two box paradox

[philllip1882]

a coin is flipped until it comes up tails.

lets call the number of flips T.

I'll randomly place 3^T in one box, and 3^(T-1) in the other.

now i give you the option of choosing one box. after seeing the contents, you can switch.

would it be to your advantage to switch?

 

[bonanova]

Assuming you meant things of value (perhaps coins?) into the boxes, then ...

Spoiler

Say the box you chose has n coins. There are (at least) four ways to look at the question.

1. Comparing probability of advantage.
   The other box has fewer or more coins, with equal probability.
   That is, I chose the box with more coins with probability of one-half.
   The decision is binary -- advantageous to switch, or not?
   There is no advantage (or penalty) to switch.
    
2. Comparing probability of advantage.
   The other box has fewer or more coins, with unequal probability.
   Fewer coins is four times more likely (one fewer initial heads flip vs. one additional initial heads flip)
   There is a substantial advantage not to switch.
    
3. Comparing expected value.
   The other box has n/3 or 3n coins, with equal probability.
   The expectation of switching is (n/3 + 3n)/2 = (5/3)n > n.
   There is an advantage to switch.
    
4. Comparing expected value.
   The other box has either n/3 or 3n coins, with unequal probability.
   If the other box has 3n coins it took an extra head flip - half as likely as your n-coin box.
   If the other box has n/3 coins it took one fewer initial head flips - twice as likely as your n-coin box.
   The expectation of switching is (2n/3 + 3n/2)/2 = (13/12)n > n.
   There is an advantage to switch.

I lean towards the first analysis, because symmetry is simple and beautiful.

Each analysis has its merits, tho, so I'll give the problem the status of a true paradox.