Sorry for making a solution thread to a locked puzzle, but I'm stumped, and he solution thread to the equivalent puzzle had me equally confused. Anyways, for reference:

The puzzle I am referring to: http://forums.xkcd.com/viewtopic.php?f=3&t=38673

The solution thread for the equivalent puzzle: viewtopic.php?f=3&t=7211

My original (apparently fallacious) logic was:

1. Person 1 chooses two boxes. Odds 50%.

2. Person 2 acts under the assumption that Person 1 succeeds. Best strategy seems to be to pick 2 different boxes so that he can guarantee not taking Person 1's name. Person 2 sees 2/3 of the possible boxes that could contain his name.

3. 1/2 * 2/3 < .4

Sooo, it seemed not possible, even if only the first 2 had to get it right.

After reading the solution thread, I tried thinking of the problem in terms of nodes. I don't see how this helps though. Person 1 has to pick either:

(a). His name; or

(b). The name previous to his

Out of 4 names, the odds are still 50%.

And I'm stumped as to how Person 2 can increase his odds above 2/3, even if he follows the nodes strategy. If anything, it seems to lower his odds because he might pick Person 1's name.

So, can someone explain how you get odds better than 40%? Maybe I'm missing something important about the nodes, but rereading the solutions just makes me more confused.

## Solution for 4 students, 4 boxes

**Moderators:** jestingrabbit, Moderators General, Prelates

### Re: Solution for 4 students, 4 boxes

Your reasoning is a bit off. Suppose person 1 succeeded with the "nodes" strategy, and person 2 picks person 1's name. Since person 1 succeeded, he must have opened box 2. The only way this would happen is if box 1 had person 2's name. This is the second box which person 2 will open, so he will find his name too.

In fact, the only way for person 2 to fail after person 1 succeeds is if person 1 found his own name in box 1. In such a case, person 2 has a (conditional) 2/3 chance of success.

In fact, the only way for person 2 to fail after person 1 succeeds is if person 1 found his own name in box 1. In such a case, person 2 has a (conditional) 2/3 chance of success.

### Re: Solution for 4 students, 4 boxes

Remember, too, that the group fails together or succeeds together. If the first person fails, at least two others will also fail, but if the first person succeeds it's fairly likely that the others will succeed. This game is bad news if the probabilities are independent and multiplied out - the later players' successes are all quite likely when conditioned on the first success.

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