OK, smarty-pants, what's the best move?

[boost]

I know I've heard this before, and I remember, "you should always switch." But no matter what, I cannot understand why.

The logic seems like picking tails, because last the last coin flip resulted in heads.

You have a 1/3 chance to pick the right box. Once you make your initial pick, one of the, or the, empty box is revealed and removed. You now get to chose between the two remaining boxes.

What information has been revealed about either of the remaining boxes by removing the empty one from play? The initial pick was left in because it was the prize, or because it was empty. The alternative box, the one not revealed to be empty was left in because it is either the prize or it is empty.

The revealed box is always empty, so the 2/3's chance of containing the money has collapsed into one box.

wait. wat. why?

[givememyleg]

I know I've heard this before, and I remember, "you should always switch." But no matter what, I cannot understand why.

you have to think of it in two separate parts.

1) 1 out of 3 boxes is $300. 2 out of 3 boxes is $0.

2) 1 out of 2 boxes is $300. 1 out of 2 boxes is $0.

if you have your choice between the two, obviously you would pick #2.

The first "coin toss" was three side, whereas the next one is two sided.

Also, if it helps, try replace 3 boxes with 300 boxes. Then go through the problem again:

- Choose 1 box out of 300
- Of the remaining 299 boxes, 298 boxes are revealed as $0
- Do you stick with your box (at 1/300) or pick the new box (at 1/2)?

[jackvance]

you have to think of it in two separate parts.

1) 1 out of 3 boxes is $300. 2 out of 3 boxes is $0.

2) 1 out of 2 boxes is $300. 1 out of 2 boxes is $0.

if you have your choice between the two, obviously you would pick #2.

The first "coin toss" was three side, whereas the next one is two sided.

Also, if it helps, try replace 3 boxes with 300 boxes. Then go through the problem again:

- Choose 1 box out of 300
- Of the remaining 299 boxes, 298 boxes are revealed as $0
- Do you stick with your box (at 1/300) or pick the new box (at 1/2)?

Math is still wrong here. It's still a 3-side cointoss, but now two of them net you $300. You basically get to choose between your first choice, or 'both of the others'.

And in the example of 300 boxes, you go from 1/300 to 299/300 chance.

[MadMojoMonkey]

You had a 3/1 shot the first time. now its 2/1.

It is not 50/50 on the 2nd choice; it's 33/67. This is a common mis-assessment of the effect of the revealed information.

In this case, the revealed box is ALWAYS empty. There is 0% chance the revealed box contains the money.

And in the example of 300 boxes, you go from 1/300 to 299/300 chance.

This is correct. You do NOT go to 50/50 on the 2nd choice.

[kiwiMark]

you have to think of it in two separate parts.

1) 1 out of 3 boxes is $300. 2 out of 3 boxes is $0.

2) 1 out of 2 boxes is $300. 1 out of 2 boxes is $0.

if you have your choice between the two, obviously you would pick #2.

The first "coin toss" was three side, whereas the next one is two sided.

In your second part, there's still no reason to pick Box B over Box A. Yes it's a better situation to be in than Situation One, but you're in this new better situation 100% of the time and your choice comes now, your choice doesn't come first and put you in the new better situation.