The paradox is resolved by realising you are doubling a smaller amount than you are halving.
Two envelopes problem
81–90 of 318 posts
Re: Two envelopes problem
#82And what is the solution? I would just take the money at point A, i.e. at the very beginning, you never know when the experiment supervisor decides that he has enough of potentially giving even more free money (in case I'm smarter compared to him/her). Plus, free money at moment t is better than potentially double free money at moment t+t1
At some before point memorise enough random generated entropy so that when you enter this situation you combine that with the Schelling point (the left hand envelope for example) to pick one at random.
Then whatever trick is being played you still have a 50-50 chance.
Re: Two envelopes problem
#83Re: Two envelopes problem
#84If you treat this as a Bayesian problem, your context and experiences could lead you to specify a prior which has a finite mean, in which case for some values you will have a posterior probability of > 2/3 of it being the larger value.
E.g. if the maximum volume of currency that could fit in an envelope is $10k, it would be possible to have a uniform prior between 0-10k, at which point I would stick if my envelope had $5.01k.
A more extreme prior could be one capped at the largest ever academic research grant (I bet it's not more than $100m).
It could also be that you have a prior that has infinite mean, in which case it is not surprising that you will always want to switch if you draw a specific value.
If you have risk aversion, this further increases the range that you would like to stick, since switching could decrease your expected utility, even if it increases your expected return. There is a separate paradox in decision theory that we should be approximately risk-neural for 'small' amounts.
Re: Two envelopes problem
#85I must admit I don't find the "compelling line of reasoning" all that compelling; to me it seems the "paradox" has more to do with the meaning of probability than with its calculation. That is, the probability is 1/2. That's all that matters to the decision making. Calculating an "expected value" at all is completely useless, whether or not you do it "correctly". Am I missing something?
Calculating an expected value is meaningful and regular and does affect decisions. If you were told that you could pay £1 to flip a coin and heads would pay out 50p and tails would pay out £2, then playing it you would expect to make £1.25 (£0.25 profit) per play, on average. Many people interpret the envelope problem as being the same situation.
Re: Two envelopes problem
#86One mental model I apply to this and other similar problems is tracking what 'universe' you're in. In the flawed solution, the problem is that there's a universe in which the amounts are A and 2A and a universe in which the amounts are A and A/2, but they aren't the same universe, and the two possibilities will have different values for A and so you can't easily average A/2 from one universe with 2A from the other.
(The usual term is sample space; it's just more fun to split and eliminate universes than states.)
Re: Two envelopes problem
#87This line of reasoning isn't possible if you have specific amounts (eg £50 and £100) for the envelopes, which suggests that the variable A is being misused somehow.
`B = (2A if A=50, A/2 if A=100)`. Simplifying this to `B = 2A or A/2` loses important information: namely that when B is smaller, you expect A to be larger. Or alternatively, treating A as fixed (say A=100) conflates two different situations: one where the amounts are `100, 200` and one where they are `100, 50`. So you end up thinking B must have (200+50)/2 in it, which is incoherent. If you compute the expected B correctly then `A` is different in each branch.
It is subtle though. If this appeared in a paper, arguing something not obviously wrong, it'd be hard to convince everyone that the reasoning is bad.
Re: Two envelopes problem
#88> Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch? Normally we would assume there is someone who owns the envelopes of money. Now their incentive could be to offer to switch only if the envelope you chose contains more money (so they have a chance of keeping more). More like Monty Hall. Even without that, after switching, should you switch b…
> Even without that, after switching, should you switch back? The equation given argues you should switch back, so obviously the equation must be wrong.
Ah, that observation is very clever! Even if you let yourself be bamboozled by the logic for switching, there is no reason then not to switch again by using the same logic. And again.. and again!
Re: Two envelopes problem
#89This line of reasoning isn't possible if you have specific amounts (eg £50 and £100) for the envelopes, which suggests that the variable A is being misused somehow. `B = (2A if A=50, A/2 if A=100)`. Simplifying this to `B = 2A or A/2` loses important information: namely that when B is smaller, you expect A to be larger. Or alternatively, treating A as fixed (say A=100) conflates two different situations: one where th…
The correct answer is described here:
https://www.youtube.com/watch?v=_NGPncypY68
TL;DR (Spoiler alert): the expected value of the amount of money you end up with is an infinite series whose sum changes depending on the order in which you add up the terms, and so you can choose an order that makes this value come out to be positive, negative, or zero.
Re: Two envelopes problem
#90> Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch? Normally we would assume there is someone who owns the envelopes of money. Now their incentive could be to offer to switch only if the envelope you chose contains more money (so they have a chance of keeping more). More like Monty Hall. Even without that, after switching, should you switch b…
hmm, I don't really see where the probability of 50% comes from? even if we assume that we don't have any contextual knowledge about what amounts would typically be carried in wallets, the probability of winning would surely have to depend in some way on the amount currently in my wallet? in the extreme case, if I had zero cash, my probability of winning (not just drawing) would be zero, since we can't carry negative amounts of cash.