Y is 10, therefore the sample space for X is [5, 10] — Michael
I know it seems that way to you. That is your subjective view of the situation. But we are told no such thing. We are not told X is selected by some random process or even by a choice. There is no such thing, so far as we know, as a "sample space for X". — Srap Tasmaner
But your own argument made this same assumption. — Michael
That's not an argument, it's just a calculation, and it didn't. It weights the possible values known to be in the sample space for , namely and , by the probability of the event of containing the larger of the two values, given that the value of one envelope is known to be 10, and the probability of the event of containing the smaller of the two values, given that one envelope is known to contain 10. — Srap Tasmaner
Knowing that one envelope contains 10 does not tell you whether UU is the larger or the smaller envelope, and does not tell you whether 5 or 20 are in the sample space. Not knowing whether 5 or 20 are in the sample space for UU we cannot use them. We could use 10, but 10 is the value of the envelope we've already got. — Srap Tasmaner
In this expression
is a dollar value — Srap Tasmaner
You can't condition on Y=X and then multiply by 2X where 2X≠2Y. That's just a contradiction. — Michael
no matter what you imagine it might be or how much you know about the envelopes. — Jeremiah
Since we don't know which one has to go, the only thing to do is continue, reluctantly perhaps and with some disappointment, using X and 2X. And what you find is that your expectation for U, in terms of X, does not change when you learn that Y = 10. Knowing only one of the two values is just not enough information. — Srap Tasmaner
That it is not enough information should be obvious. Look again at this example: — Srap Tasmaner
If Y<L/2 she should switch, otherwise she should not. — andrewk
I predict that, if that sim is run with a strategy that only switches when Y<num_trials / 2, it will converge on a gain that averages L/4 across those trials and 3L/16 across all trials. — andrewk
That's a different metric than the ones I quoted. Mine is in dollars while yours, if I'm reading the PHP correctly, is the dollars divided by the total of all Ys observed, minus 1.It's a .25 gain in my program. — Michael
# The amount earned if following the switching strategy switch <- 0 # The amount earned if never switching no_switch <- 0 # The highest value seen max_2x <- 0 # Play 10,000 games for (i in 1:10000) { # Select a value of X at random, multiples of 4 between 4 and 400. # We do this because we don't want odd numbers or the chosen envelope # to be double an odd number x <- sample(1:100, 1) * 4 # Put X in one envelope and 2X in the other envelopes <- c(x, x * 2) # Select an envelope at random n <- sample(1:2, 1) choice = envelopes[n] # Get the other envelope if (1 == n) { other <- envelopes[2] } else { other <- envelopes[1] } # Increase the amount if never switching by the amount in the chosen envelope no_switch <- no_switch + choice # If the first game or the amount in the chosen envelope is less than or equal to # half the highest value seen then switch and increase the switching strategy amount # by the amount in the new envelope if ( 0 == max_2x || choice <= max_2x / 2 ) { switch <- switch + other } # Otherwise increase the switching strategy amount by the amount in the chosen # envelope else { switch <- switch + choice } # Record the highest value seen max_2x <- max(c(choice, other, max_2x)) } # Output the gain that the switching strategy has over never switching paste('Gain: ', (switch / no_switch) - 1)
That's a different metric than the ones I quoted. Mine is in dollars while yours, if I'm reading the PHP correctly, is the dollars divided by the total of all Ys observed, minus 1. — andrewk
You have two envelopes, A and B.
There are two possible amounts for the contents of the two envelopes either X or 2X. You don't know which is which.
We'll call amount X case R, and and amount 2X case S.
You are handed envelope A.
Before you open it, you know that A could be case R or it could be case S. You have no clue which it is, so as an uninformative prior you give each case a 50% likelihood.
Now for the tricky part, you open envelope A and see it has 10 bucks in it.
Intuitively this seems like new information which would call for you to update your prior; however, it still does not tell you if are you in case R or case S. You don't know if that is 10=X or 10=2X. So in truth you can't really update your prior, as your prior was based on the uncertainty of being in case R or case S.
So your prior stands.
You consider swapping. You look at envelope B and realize it could be in case R or case S. You don't know which, so just as before you give it an uninformative prior and you give each case a 50% likelihood.
Now you could swap, but really probabilistically it changes nothing as your initial uncertainty would still stand. You would just trade the uncertainty of A for the uncertainty of B which is the same uncertainty. And the math? Well the math is indifferent to all options. So the decision to swap really comes down to if you feel lucky or not. — Jeremiah
I mostly use R, and am toying with Python in case that turns out to be useful if I want to get into 'data science' — andrewk
so please point out exactly which step it is you disagree with — Michael
You are taking a weighted average of one value which is not only possible but actual, and one value that is not possible. — Srap Tasmaner
You saying that there's a 50% chance that U is 2X given that Y is £10 is you saying that there's a 50% chance that U is £20.
You saying that there's a 50% chance that U is X given that Y is £10 is you saying that there's a 50% chance that U is £5. — Michael
Please Nooooooo! Let's not do that. I reserved the symbol L many pages ago for the maximum possible value ('Limit') in the player's prior distribution for 2X. If we start using it for something else, like one of the envelopes, we'll end up a terrible muddle. :joke:We'll label the envelopes L and R. — Srap Tasmaner
Nooooooo — andrewk
the player's expected gain from blind switching is zero, but her expected gain from switching based on comparing the value of Y to L/2 is strongly positive. — andrewk
Why? It's not about helping, it's about her expected gain, which is a pure calculation and nothing to do with luck. And my analysis is for a single trial.for a single trial, your guess would have to be awfully lucky to be any help. — Srap Tasmaner
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