Okay, I need help from all my fellow math geeks with an apparent paradox. I'll lay out the puzzle without presenting my analysis; I want to see what other people do with it first
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This reeks of being a Monty Hall problem, but given that there is no dud, the chances remain 50/50, so that's out.
The only thing we have to work on here is the exponential difference in amounts.
Now, since N can be 0, there is a chance that one box has 0 dollars, so by telling me that the box I've opened has SOME money, we know that for N=0, this is the high box, where for all other boxes, this may or may not be the high box, so I'm saying, given at least one case where this is the known winning box, and none I can think of where it's the known losing box, we keep what's in box #1.
How'd I do?
Edit: By "some amount of money" am I to presume that ammount is > 0? If not, obviously, I fail.
If the box is empty, the other box has $3 in it, and you should trade.
If the box has $3, the other box has either $0 or $9 in it. You have 50/50 odds of either losing $3 or gaining $6 (double your maximum losses).
If the box has $9, the other box has either $3 or $27. You have 50/50 odds of either losing $6 or gaining $18 (triple your maximum losses).
If the box has $27, the other box has either $9 or $81. You have 50/50 odds of either losing $18 or gaining $54 (triple your maximum losses).
etc.
Another way to look at it is that someone says, "I will give you $x+n and then flip a coin. You can either keep all $x+n or bet $x on the outcome. If it comes up heads, you lose. If it comes up tails, you win at least $2x and probably $3x. Either way, you keep $n." I'd take that bet, especially if we played the game multiple times.
There is one other factor: This being based on a fair coin toss rather than truly random, the higher the amount in the box, the more likely that it's the higher of the two. e.g. if you find 3^8, the odds that there were 7 coin tosses before hitting tails and this is the N+1 box are (I think) higher than the odds that there were 8 coin tosses before hitting tails and this is the N box. I just don't feel like crunching those numbers right now, mostly because I can't remember whether it makes sense to calculate odds this way for coin toss problems.
Oh, I just meant money available for putting in boxes. I expected that it would never reach the point where I got to choose; you would have to cancel the game for lack of funds first.
Yes, but consider that the "put money in box" action is triggered by tossing a tails...which can never happen. I might end up knowing I could never cover the required amount were I to toss tails, but that would never actually happen. :)
This all revolves around how much money in the box you choose.
In the case where the box has $1, always switching is a no brainer.
For all other cases, if you always take the other box there is a 1 in 3 chance of winning of tripling your money and a 2 in 3 chance of losing 2/3 of it. For example if the amount in hand is $3 there is a 1/2 chance that the other box contains $1 and a 1/4 chance it contains $9.
But doesn't that strike you as paradoxical? How can choosing a box at random, and then, regardless of what you see (other than the $1 case) mechanically switching to the other, improve your expected return?
The knowledge of money in hand changes the first choice. You've setup a complicated system for determining what is in the boxes. Right off the bat 1/4 of the time you will improve your money from $1 to $3.
As in the example I gave before, if the amount in hand is $3 there is a 1/2 chance that the other box contains $1 and a 1/4 chance it contains $9.
1/2 x 3 + 1/4 x 9 = 1.5 + 2.25 = 3.75
$3.75 is greater than $3, for each case, the probabilities that we have setup simply favor switching. As in all gambling, the odds don't favor the short term, you only come out ahead if you can stand to lose X number of games while sticking it out for the long run.
As in the example I gave before, if the amount in hand is $3 there is a 1/2 chance that the other box contains $1 and a 1/4 chance it contains $9. Adding these two percentages together gives the total probability of being in the situation where $3 is showing.
(1/2) / (1/2+1/4) = 2/3 (1/4) / (1/2+1/4) = 1/3
Therefore there is a 1/3 of the other box containing $9 and 2/3 chance of the box containing $1.
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The only thing we have to work on here is the exponential difference in amounts.
Now, since N can be 0, there is a chance that one box has 0 dollars, so by telling me that the box I've opened has SOME money, we know that for N=0, this is the high box, where for all other boxes, this may or may not be the high box, so I'm saying, given at least one case where this is the known winning box, and none I can think of where it's the known losing box, we keep what's in box #1.
How'd I do?
Edit: By "some amount of money" am I to presume that ammount is > 0?
If not, obviously, I fail.
-- James
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If the box is empty, the other box has $3 in it, and you should trade.
If the box has $3, the other box has either $0 or $9 in it. You have 50/50 odds of either losing $3 or gaining $6 (double your maximum losses).
If the box has $9, the other box has either $3 or $27. You have 50/50 odds of either losing $6 or gaining $18 (triple your maximum losses).
If the box has $27, the other box has either $9 or $81. You have 50/50 odds of either losing $18 or gaining $54 (triple your maximum losses).
etc.
Another way to look at it is that someone says, "I will give you $x+n and then flip a coin. You can either keep all $x+n or bet $x on the outcome. If it comes up heads, you lose. If it comes up tails, you win at least $2x and probably $3x. Either way, you keep $n." I'd take that bet, especially if we played the game multiple times.
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-- James
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In the case where the box has $1, always switching is a no brainer.
For all other cases, if you always take the other box there is a 1 in 3 chance of winning of tripling your money and a 2 in 3 chance of losing 2/3 of it. For example if the amount in hand is $3 there is a 1/2 chance that the other box contains $1 and a 1/4 chance it contains $9.
0.50 / (0.50 + 0.25) = 2/3
0.25 / (0.50 + 0.25) = 1/3
3x(1/3) + (2/3)x(2/3) = 1.222
Your payoff will be 1.222 times higher if you always swap.
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As in the example I gave before, if the amount in hand is $3 there is a 1/2 chance that the other box contains $1 and a 1/4 chance it contains $9.
1/2 x 3 + 1/4 x 9 = 1.5 + 2.25 = 3.75
$3.75 is greater than $3, for each case, the probabilities that we have setup simply favor switching. As in all gambling, the odds don't favor the short term, you only come out ahead if you can stand to lose X number of games while sticking it out for the long run.
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As in the example I gave before, if the amount in hand is $3 there is a 1/2 chance that the other box contains $1 and a 1/4 chance it contains $9. Adding these two percentages together gives the total probability of being in the situation where $3 is showing.
(1/2) / (1/2+1/4) = 2/3
(1/4) / (1/2+1/4) = 1/3
Therefore there is a 1/3 of the other box containing $9 and 2/3 chance of the box containing $1.
2/3 x $1 + 1/3 x $9 = $3.66
Not switching gives:
1/1 x $3 = $3
$3.66 > $3
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