The solution

[yendi]

Warning: This is the solution as presented by a liberal-arts person who hasn't taken a math class in over twelve years. So I may have left gaps unfilled.



Okay, let's go over the one fact that's not likely to be disputed:

After picking your initial box (which we'll call Box A), you have a 1/3 (or roughly 33.3%) chance of having picked the $5000.

That means, of course, that there is a 2/3 or ~67% chance of that money being in one of Box B or Box C.

Now, remember (as trochee emphasized), the Prizemaster knows where the money is.

He removes a box (we'll call it Box B) that was known not to contain any money, and never did.

Box A still retains a 33% chance of being the correct box. That's because it was picked before the state of Box B was known.

But remember -- the chances of the money being in one of the other two boxes were 2/3. Box B is out of the way. That means that the odds for the money being in Box C (which is, of course, "one of Box B or Box C" remain 2/3.

Find this troubling?

The argument that most folks have with this is best articulated by force_of_will in this comment and the follow up. And had the Prizemaster been choosing whether to remove a box, or if the PM was choosing randomly and happened to pick an empty one, this logic would be correct.

Let's pretend there are 100 boxes. You pick one. Then the Prizemaster, knowing where the money is, removes 98 of the remaining boxes which contain no money. He again offers you the chance to switch. Still think the box in your hand, at 1%, is the way to go?

(that last example is swiped from Raymond Smullyan, btw)

Comments

if I haven't done this already...

[dopple.livejournal.com]

You have been judged significantly geeky to join de (http://www.livejournal.com/userinfo.bml?user=multi_genre_fan)Multi-Genre Fandom (http://www.livejournal.com/community/multi_genre_fan/). Come on over and bask in de dorky afterglow!

Re: if I haven't done this already...

[yendi]

Woohoo! Dorky afterglow!

[mfree.livejournal.com]

If the prizemaster has knowingly removed boxes containing no money, and you know now of two boxes, one that you are holding and one that is on the table, and either could contain money, how is it that the odds are not .5?

[yendi]

Because the prizemaster removes the empty boxes after you've made your initial 1/3 (or 1/100) choice.

[trochee.livejournal.com]

The Smullyan example doesn't do it for ya?

[mfree.livejournal.com]

But that's irrelevant... there are two boxes to choose from, and you *know* one of the two contains the money because the removed boxes were shown to be empty.

[photognome.livejournal.com]

(polishes fingernails) its hard work being this right all the time...... ;-)

[yendi]

But the two boxes are not equal. Box #1 (the one you already chose) was chosen with different information available, and before the empties had been removed.

[mfree.livejournal.com]

No, because once the removed boxes are shown to have no money they are removed from the pile.

The box you are holding is no longer one of 100 unknowns, it's one of two unknowns and 98 knowns. Knowns aren't considered as chance.

[mfree.livejournal.com]

But this is an entirely different question, the second choice. It's two problems sequentially, you're just shown to be in the right on the first one.

The *total* odds aren't 1/2, but the odds in question, whether you have a box with money or not, is 1/2. two boxes in question, one contains money.

[terracinque.livejournal.com]

Oh right, right! I remembered wrong. What you have above is what my friend and I demonstrated.

[theferrett.livejournal.com]

Not particularly, no. He has removed 98 boxes, none of which has a prize. All that proves to me is that the money box is now either my box or his. That's 50%.

There's obviously a greater chance after he removes the box in the initial best-of-ninety-nine that one of the remaining boxes is the money getter, but I fail to see how this translates to, "So the odds of the money being in his box are greater" when you're down to a choice of two.

I've tried to understand this logic before, but it feels like trying to understand quantum physics. It may be right, but I'm not convinced. If Stephen Hawking says yes, I'll believe it.

[mfree.livejournal.com]

Holy shit, I think my eyes just opened. The initial boxes *are* relevant.

Ok, you grabbed one of 100, you have a 1% chance of holding the money.

Forget numbers at this point.

Your holding has a 1% chance of containing money. His holding has a 99% chance of containing money. he narrows the field down and gives you an offer, your holding or his. His holding is still 99%, and yours is 1%. The logical thing to do then is switch, right?

----

Using three boxes, you're at 33% and he's at 66%. There's still a 66% chance that his holding contains money, not 50%.

Damn, it's been so long my logic centers are smoking from all the burning dust.

[beowabbit]

The little piece of information that’s important is that it’s not random that the prizemaster is picking an empty box. If the prizemaster picked one of the remaining two boxes at random (with a 1:3 chance of revealing the prize), then everybody’s first intuitions about this problem would be correct.

[yendi]

*dingdingding*

Exactly!

[mfree.livejournal.com]

Scratch this, I see my error now.

[shawnj.livejournal.com]

Try it for yourself. (http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html)

[yendi]

If I ever see him, I'll ask him. I suspect he'd agree.

[mfree.livejournal.com]

May be easier to think of it this way, as not removing the empty boxes, but shrinking them and stuffing them into one box, and then offering to switch.

[robyn-ma.livejournal.com]

Or there's my solution:

You hit the Prizemaster in the head with a broom, open all the boxes, take the money, and run like hell. :)

[mfree.livejournal.com]

Sweet. I gave myself a sample size of 20 per attempt. Switching got me 14 wins and keeping got me 5. Looks good to me :)

[force-of-will.livejournal.com]

It "works" that way as long as you play from the point of view of the "player". The Prizemaster has a whole different problem. Then there is the perspective outside of the game...

If you believe that you get to "keep" that 1% chance, you should choose to switch boxes. If you believe that you get a whole new problem, it doesn't matter to switch. For the Prizemasters part, he knows that you'll never get to keep the 1% chance, you'll wind up having to choose from one of two boxes with everything that came before being thrown out.

I don't believe that you get to keep the 1% chance packed up in the box. It was had, is gone, and you were never going to get that particular shot to win the prize because the Prizemaster throws it out. You were always going to get a choice that he gives you in the end, one of two boxes.

The problem reminds me of those sorts of riddles that give you a lot of extraneous things only to confuse you with them.

Will

[lubedpumpkin.livejournal.com]

OHH now I get it.

*deep breath*

[brujah.livejournal.com]

My. Head. Just. Exploded. (thud)

[yendi]

Thinking outside the box there. Or inside the boxes, as the case might be.