An old but still useful logic puzzle

[yendi]

This came up in a discussion at work today.

[Poll #403118]

Yes, there is a correct answer to this. It's not a matter of opinion. But a lot of people insist on arguing over it. I'll post an answer in a bit, although I suspect a lot of you (tablesaw, pbmath, jet_li_wannabe, etc) are probably blinking at this, wondering how anyone could not know.

Comments

[trochee.livejournal.com]

The critical piece of the puzzle (though it may not be obvious to readers who don't already know the answer) is obfuscated below:



But I often have trouble convincing people to believe the answer.

[thornsilver]

Does the Prizemaster get to keep the money if you lose?

[yendi]

Of course! :-)

[thornsilver]

Then, I think I know what the right answer is. Except one has to wonder how wiley both players are. :P

[jet-li-wannabe.livejournal.com]

I believe that people don't know. I still make my student's heads spin with this one from time to time.

I recall...

[wishiwasnt.livejournal.com]

...you fighting hard against the right answer when we first discussed it in college.

I got into a huge fight at work when I posed this question. This one guy would not believe it and kept shouting at me. Ah, good times.

[ketzl.livejournal.com]

Ahh it takes me back to probability class, it does.

[tofu-cat.livejournal.com]

Heh, my husband got asked the following question at a job interview:

What is the probability that you would have 2 girls and 2 boys if you had four children.

Amazing, really, the number of people who say 50%.

[thornsilver]

What is the correct probability? I know it's not 50%, but I don't remember how to correctly calculate it. *pout*

[tofu-cat.livejournal.com]

basically, the easy way is to count up all the probabilities:
there are 16 total permutations of children possible, involving boy/girl and 4 total.

there are 6 possible permutations of 2 boys and 2 girls

ergo, 6/16 is the probability.

[force-of-will.livejournal.com]

When you condense the problem mathematically the problem is a loss in the change of state.

This is to say that because he is going to show you an empty box between choices, well, it breaks the problem. Or what I'm saying is, your chance of guessing right changes with the choices made, a situation that the math does not collapse. Or the box that he is going to show you, in effect, does not matter, although the math is going to "count" it. The choice probability in each problem is rendered correctly but the idea that our prizemaster has better knowledge splits the problem. You pick with 1/3 chance, and then one half chance, but, since the choices aren't all yours, it doesn't matter to switch after his revelation. It is his action, not yours which collapses the problem from 1/3 to 1/2, and thus your choice doesn't matter. Once he makes his revelation, and prior to your choice, the odds have switched. His revelation, not your choice. The choice is moot.

Will

[force-of-will.livejournal.com]

Gads what a horrible explnation.

Basically his given option to choose forces you to choose and, coupled with the fact that he has a perfect opportunity to show you an empty box, collapses the initial 1/3 option. The initial part of the problem disappears and you are forced to choose between one of two boxes at a 50% chance. Or you don't get the option of keeping your 1/3 chance...

Will

[jmfunnyface.livejournal.com]

I've heard/seen this riddle before and the answer I've been given was that you stick with the one you picked initially. I don't remember where I've seen it or the logic behind it, so I can't back it up.

[force-of-will.livejournal.com]

As I remember it the solution is that the first choice is at 1/3 and the second at 1/2 so that you always switch boxes at the second opportunity for a 1/2 chance is better than a 1/3 chance.

My belief is that the fallacy is that the events are dependant, but they aren't. Your first choice is at 1 of 3 but his action doesn't allow you to "keep" that choice. His action presents basically a whole new problem at 50%.

If you look at the problem from the frontside with his certain option to show you an empty box it becomes clearer. In the end the only option you really have is to select one of two boxes.

Wiil

[slipjig.livejournal.com]

But here's the thing:

After your initial choice, you have a 1/3 chance of being right. Nothing the Master does will change the fact that you will have initially chosen the right box 1/3 third of the time, and the wrong box 2/3 of the time.

Now he's done his thing, and there's only two boxes in front of you. So what happens if you swap?

If you had originally chosen the right box, you'd switch to a wrong one. If you'd originally chosen a wrong box, you'd switch to the right one, the only one that's left.

In other words, if you were right before, now you're wrong. And if you were wrong before, now you're right.

Do you see where I'm going with this? Originally, you were right 1/3 of the time, and wrong 2/3 of the time. But when you switch boxes after he removes a wrong one, the odds flip-flop: you switch over to the wrong one 1/3 of the time, but switch over to the right one 2/3 of the time.

Therefore, you should switch, becuase you double your chances in doing so.

(I was an it-doesn't-matter guy for ages, and could not be dissuaded, until my dad chaged my mind in five seconds flat.)

In fairness

[wishiwasnt.livejournal.com]

Very few people get this right the first time they hear it. I've never met anyone who has, though I'm sure such persons exist. This problem is designed to start fights.

Re: In fairness

[yendi]

*nod* I freely admit I got it wrong the first time, and probably the second. I think it wasn't until I read Smullyan's Satan, Cantor, and Infinity, that it finally took.

[ian-gunn.livejournal.com]

I got this wrong the first time I ran into it. I finally banged my head on it till the correct answer made sense to me. I'm so tempted to explain it now but I'll sit on my hands and get back to work.

[lubedpumpkin.livejournal.com]

Ugh, I'm confused. I want to say it doesn't matter.. because at that point, you have a 50% chance with either box you pick.. right? Right? Bleh. Mathy things escape me..

[terracinque.livejournal.com]

Wrong. You have a two-thirds chance if you stay with your original box, but only one-third if you switch.

[lubedpumpkin.livejournal.com]

Ohh I get it. The probably changes when you switch, but if you stay its still the original probability? Maybe?

[beowabbit]

To be fair, there is a tiny little piece of information that’s not quite explicit in the question, although it’s obvious to anybody who’s watched game shows.

And now that I’ve made this harder and more frustrating for everybody else, I’ll bow and walk offstage. :-)

[trochee.livejournal.com]

uh, it's that the puzzlemaster always does this, regardless of which one you pick, right?

that is what you had in mind?

[terracinque.livejournal.com]

My friend electricrocket and I proved this experimentally several years ago, with a small ball and three coffee cups.

[trochee.livejournal.com]

you proved experimentally that you have a 2/3 chance if you stay?

okay, now I'm confused and want to go try it.

[terracinque.livejournal.com]

Umm, it was a few years ago, but I think what was proved was that you win a third of the time more often if you stick.

[yendi]

I'd say that as experiements go, this one proved about as much as Guildenstern's coin flips did.

[trochee.livejournal.com]

Spoiler below:

I am pretty convinced that you win 2/3 of the time when you switch, because you the contestant are taking advantage of the host's knowledge.

Switching becomes a choice between the cup I picked and the right one -- if one of them is -- of the other two.

Thus switching is 2/3, sticking is 1/3. But maybe I'll write a simulation and do it.

[terracinque.livejournal.com]

Whoa, my bad! I remembered it backwards. What we proved by experiment is just as you report there.

[trochee.livejournal.com]

w00t!

Now I don't have to write a simulation.

But I'll just include the first few lines of the text file I was aiming for: capitalization reflects the prize, brackets indicate my choice. 1 means you win the money, 0 means you lose.

        after PM  switch   stay
[A]b c   [A] b      0       1
[a]B c   [a] B      1       0
[a]b C   [a] C      1       0
 A[b]c    A [b]     1       0
 a[B]c    a [B]     0       1
 a[b]C   [b] C      1       0

[photognome.livejournal.com]

I know, I know!

[xforge.livejournal.com]

I was wondering when I'd see the day that someone had Avatar for an avatar.

[photognome.livejournal.com]

:-)

tattoo as well!

http://www.livejournal.com/users/photognome/105923.html?mode=reply

Ya stick with your choice.

[softlywhispered.livejournal.com]

If the PM knows, then he's going to obfuscate and offer you the one without the money in it. Human instinct. Of course, I'm probably wrong.

[chiara607.livejournal.com]

...and I'm just staring at this, going "What the hell?" Math makes my brain hurt.

[unwilly.livejournal.com]

This is not a 1/3 to 1/2 or 1/3 to 2/3 question for me, but a question of real world motivation.

Since the PrizeMaster knows which box has the money, I must question his motivation.

IF he wants me to win the money, than by showing me the empty box, and allowing me to choose again, THEN he wants me to take the remaining box, wining the money.

IF he wants me not to win the money, than by showing me the empty box, and allowing me to choose again, he is trying to make me change my mind about my choice, and I already have the money, thus changing my mind loses me the money.

Being the cynical person I am, I think by offering me the second choice he is trying to make me lose the money, thus I will keep the box I first picked. If he really wanted me to win the money, and he knew I picked right the first time, he never would have offered me the second choice.

I'll keep the first box.

[yendi]

*nod* One problem with any strict logic puzzle is that it's affected, in the real world, but knowledge of that puzzle. Although without cheating, there's only so much that the PM could do to affect things.

[unwilly.livejournal.com]

It's all a head game to me. The PM is just trying to plant the seed of doubt in the person's head, hoping that they will change their mind.

[napalmmk9.livejournal.com]

"but it has worked, don't you see? You've given everything away! I know where the poison is!"

[trouvera.livejournal.com]

Thank you!! That is EXACTLY what I've been thinking!!

[unwilly.livejournal.com]

AHHAHAHAHA, AHAHAHAHAH, thud.

[amokk]

God, I hate this one. Every time I see it. Everyone thinks it goes from 1/3 chance to 50/50 and it doesn't.

[lovingboth]

Without looking at other people's comments :)

Two possibilities: you've got the money box or you haven't.

If you have (1/3 chance), you'd be foolish to swap the box - you're certain to lose.

If you haven't (2/3 chance), if you swap the box, you're certain to get the money box (because you're not going to pick the one you've just been shown is empty!)

So you swap, because 2/3 is greater than 1/3.

[yendi]

Bingo!

[dadandgirl.livejournal.com]

Wrong! Everybody seems to be missing the obvious here. There are two choices in the puzzle as presented. The first choice is "Pick one of three boxes", but it's a meaningless choice because of the secondary stage of the puzzle.

Assuming that the rules of the puzzle are fixed, and the puzzlemaster will remove an empty box whether your original choice is correct or not, the first decision becomes moot. The second choice is "Pick one of two boxes", which is a 50% probability.