 1/3 chance you have the right egg. There's a 2/3 chance that Monty has it.

If Monty paints one egg red, there's still a 2/3 chance that he has the right egg. If he writes "Wrong egg" on one, there's still a 2/3 chance that he has the right egg. If he smashes one, there's still a 2/3 chance that he has the right one. What Monty does with the egg is irrelevant.

This is not quantum mechanics or something, where the probabilities mysteriously reshuffle evenly over all the possibilities. The eggs are physical macro objects. The probabilities are fixed on those objects, and the probabilities only change when the eggs move. If Monty had a 2/3 chance of having the right egg, he will always have that chance until one of the eggs changes ownership, or he reveals the true egg.

 

[Pielorinho]

Jericho, you say you don't know how to make it any simpler. The problem is, you're oversimplifying the question: you need to make it more complex, so you don't lose information in the course of the simplification.

I know you believe you're reasoning it through correctly, but this is a well-known mathematical problem, well-known because at first blush it flies in the face of intuition. If you ask anyone with a 4-year-degree in mathematics, they should be able to confirm the solution as it's been given to you in this thread: amongst mathematicians, the answer to this puzzle is uncontroversial (assuming, as we are, that Monty knows the location of the prize, and that his revealing of one door is affected by his knowledge).

If you are consistently getting 1/2 results, then I suspect you're not doing the switch the way I'm doing it. I played it by choosing door #1 to begin with every single time, and then switching to the other door every single time. As expected, I won 2/3 of the time, and lost only when the prize started behind door #1, as it did 1/3 of the time.

Try it that way, maybe: start always with door #1, and always switch on your second guess to the other door. Try it, I dunno, 100 times, and see what your results are.

Daniel

 

[Doug McCrae]

To Umbran and Jericho: Play the simulators, keep playing them. Make your own, whatever. Eventually you'll find that switching wins two thirds of the time.

 

[Doug McCrae]

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.

No. Door #2 still only has a 1 in 4 chance of being the prize. There's a 3 in 4 chance that the prize is behind door #3 or #4, so if you choose either #3 or #4 you've got a 3 in 8 chance of winning or 37.5%, better than the 25% chance if you stick with your original choice.

Even with a hundred doors and Monty only revealing one you're still better off switching but only very, very slightly.

 

[Umbran]

To Umbran and Jericho: Play the simulators, keep playing them. Make your own, whatever. Eventually you'll find that switching wins two thirds of the time.

Uh, I accepted that switching wins back before I played the numbers game (I did it not to be convinced, but because a straight man was needed for the exercise). If you read my latest post above, you'll note that I'm arguing for the 2/3, not the 1/2.

 

[Thotas]

I first ran across this one in Marilyn vos Savants column (the one and only reason to touch a copy of Parade magazine in the paper). It's amazing how stubborn the 50/50 contingent is on this one when the actual test is so easy to do.

Of course, I also saw someone write into said column asking how it can be weather casters will say there's a 25% chance of rain on a given day. After all, either it rains or it doesn't; that's two possibilities so shouldn't the odds always be 50/50?

 

[Hypersmurf]

Of course, I also saw someone write into said column asking how it can be weather casters will say there's a 25% chance of rain on a given day. After all, either it rains or it doesn't; that's two possibilities so shouldn't the odds always be 50/50?

Damn it, you beat me to it!

It gets better, though.

Will it rain between noon and 1pm? Well, it will or it won't; that's two possibilities, so a 50% chance.

Will it rain between 1pm and 2pm? It will or it won't; that's two possibilities, so a 50% chance.

Will it rain between 2pm and 3pm? It will or it won't; that's two possibilities, so a 50% chance.

So, obviously, there's a 150% chance that it will rain between noon and 3pm, and your odds of picking the right door by switching are 50%.

Simple, really

-Hyp.

 

[Thanee]

When looking at the second choice only, the chance is, of course, 1/2 for either door. But that is completely irrelevant for this problem, since you are not looking at the second choice only. You have information from the first choice already, which is still valid.

In the first choice it was only 1/3 to get the right door out of three, so you have picked one of the wrong doors with a 2/3 chance, which doesn't change, just because one of the wrong doors is being opened. It still is a 2/3 chance to have picked a wrong door, which makes the other door a 1/3 chance to lose only, since the remaining probability does now completely rest upon that door, as there is no other choice remaining.

So picking the other door gives you better odds, since it turns the situation upside down, since the 1/3 chance to lose of the other door equals a 2/3 chance to win.

Bye
Thanee

 

[jerichothebard]

OK - I concede the point, based on both all the excellent explanations, and also some empirical results.


I modified my script to play the game automatically. Instead of asking you to choose a door, and then to switch or stay, it now asks you to choose a strategy - switch or hold - and a number of repetitions.

Then I ran the simulation 10000 times each way. I suspect we can all agree that 20000 total is a statistically valid trial.

The results are consistently within 100 wins of 2/3 for switching, 1/3 for holding - or, in other words, a 1% margin of error.

Very counter-intuitive, but ya can't argue with consistent results!


If you are interested, try it here:
http://www.maestrakara.com/srjosh/autogame.php


jtb

PS - I set the script to only allow 1000 at a time - please don't abuse my server!

 

[Pielorinho]

If it makes you feel any better, you conceded the point a lot faster than I did, jericho: it was years after I first heard this conundrum that I accepted its truth, and years more before I understood WHY it was true (as I said earlier in this thread, it was the hundred-doors explanation that finally worked for me). Your scripts are things of beauty, and I'll hang onto this thread to show them to other folks in case this puzzle ever comes up again.

Daniel