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

[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.

[yendi]

*dingdingding*

Exactly!

[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.

[thryn.livejournal.com]

I've got to try this with cups and a ball, 'cause I *still* can't wrap my brain around it. I understand the math, but I still can't see it playing out in practice.

[lubedpumpkin.livejournal.com]

OHH now I get it.

*deep breath*

[cscottd.livejournal.com]

*LIGHT-BULB*

I just got it.

After reading yendi's solution post, I realised what the answer was, and I even knew (more or less) why it was the answer, but I didn't truly get it, until I read your comment.

When you applied the statistics to the two groups of boxes (yours vs the Prizemasters), that's when it all became crystal clear to me.

Thanks!

[clawfoot.livejournal.com]

I don't get it.

Okay, so if you choose one box out of 100, you've got a 1% chance of guessing right. I got that part.

And when the Prizemaster reveals 98 boxes to be empty, his holdings still have a 99% chance, and you still have a 1% chance. I got that part too.

But what happens if the Prizemaster then gives you all 98 empty boxes? And then asks if you want to trade all 99 boxes for his one box? Do you now have a 99% chance of having the right box? Or does his one box still have that 99% chance, and your 99 boxes still only have 1%?

Or do all 98 empty boxes disappear in a puff of logic, and you're left with the 50/50 chance?

*wince* My brain's all melty.

[mfree.livejournal.com]

Well, you're obviously not going to pick an empty box, but this is how that works. He had a 99% chance of having the money in his boxes. He had 99 boxes.

He removes a box, he still has a 99% chance of having the money, just in 98 boxes.

Remove another box, 99%, 97 boxes.

Get down to one box. 99% chance the money is in that box. Forget whee all the empties went, it's unimportant. It's your holdings versus his, and his holding is 99% sure to contain the money.

[clawfoot.livejournal.com]

OOOOOOOOOOOOoooooooooooooooooooooh.


Okay.

:)

[demetria23.livejournal.com]

Ahhh! OK, that's the comment that finally made me get it.

[mfree.livejournal.com]

*snicker* this is the one thing I'm good at.... make *me* understand, and I'll make sure everyone else does too :)

[shawnj.livejournal.com]

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

[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 :)

[gutman.livejournal.com]

I finally get it!
Sorry Yendi, but the web page explanation is a lot clearer.

[thryn.livejournal.com]

I did each 50 times.

Switching got me 31 wins.
Not switching got me 28.

The math isn't holding up. Maybe I'm just lucky.

[thryn.livejournal.com]

Larger sample size got me way closer to probability.

101 of each:

Not switching, 37 wins. Switching, 68 wins.

I'm still beating the odds by a smidge, though. ;)

[yendi]

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