If there's £10 in my envelope then the expected value for switching is £12.50, and the expected value for switching back is £10. — Michael
That's only true if £10 isn't at the top of the distribution. When the distribution for single envelope contents is for instance (£10, £5, £2.5, £1.25 ...) then the expected value for switching from £10 (which is also a unique outcome) is minus £5 and, when it occurs, it tends to wipe out all of the gains that you made when you switched from smaller amounts. Even if you play the game only once, this mere possibility also nullifies your expected gain. — Pierre-Normand
Assuming your goal merely is to maximise your expected value, you have not reason to favor switching over sticking. — Pierre-Normand
There's also the possibility that £10 is the bottom of the distribution, in which case the expected value for switching is £20. — Michael
Assuming your goal merely is to maximise your expected value, you have not reason to favor switching over sticking.
— Pierre-Normand
Which, as I said before, is equivalent to treating it as equally likely that the other envelope contains the smaller amount as the larger amount, and so it is rational to switch.
The issue is this: are you prepared to apply the principle of indifference, and hence to rely on an expected value of 1.25X, for switching for any value in the (£10,£20,£40,£80) range? — Pierre-Normand
It is not rational since you are ignoring the errors that you are making when the envelope contents are situated at both ends of the distribution, and the negative error at the top wipes out all of the expected gains from the bottom and the middle of the distribution. When this is accounted for, the principle of indifference only tells you that while it is most likely that your expected gain is 1.25X, and it might occasionally be 2X, in the cases where it is 0.5X, the loss is so large that, on average, you expected gain from switching still remains exactly X. — Pierre-Normand
then I am effectively treating both cases as being equally likely, and if I am treating both cases as being equally likely then it is rational to switch. — Michael
That would only be rationally justified if there weren't unlikely cases (namely, being dealt the value at the top of the distribution) where the conditional loss from switching (0.5X) is so large as to nullify the cumulative expected gains from all the other cases. — Pierre-Normand
If there is no reason to favour sticking then ipso facto there is reason to favour switching. — Michael
I've noticed you express this sentiment a couple of times in the thread; could you elaborate further? I'm not sure how you're arriving at "A isn't favourable so B must be" - bypassing "neither A nor B is favourable."
If I were to borrow your logic, I could say that if given the choice between being shot in the head and having my head cut off, if I conclude that there's no advantage to having my head cut off, I must surely want to be shot in the head? — Efram
When all you consider is the relative chances of "low" compared to "high," this is true. When you also consider a value v, you need to use the relative chances of "v is the lower value" compared to "v is the higher value." This requires you to know the distribution of possible values in the envelopes. Since the OP doesn't provide this information, you can't use your solution. and no matter how strongly you feel that there must be a way to get around this, you can't.I'm not really sure that this addresses my main question. ... There's a 50% chance of picking the lower-value envelope, and so after having picked an envelope it's in an "unknown state" that has a 50% chance of being either the lower- or the higher-value envelope? — Michael
Statistics uses repeated observations of outcomes from a defined sample space, to make inference about the probability space associated with that sample space.Statistics is a data science and uses repeated random events to make inference about an unknown distribution. We don't have repeated random events, we have one event. Seems like a clear divide to me. You can't learn much of anything about an unknown distribution with just one event. — Jeremiah
Statistics uses repeated observations of outcomes from a defined sample space, to make inference about the probability space associated with that sample space. — JeffJo
Not even "one event," if that term were correct to use in your context. — JeffJo
The OP deals with a conceptual probability problem. There is no no observational data possible. "Statistics" does not belong in any discussion about it. — JeffJo
The only difference in the theory behind these two solutions, is that #1 uses an approach that implies different sets of values in the two envelopes, where #2 uses an approach that implies the same set of values. — JeffJo
Statistics uses repeated observations of outcomes from a defined sample space, to make inference about the probability space associated with that sample space. — JeffJo
I just said that. That is exactly what I said. — Jeremiah
You may well have. If you did, I accepted it as correct and have forgotten it. If you want to debate what it means, and why that isn't what you said above, "refer it to me over" again. Whatever that means.I already posted the definition of an event from one of my books, which I will refer to you over. I will always go with my training over you.
Maybe true in some cases. But "event" is not one of them. Look it up again, and compare it to what I said.One thing I was taught in my first stats class was that the lexicon was not standardized.
The other way to phrase the difference is that my solution uses the same value for the chosen envelope — Michael
And you are ignoring my comparison of two different ways we can know something about the values.It doesn’t make sense to consider those situations where the chosen envelope doesn’t contain 10.
True. But when that variable is a random variable, we must consider the probability that the variable has the value we are using, at each point where we use one.We should treat what we know as a constant and what we don’t know as a variable.
The other way to phrase the difference is that my solution uses the same value for the chosen envelope (10) and your solution uses different values for the chosen envelope (sometimes 10 and sometimes 20 (or 5)). — Michael
But we're just talking about a single game, so whether or not there is a cumulative expected gain for this strategy is irrelevant. — Michael
If it's more likely that the expected gain for my single game is > X than < X then it is rational to switch.
Or if I have no reason to believe that it's more likely that the other envelope contains the smaller amount then it is rational to switch, as I am effectively treating a gain (of X) as at least as likely as a loss (of 0.5X).
By that argument, it would be irrational to purchase for $1 a lottery ticket that gives you a 1/3 chance to win one million dollars since it is more likely that you will lose $1 (two chances in three) than it is that you will gain $999999. — Pierre-Normand
No, because the expected gain is $333,334, which is more than my 333,334X. — Michael
But then, if you agree not to disregard the amount of the improbable jackpot while calculating the expected value of the lottery ticket purchase, then, likewise, you can't disregard the improbable loss incurred in the case where v is it the top of the bounded distribution while calculating the expected value of the switching decision. — Pierre-Normand
But I don't know if my envelope contains the upper bound. Why would I play as if it is, if I have no reason to believe so? — Michael
Sure, most of the time, the conditional expected value will be 1.25v. But some of the times it will be 0.5v. — Pierre-Normand
And some of the time it will be 2v, because it could also be the lower bound. So given that v = 10, the expected value is one of 20, 12.5, or 5. We can be indifferent about this too, in which case we have 1/3 * 20 + 1/3 * 12.5 + 1/3 * 5 = 12.5. — Michael
we nevertheless know that, on average, for all the possible value of v in the bounded distribution, their weighted sum is zero rather than 1.25v. — Pierre-Normand
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