Interestingly, if you go a step back (before opening box A), you can do better then a 50% winning chance:
Chose a number at random, [math] X [/math]. If you then pick a box, keep it if [math] X [/math] is smaller then the amount in the box. There are 3 cases: I) [math] X [/math] is smaller then [math] A [/math] and [math] B [/math], with probability [math] P = x_1 [/math] II) [math] X [/math] lies between [math] A [/math] and [math] B [/math], with [math] P = x_2 [/math] III) [math] X [/math] is bigger then [math] A [/math] and [math] B [/math], with [math] P = x_3 [/math]
In case i) and III), it is a 50% to pick the better box, so your "win" probability would be [math] \frac{x_1 + x_3}{2} + x_2 [/math] and since [math] x_1 + x_2 + x_3 = 1[/math] [math] \frac{1}{2} + \frac{x_2}{2} [/math] Because [math] x_2 > 0 [/math] you get a better then 50% probability.
Nicholas Ross
You accept a bet if the possible gain times probability of winning exceeds possible loss times probability of losing. If you take box B there's a 50% chance of gaining $9000 vs. a 50% chance of losing $900. That says you should switch.
Something wrong with the logic here since the same would apply if you'd chosen box B first. After all, if one person chose A and another person chose B, they can't BOTH have a winning strategy by switching.
It's a good problem and I hope someone can explain the correct solution clearly.
Yes. Risk reward says the choice is always profitable.
Logan Robinson
>game theory goyim believe that loss and gain have equal weight
James Cox
I posted but recognized something was wrong with my logic. So I read ps://en.wikipedia.org/wiki/Two_envelopes_problem particularly the "simple solution" section. There, the envelopes contain 1 and 2 dollars. If you have 1 and switch, you gain 1 If you have 2 and switch, you lose 1 Since the expected gains and losses are equal, there's no point in switching. That's clear. But here we have a TEN to one ratio. If your envelope contains 10, then you could gain 90 or lose 9 by switching. Switching again seems advantageous. So your links really aren't clearing this up for me.
Brayden Barnes
Being in this situation, I know for sure no one would have put 10k as a prize for a random stranger like me. I pick my 1k and get the fuck outta here
Evan Campbell
>$5050 in average Literally lies, damned lies and statistics.
Joshua Russell
>But here we have a TEN to one ratio. The ratio makes no difference:
>If you have 1 and switch, you gain 1 >If you have 2 and switch, you lose 1 If you have 1000 and switch, you gain 9000 If you have 10000 and switch, you lose 9000
Xavier Reed
I see what you're saying but it still seem asymmetric. Your numbers are right but you don't know if you have the larger or smaller amount. You just know you have X and the possibility of gaining 9X or losing 0.9X I'm afraid I still don't understand the flaw -- though your example helps.
