EV/Random Events (1 Viewer)

[Rorix]

This , adapted from a poker forum, is really messing with my head


You are on a game show where the host presents two envelopes for you to choose from - they both contain a cheque. One envelope contains a cheque for the value of half as much as the other one. Since the envelopes are indentical you just go ahead and pick one randomly. You open a cheque for N dollars.

Then, the host gives you the chance to switch envelopes. The EV of switching envelopes is obviously N/4. How do I get N/4 EV out of randomly choosing an envelope? In fact, since I know that it will be +EV to switch regardless of what I open, how is it any different to just picking the other envelope in the first place? In fact, it doesn't even matter what N is, because I know I will switch (unless N=0 but that's trivial) - however, when I then switch do I not get a further N/4 EV from switching back?

 

[ngai]

This , adapted from a poker forum, is really messing with my head


You are on a game show where the host presents two envelopes for you to choose from - they both contain a cheque. One envelope contains a cheque for the value of half as much as the other one. Since the envelopes are indentical you just go ahead and pick one randomly. You open a cheque for N dollars.

Then, the host gives you the chance to switch envelopes. The EV of switching envelopes is obviously N/4. How do I get N/4 EV out of randomly choosing an envelope? In fact, since I know that it will be +EV to switch regardless of what I open, how is it any different to just picking the other envelope in the first place? In fact, it doesn't even matter what N is, because I know I will switch (unless N=0 but that's trivial) - however, when I then switch do I not get a further N/4 EV from switching back?

hi
long time no see

i think its coz the 'N' depends on which one u picked in the first place
lets say the N is actually the lower amount
then by switching, you +N/4
but originally, before u picked anything, the EV was 3N/2, ie. +N/2
so in fact the new EV is lower

so why choosing to switch is always better?
because theres an equal chance u win or lose, but if u win, u gain more dollars than if u lose since one money is double the other
if they said one money is 100 more than the other, then theres no advantage to switch

 

[Rorix]

i think its coz the 'N' depends on which one u picked in the first place
lets say the N is actually the lower amount
then by switching, you +N/4
but originally, before u picked anything, the EV was 3N/2, ie. +N/2
so in fact the new EV is lower

Even if the new EV is lower you still have the same problem. If my EV from getting a shot at the game is 3N/2 and my EV from switching, after I've picked an envelope, is 5N/4 (my EV from sticking with the same envelope is obviously N) - how is it that by picking an envelope, regardless of what it contains, my EV in either scenario is lower than the EV from playing the game?

because theres an equal chance u win or lose, but if u win, u gain more dollars than if u lose since one money is double the other
if they said one money is 100 more than the other, then theres no advantage to switch

Yes, but the problem is that the following scenario seems logically impossible:
By choosing an envelope, I know that the EV of switching will be N/4. Suppose I don't even bother opening it, because I know it will be worthwhile switching - how is this any different to just choosing the other envelope in the first place? Surely, it is not, but the actions have different EVs?

By the same token, if I switch to the second envelope switching back to the first would have EV of N/4 if I had the possibility from the game, assuming I hadn't yet opened the first one.

What if the game had two players, and each had an envelope? Is it not +EV for both of them to decide to switch?

 

[who_loves_maths]

sorry for my stupidity, but what is "EV"?

 

[martin]

EV is expected value, the sum of probabilities time payoffs (for discrete distributions). For example the expected value of a dice is 1/6*1+1/6*2 + ... + 1/6*6 = 3.5

Its my feeling that it doesn't matter whether you switch or not your expected value will be the same. This is because of the argument above about not looking at the envelopes, then why wouldn't you switch multiple times. I think that what is leading people astray is using N for two different concepts - the smaller amount and some uncertain amount (the N that you look at its a random variable).

Say the envelopes contain x and 2x dollars and you open up an envelope with N dollars. (N=x or 2x both probability 1/2).

EV(staying) = EV(N) = 1/2*x + 1/2*2x = 3/2 x

If you switch envelopes you can go up or down with probability a half so

EV(switching)
= 1/2*EV(switching given N=x) + 1/2*EV(switching given N=2x)
= 1/2*2x + 1/2*x
= 3/2 x

So it makes no difference. I haven't done any probability for a few years so I'm not sure on my notation and definitions but I think that's the crux of the problem.

 

[acmilan]

Haha and here i was sitting here trying to work out what EV was when I was using the exact concept earlier today. It was just a notation difference, we use E(X) for expected value.

 

[Rorix]

I think that what is leading people astray is using N for two different concepts - the smaller amount and some uncertain amount (the N that you look at its a random variable).

N is always being used as the amount that you open. From me and ngai, at least.

Say the envelopes contain x and 2x dollars and you open up an envelope with N dollars. (N=x or 2x both probability 1/2).

EV(staying) = EV(N) = 1/2*x + 1/2*2x = 3/2 x

If you switch envelopes you can go up or down with probability a half so

EV(switching)
= 1/2*EV(switching given N=x) + 1/2*EV(switching given N=2x)
= 1/2*2x + 1/2*x
= 3/2 x

I don't dispute the validity of your calculations but suppose we do open the envelope:
If we switch, there is a 50/50 chance that its either N/2 or 2N
i.e. EV of switching: 1/2 * 2N + 1/2 * N/2 - N = N/4.
The only difference between these two scenarios is that I know what N is. I don't see how, logically, knowing what N is is worth N/4 in EV. The value of N is totally irrelevant as for any non-zero (or non negative of course) N you will switch, if you are interested in maximising profit. If someone can explain why knowing N is worth N/4....

Haha and here i was sitting here trying to work out what EV was when I was using the exact concept earlier today. It was just a notation difference, we use E(X) for expected value.

Gamblers like to use EV

 

[martin]

We have two very similar problems here but they are not equivalent. In the spirit of concreteness I've written 2 computer programs (in Matlab but they are pretty close to pseudocode).

Problem 1: Given one of two envelopes with one double the other should you switch

N=0; Nswitch=0; totalN=0; totalNswitch=0;
for i=1:100
    if rand > 0.5
        N=1;
    else
        N=2;
    end
    if N == 1
        Nswitch = 2;
    else
        Nswitch = 1;
    end
    totalN = totalN + N;
    totalNswitch = totalNswitch + Nswitch;
end

Problem 2: Given a number N should you take the option of doubling/halving it

N=0; Nswitch=0; TotalN=0; TotalNswitch=0;
for i=1:100
    N=2;
    if rand > 0.5
        Nswitch = 4;
    else
        Nswitch = 1;
    end
    TotalN = TotalN + N;
    TotalNswitch = TotalNswitch + Nswitch;
end

Now running these programs shows that in the first case it makes no difference and in the second you should switch.

So the question is why is the first problem not equivalent to the second? I have to admit that Rorix's argument seems convincing to me but there must be some problem with it. Are there any probability theorists about?

 

[Rorix]

So the question is why is the first problem not equivalent to the second? I have to admit that Rorix's argument seems convincing to me but there must be some problem with it. Are there any probability theorists about?

To clarify I know there must be a problem or fallacy somewhere (or a logical explanation) and am defending the view in terms of the adversarial system.

 

[who_loves_maths]

Originally Posted by martin
EV is expected value, the sum of probabilities time payoffs (for discrete distributions). For example the expected value of a dice is 1/6*1+1/6*2 + ... + 1/6*6 = 3.5

Thankyou martin for showing me what "EV" means, and your example with the dice is very helpful

even though i don't have any previous experience with the meaning and use of EV, i think i can gather at least an intuitive appreciation of what EV is a measure of from martin's kind example with the dice... and i think, from that, there might be a "logical" way out of Rorix's 'paradox'.

this is just a "theory" of mine, and may be very well wrong, but here it is anyways:

i think maybe the problem with this question is in the application and definition of EV and what exactly it represents... the 'definition' of EV that i have so far inferred from martin's example with the dice comprises of two parts: [and plz correct me if i am wrong below.]

i) the Expected Value of an event is the sum of the product of the probabilities of all possible outcomes of that event and the value of the outcome of their corresponding events (payoff).

ii) since the EV incorporates all possible outcomes of an event - then it is only applicable when all possible outcomes are still viable and intact, and where none are eliminated (otherwise the EV would be inaccurate and meaningless). ie. the EV of an event is relevant only before a decision (between two or more paths) is made.


these two "postulates" over the EV probably seem dead-obvious and logically simple (unless they are wrong), but may be the solution to this conundrum as follows:

Let us first calculate the EV of playing the game - ie. the first pick of envelopes.

the EV must thus incorporate all possible outcomes of the first pick:
namely, 1) the amount N in the envelope is the smaller payment, OR, 2) the amount N in the envelope is the larger payment.

if 1) is the case, then the two amounts are N, and 2N - making the partial EV for this case = N/2 + N = 3N/2

if 2) is the case, then the two amounts are N, and N/2 - making the partial EV for this case = N/2 + N/4 = 3N/4

since both cases are equally likely outcomes, then both has a probability of (1/2) of happening. hence, the total EV for the entire first pick is:
EV(total) = (1/2)(3N/2) + (1/2)(3N/4) = 3N/4 + 3N/8 = 9N/8

[ie. unlike what ppl have said before, EV for the first pick is not just 3N/2 - that's just a partial EV that only takes into account of case 1), the total EV is 9N/8]


OK, with that done, let's now calculate the EV of the second round - the time where a decision has to be made between either switching or not switching:

[this is where definition ii) above for the EV comes into play in solving the problem - remember we are finding EV total, not EV partial, because EV applies only to occasions in which a decision has not yet been made...]

in this second round, once again, there are two possible outcomes:
namely, 1) you decide to switch, OR, 2) you decide not to switch.

if 1) is the case, then the envelope you decide to switch to has either an amount of N/2, or, 2N - making the partial EV for this case = N/4 + N = 5N/4

if 2) is the case, then you retain the envelope you've already chosen - making the partial EV for this case = N

since both cases are equally likely outcomes, then both has a probability of (1/2) of happening. hence, the total EV for the entire second round is:
EV(total) = (1/2)(5N/4) + (1/2)(N) = 5N/8 + N/2 = 9N/8

but this total EV for the second round of (9N/8) is the same as the total EV for the first pick - hence, there is no paradox here, and resolves your conundrum Rorix: there is indeed NO difference (as you suspected) between choosing the first time, or choosing from the same two in the second round (or third, or fourth, etc...) - a decision of switching or not-switching, or the offer of choosing again, does not affect or increase the total EV of any round or ordered choices.


so how does that solve you're problem? this is what i think:
now that we know the total EV of both rounds are same, then the answer to you problem becomes a bit clearer through definition ii) :

you keep getting an EV of 5N/4 because it is only in fact a partial EV of the entire second-round. it does not represent the total EV of all possible outcomes - it is only the EV for one particular case, the case that you choose to switch. more specifically, the case that you have already chosen to switch.
Definition ii) is about that a total EV is only meaningful when you are using it to describe an event or decision that has all its possible outcomes still intact - ie. before a decision is made (otherwise you skew the probability of one possible outcome to always =1, which is what i suspect is done in your case of considering only the EV of the switching option).

By only considering one case, you are only looking at only one possible outcome - effectively saying that you have already made the decision to switch... in which case why do you need EV at all? your decision has already been made... its equivalent to that any talk of EV of an event after it has already occurred and taken place is irrelevant and meaningless - thus becoming a logical fallacy.


anyways, that's my 2-cents Rorix... like i said, i'm not very familiar with the common use of EV, so all i have said above is from pure imagination and "logical" interpretation of martin's dice example where EV was employed

but i do hope i've at least put something here that might stimulate a bit of thought

 

[martin]

Alright, I think the problem is this statement by Rorix:

If we switch, there is a 50/50 chance that its either N/2 or 2N

I can see no justification for it. It is not possible to have a uniform probability distribution on the positive integers. Consider all the integers from 1 to 1000, we can choose each with probability 1/1000, but if we have infinitely many positive integers each chosen with the same (finite) probability then finite times infinite equals 1 which isn't possible.

It is possible I think to have a nonuniform distribution on the positive integers, something like probability of choosing k is (1/2)^k. Then 1/2 + 1/4 + 1/8 + ... = 1 so it works as a probability. But in our case if you see N in the envelope then N/2 is more likely than 2N so the statement isn't true.

The original statement wasn't in terms of integers but I'm just assuming. Working with real valued currency would be difficult.

 

[Rorix]

mr. maths,

you are correct regarding the concept of EV ( E(X) = sigma Xi P(Xi)). When talking about situations with money, we can think of it in terms of we earn or lose a certain amount of money, whenever we take a certain action. To take a coinflip example, if I bet you $1 that a coin will turn up heads, and you give me $3 if I'm right (i.e. 2 dollar return), my EV is $3* 1/2 + (-$1)*1/2 = $1. In other words, every time I play the game, I 'earn' $1. Note that I can't earn $1 from playing the game once, as my possible outcomes are +2 or -1, but the point is that over the long run I'll win $1 per game.

However, the problem here does not turn on that issue - the thing is:

The two envelopes are identical and we chose one at random. Obviously, in this scenario, our EV is equal fwhether we choose envelope 1 or envelope 2.

HOWEVER, suppose we select envelope 1. Now, the EV of opening envelope 2 is 5N/4, and the EV of not opening envelope 2 is N.

In other words, it is always correct for us to switch, even though we chose at random and they had identical EV beforehand.

Also, consider, from your calculations you have calculated the EV of the game [without considering the further dynamic of switching] as 9N/8. I'm not sure if that is correct, as your treatment of cases and use of N seems a bit intuively wrong, but you could very well be correct - the most I use EV for now is my online poker which isn't too taxing. However, assuming you are right, in the long run we expect to earn 9N/8 every time the host gives us one of two envelopes to chose from, and we pick one.

However, when we open that envelope, our return is N. Every time. We lose N/8 of expected value by opening the envelope.

Also, the game does not seem to be always zero sum. Suppose instead, we have the same rules, but there are two contestants who have envelope 1 and 2. These men are cunning maths students, so they know that it is +EV to switch. They both switch. The first consentant gains N/4, the second contestant gains X/4 where X has a probability distribution of 50% N/2 and 50% 2N.

i.e. the sum of EV in this situation is 5N/4 + 5X/4
5N/4 + 1/2(10N/4 + 5N/8)
5N/4 + 5N/4 + 5N/16 = 45N/16

whereas the EV of not switching is N + X
N + 2N/2 + N/4 = 9N/4

i.e. from these calculations, both players will gain by switching. In other words, unless I've made a mistake, the game cannot be zero sum.

 

[Rorix]

I can see no justification for it. It is not possible to have a uniform probability distribution on the positive integers. Consider all the integers from 1 to 1000, we can choose each with probability 1/1000, but if we have infinitely many positive integers each chosen with the same (finite) probability then finite times infinite equals 1 which isn't possible.

I do not think the issue is relevant as talk of the probability distribution of the amount on the cheque is not essential for the analysis of the EV of the situation, fixed amounts would seem to work fine.

Even if I am wrong, reconstruct the game such that the amounts in the envelops are c^n and c^n+1 [c=unknown constant], where n is given by some random event where the probability distribution of amounts can be calculated (say e.g. number of odd numbers on dice in a row - a known constant could also be used but I'd have to give more thought on designing an appropriate event such that it is still +EV to switch given that lower amounts have higher probability weightings).

 

[martin]

I do not think the issue is relevant as talk of the probability distribution of the amount on the cheque is not essential for the analysis of the EV of the situation, fixed amounts would seem to work fine.

My point is that if we choose the values in the envelopes (x and 2x) beforehand then there is no paradox because you can go through each case separately and show switching makes no difference. The only problem is when you see the value N without knowing what x is.

The argument claims that the values (N,2N) and (N,N/2) are equally likely to be the amounts in the envelopes but I have shown that it is not possible to choose these values from some distribution so these two cases are equally likely. The fact that there are only two options does not make it 50-50.

Even if I am wrong, reconstruct the game such that the amounts in the envelops are c^n and c^n+1 [c=unknown constant], where n is given by some random event where the probability distribution of amounts can be calculated (say e.g. number of odd numbers on dice in a row - a known constant could also be used but I'd have to give more thought on designing an appropriate event such that it is still +EV to switch given that lower amounts have higher probability weightings).

I don't quite follow you here, maybe there is a way to reconstruct the game. However I will go to bed happy that the fallacy is that we assume that the game happens in a case where the amounts are chosen uniformly over the positive integers but in fact this is not possible.

 

[Rorix]

I don't quite follow you here, maybe there is a way to reconstruct the game. However I will go to bed happy that the fallacy is that we assume that the game happens in a case where the amounts are chosen uniformly over the positive integers but in fact this is not possible.

Suppose the envelopes contain 3^x and 3^x+1, where x is the number of odd dice numbers rolled in a row. Ignore x=0. It is still, by my calculations, +EV to switch, and your possible N amounts now obey a probability distribution.

 

[who_loves_maths]

Originally Posted by Rorix
However, the problem here does not turn on that issue - the thing is:

The two envelopes are identical and we chose one at random. Obviously, in this scenario, our EV is equal fwhether we choose envelope 1 or envelope 2.

HOWEVER, suppose we select envelope 1. Now, the EV of opening envelope 2 is 5N/4, and the EV of not opening envelope 2 is N.

In other words, it is always correct for us to switch, even though we chose at random and they had identical EV beforehand.

okay, i understand... but what i was trying to say was that it is meaningless to look at the whole situation from separate cases... when you say "the EV of opening envelope 2 is 5N/4..." and then say "and the EV of not opening envelope 2 is N." what you are doing is looking at the possible outcomes of one event one after the other, instead of considering them as a whole...

saying that "the EV of opening envelope 2 is 5N/4..." is as if you have already made the decision to switch anyways - so it's irrelevant to talk about EV from that point on since whatever the "expected" yield is it won't matter since you'll get exactly what you'll get in the 2nd envelope - you've already taken it.
then by saying that "and the EV of not opening envelope 2 is N." is, once again, as if you have already decided to not switch - in which case it makes logical sense that for a no-switch the EV is N, since you already know what's in the first envelope - the N amount!


but there's one very interesting thing that follows on from this:
you say that in theory the EV of the second round should not govern your decision to "switch no matter what", yet the paradox is that it 'seems' as if always switching is the best option - and this in turn differentiates between the first pick and the second round even though it shouldn't do that...
now i agree completely with you on this point , and i think there might be an answer to this:

the reason why in the second round, the EV for switching is higher than logic would suspect is because that the first pick and the second round are in fact not identical anymore (even though most ppl would think it still is...). they are actually different 'games' now.
why?
because in the first round, when you received your first envelope - you opened it up and looked at the amount N. i think this mere act of observing into the envelope is what 'collapses' the second round into a whole new game (lol, kinda like concepts in Quantum Physics... ).

to make myself clear, let me use an example to convey my point:
let's say we have two games - A and B.

1) in game A, after you make a first pick, you open the envelope and reveal a cheque for N dollars... then you are offered the choice of either to switch or not to switch, which one would you choose?

2) in game B, after you make a first pick, you are given the envelope but are not allowed to open it and see what's inside ... then you are offered the choice of either to change your pick or not, what would you do?

[actually try and answer (or compare) these two scenarios before reading on if you will.]


Our problem here is of course scenario A, but in scenario B the first pick and the second pick is indeed identical and have the same EV because you don't know what's in the first envelope you picked... yet in game A, by you looking into the envelope to see the N amounts, you 'collapse' the situation of the second round into giving the other envelope the possibility of having either N/2 or 2N amounts - but this changes the second round completely now, it is no longer identical/equivalent to the first pick where the possible amounts were either N/2 & N, or, N & 2N.
whereas in game B, both the first pick and the second round all have possibilities of either N/2 & N, or, N & 2N in the envelopes since you have not looked inside after the first pick.

this 'uniqueness' of the second round in game A is probably evident when you consider whether or not it's possible to make a similar game C such that the first pick of game C exactly mimics the second round of game A ... and the answer is you can't.
because in order to make such a game C, you need to maintain the randomness of the first pick of game A.
eg. say that the initial condition in this game C is that "one of the envelopes has four times as much money as the other." - then you can have N/2 & 2N as the amounts, but you wouldn't. because that wouldn't be random... to have N/2 and 2N, you need to define an absolute fixed value of N - here, N becomes significant as a number (it's like a fixed "frame of reference" now), but in doing so, you alter the randomness of the game by introducing a new variable N into it...
in normal circumstance, where everything is random, you would intepret the envelopes as having either N & 4N, or, N & N/4 - but those two sets of values give different EVs than using the set N/2 & 2N !
this clears shows the dependence of the second round of game A on the first pick of the same game. more so, it is dependent on defining a fixed number N whereas the first pick of the game does not. and yet, the N only becomes defined when you looked into the first envelope you picked in the first round...

Therefore, the act of opening the envelope and looking inside to fix an N in the first pick is the factor that changes and 'collapses' the second round into a whole new ball game.

think about it this way: the EV for both the first round and second rounds can only be equal IF all conditions are maintained the same for both rounds... but for the second round, you have a knowledge of a fixed amount N. So the question is, who was there to define a similar amount for the first pick before the game started?
the answer is no-one! it wasn't done. so the conditions for both rounds are different... hence it would be logical to expect different EVs for the two consecutive rounds.


okay, i hope i made sense in my explanation - and i hope you can see what i'm trying to say Rorix...


P.S. the above explanation can also solve your dilemma of continuing rounds and continuing the game to rounds 3, 4, 5, etc...
Why? because the trend of increasing EV for the option of 'switching' will only continue when the conditions of round 2 are emulated in rounds 3, 4, 5, etc... but this condition is that you formulate a fixed amount in a previous round to carry on to the next round. But this cannot happen beyond the second round, because IF you decide to switch envelopes in the second round, then you MUST open the switched envelope to look inside (as you did for 1st pick) before going on to subsequent rounds... but when you do that at the completion of the second round, you have effectively already seen both amounts in both envelope! ie. you now know what each envelope contains, so there would be no need/use for a round 3, or 4, etc..., because the EV in those subsequent rounds are no longer governed by probability - you will simply choose the envelope with the higher amount since you have already seen both envelopes ! so no... EVs won't change continuously by performing more and more rounds, simply because it can't happened anymore under the same conditions.
the "buck" stops at the second round!

 

[Rorix]

okay, i understand... but what i was trying to say was that it is meaningless to look at the whole situation from separate cases... when you say "the EV of opening envelope 2 is 5N/4..." and then say "and the EV of not opening envelope 2 is N." what you are doing is looking at the possible outcomes of one event one after the other, instead of considering them as a whole...

Not so. The only real point in looking at EV is to compare different scenarios and see which is more profitable. We should always do what is the most profitable in the long run (ignoring aspects of risk and money management). It is the same as considering whether to call 1-4 or 5-6 on a die.

you say that in theory the EV of the second round should not govern your decision to "switch no matter what",

??? you should always switch, in either game proposed. The problem is that in theory the EV of picking either envelope should have equal EV.

Therefore, the act of opening the envelope and looking inside to fix an N in the first pick is the factor that changes and 'collapses' the second round into a whole new ball game.

I understand what you are saying, indeed, said the similar in this thread, but can you give me a good reason why knowing N is worth N/4?

 

[Rorix]

okay, i understand... but what i was trying to say was that it is meaningless to look at the whole situation from separate cases... when you say "the EV of opening envelope 2 is 5N/4..." and then say "and the EV of not opening envelope 2 is N." what you are doing is looking at the possible outcomes of one event one after the other, instead of considering them as a whole...

Not so. The only real point in looking at EV is to compare different scenarios and see which is more profitable. We should always do what is the most profitable in the long run (ignoring aspects of risk and money management). It is the same as considering whether to call 1-4 or 5-6 on a die.

you say that in theory the EV of the second round should not govern your decision to "switch no matter what",

??? you should always switch, in either game proposed. The problem is that in theory the EV of picking either envelope should have equal EV.

Therefore, the act of opening the envelope and looking inside to fix an N in the first pick is the factor that changes and 'collapses' the second round into a whole new ball game.

I understand what you are saying, indeed, said the similar in this thread, but can you give me a good reason why knowing N is worth N/4?

 

[Rorix]

okay, i understand... but what i was trying to say was that it is meaningless to look at the whole situation from separate cases... when you say "the EV of opening envelope 2 is 5N/4..." and then say "and the EV of not opening envelope 2 is N." what you are doing is looking at the possible outcomes of one event one after the other, instead of considering them as a whole...

Not so. The only real point in looking at EV is to compare different scenarios and see which is more profitable. We should always do what is the most profitable in the long run (ignoring aspects of risk and money management). It is the same as considering whether to call 1-4 or 5-6 on a die.

you say that in theory the EV of the second round should not govern your decision to "switch no matter what",

??? you should always switch, in either game proposed. The problem is that in theory the EV of picking either envelope should have equal EV.

Therefore, the act of opening the envelope and looking inside to fix an N in the first pick is the factor that changes and 'collapses' the second round into a whole new ball game.

I understand what you are saying, indeed, said the similar in this thread, but can you give me a good reason why knowing N is worth N/4?

 

[who_loves_maths]

Originally Posted by Rorix

you say that in theory the EV of the second round should not govern your decision to "switch no matter what"

??? you should always switch, in either game proposed. The problem is that in theory the EV of picking either envelope should have equal EV.

- yes you are right but you seem to be picking on that sentence by taking it out of context??? i'm sure you know what i mean by that - and had you read on for more than one sentence or so you would see clearly what i was saying... to the effect of "in theory the EV of picking either envelope should have equal EV..."

i don't think there's any obscurity about this.

Originally Posted by Rorix
I understand what you are saying, indeed, said the similar in this thread, but can you give me a good reason why knowing N is worth N/4?

- i think i tried to explain this and gave an example in everything else that followed that statement... you probably should have finished reading my entire post before replying to bits and pieces.

generally, the reason i gave is that because by looking at N you changed the nature of the second round - it a different game now to that of the first pick, since you have the set N/2 & 2N as possible outcomes now rather than just N & 2N, or, N & N/2 .
eg. a game where the 1st pick gives possible outcomes of N or 2N, would be a different game to one where the 1st pick gave possible outcomes of N/2 or 2N...

so they're just different games now, and the rounds aren't identical anymore... so it follows that the EV would also differ.

that reason i think is enough, there needn't be a specific reason for why the EV increased... the reason for the increase is more about how the game has changed, and the fact that the set is now N/2 & 2N is a quantative measure for this increase - they provide the quantative reason.
so the mere fact that it has changed should suffice as a qualitative explanation.