What if I was to give you a sheet with 3^N on it, N>1.

If I told you that if you chose to gamble then I'd roll a fair dice and if it landed:
5 or 6 - I'd give you triple the amount on the sheet.
1, 2, 3 or 4 - I'd give you a third of the amount on the sheet.

Or you could refuse the gamble and take the amount on the sheet.

What would you do then?
(Gamble surely.)


Is this gamble any different to getting a sheet in one of your 10+ times with a number greater than 1 on it and choosing whether to switch?


(I'm actually a little unsure whether it is or not, but I think they are equivalent.)

Cool problem. Obviously very similar to the two envelopes paradox, but slightly less mysterious since you've actually defined how the amounts get chosen.

The EV of switching given the first envelope has N is indeed always positive for any fixed value of N

However, since there are an infinite amount of possible values for N and the game has an infinite EV, this doesn't actually contradict the fact that blindly switching has to have the "same" EV as blindly sticking.

What you have actually computed is the conditional expectation of the random variable 3^T given that T = 3.

More generally, for n > 0, E[3^T | you see 3^n] = (11/9)3^n > 3^n.

(the case n = 0 requires a (very easy) special treatment).

If we ask whether we should always switch, then our interest is not the just computed conditional expectation conditioned on having seen a particular number. Rather it is the expectation of 3^T conditioned on switching, which is infinite, compared with the expectation of 3^T conditioned on not switching, which is also infinite.

The questions of whether I should switch having seen a given number and whether I should switch always are distinct questions, with different answers.

As some other poster has commented, this is a version of a problem well known as the "two envelopes problem". There is lots of technical literature about this problem, not all of it equally good. An example that seems reasonable to me is: http://www.math.utk.edu/~wagner/pape...adventures.pdf

At 1st I wrote a Monte Carlo program, that seemed sort of even, but the variance was terrible making the result unreliable and inconclusive despite 10,000,000 iterations. I decided on a manual enumerative approach and got the following:

The relative EV of always switching is slightly worse. Seems counter-intuitive, but what is happening is that the always switch strategy (s1) has the possibility of selecting the slip with $1 on it. This happens often and is the reason, that s1 is slightly worse. Otherwise it would be a dead heat. Note that s2 only switches, when the slip shows N = 1. The rest of the time, it stays with the original slip, so it can never select the slip with $1 on it.

I should add that the EV of both s1 and s2 approach infinity. The Relative EV shows that s2 is slightly faster in its approach to infinity, but in theory, it is a tie.

I came up with something similar in another post. I said that from the viewpoint of the person playing the game who sees the number N

EV(switch | N) - EV(stay | N) = 2/9N

But from the viewpoint of the person writing down the numbers, he knows T but does not know N

EV(switch | T) - EV(stay | T) = 0

And for the game

EV(switch) - EV(stay) is undefined because the infinite sum diverges.

To me this is not a satisfying answer. The player does in fact see the number and his conditional EV tells him to switch every time. This is the same advice that an "always switch" oracle would give him.

To me the more satisfying explanation is simply that the EV of switching vs staying for the whole game is undefined and it is normal to get conflicting results when we attempt to group the terms in a divergent infinite sum in different ways.

Like them sum +1 + -1 +2 + -2 +3 + -3... is diverges. But if we group (+1 -1) + (+2 -2) + (+3 -3)... we seem to get 0 + 0 + 0 + compared to grouping +1 + (-1 + +2) + (-2 + +3) where we seem to get 1 + 1 + 1 + ...

Similarly, the person observing N is calculating the EV for each N while the person observing T is calculating the EV for each T and so they get different answers. That's ok because the EV of the game is undefined.

I will take a look. The envelope game I did look at starts with writing down two numbers, one of which is 2 times the other number. But that version of the game doesn't give any probability distribution for the numbers. That seems like a not-well-defined problem compared to this version.

Your simulation is calculating the EV for each T (number of tails flipped before seeing a heads) and then comparing those EV's at each T and getting EV(switch | T) - EV(stay | T) = 0. Now try calculating the EV for each N (number observed on the paper) using a monte carlo simulation and you will get an entirely different answer!! You will get EV(switch | N) - EV(stay |N) = 2/9N for any N > 1.

For example. Imagine you are playing this game and you draw your first slip of paper and it has a 9 on it. And you say to your self "I wonder if I should switch." So you go write a monte carlo simulation that plays the game a million times and whenever it sees a 9 it tries the switching vs staying strategies and accumulates the results of each strategy. The result of that simulation will tell you to switch.

the variance of your first simulation was terrible I think because EV(switch) - EV(stay) for the game is an infinite sum that diverges.

also see my post #49 where I am contemplating two monte carlo simulations.

http://forumserver.twoplustwo.com/sh...4&postcount=49

omg thank you. This is what happens when you are already a little confused and then think a new thought late at night. i forgot about the 1.

Do you agree that monte carlo simulations will give the following results?

1. EV(switch | T) - EV(stay | T) ---> 0 for all T
2. EV(switch | N) - EV(stay | N) ---> 2/9N for all N > 1, and 2 for N = 1
3. EV(switch) - EV(stay) for the game will not converge
and the apparent "discrepancy" between (1) and (2) is only possible because (3) diverges

T = number of tails flipped
N = number observed

I'm a little lost, but are we not all in agreement that switching can't be better than staying? It would be absolutely ridiculous otherwise, so any math that leads you to believe switching is better is either incorrect or being misinterpreted. Half the time you initially pick the bigger number and half the time the smaller one, so after switching you'll end up with the smaller one half the time and the bigger one half the time, just like if you don't switch.

1) Yes assuming EV(switch | T) - EV(stay | T) = 0 - 0 = 0

2) Yes assuming EV(switch | N, N>1) = (1/3)(3N - N) + (2/3)( (N/3) - N)
= (1/3)(2N) - (2/3)(2N/3)
= 2N/9

3) I really am not sure. Sorry.

I think I agree with this, but I'm a little fuzzy and not 100% confident on some aspects of the various results.

If believe that the EV if a single trial of game is infinite whether you switch or stay. I mean that for any value M you can name EV(switch) > M and EV(stay) > M for a single trial of this game. So does it even make sense to discuss which strategy is "better" from an EV standpoint?

Also, I'm not sure, but I tend to think that EV(switch) - EV(stay) for the game is undefined (as opposed to zero). In other words if we repeat the game many times with player A choosing a random slip and player B choosing the opposite slip and then we subtract those two numbers and accumulate and divide by the number of trials, we will not get a value that converges to 0. I suspect it will fluctuate all over the place.

But if you look at a sheet and see any number 3^T then you should switch to gain value.

This can be simulated any number of ways.
(I have made an excel file simulating 100000 trials with the first and second sheet distributed according to the rules of the game.)

Pressing F9 randomises so on the last trial:

When you get 1 the value from switching is 2
When you get 3 the value from switching was 0.64 (EV 0.67)
When you get 9 the value from switching was 2.05 (EV 2)
When you get 27 the value from switching was 6.22 (EV 6)
When you get 81 the value from switching was 17.81 (EV 18)
...
When you get 19683 the value from switching was 5423 (EV 4347)
...


Of course the higher the sheet number the less often it occurs in the trials and the further away from EV the result gets for that numbers switching value.

In theory for any possible sheet value you could simulate enough trials so that the calculated switching value converges to the EV, which is (2N/9).

This can easily be seen (for your own piece of mind) by considering the results for the smaller values in my excel file.
(Which merely confirm bobf's calculations given in posts above.)


Expected value of switching when you have 3^n, e(n), is (2)(3^(n-2)), n>0
e(0)=2

Probability of you getting 3^n on the sheet, p(n) is 3/(2^(n+2)), n>0
p(0)=1/4


This gives an expected value of the game over all possible sheet values of:

sum[n=0 to inf] (e(n)*p(n))

= (1/2) + (1/4)( sum[n=1 to inf] ((3/2)^(n-1)) )

Which is insane.

Thus the paradox.

Better minds than my own can explain the paradox and have hinted at it above.

I think that it has something to do with the underlying distribution of possible sheet values over the infinite set and the corresponding conditional probabilities of the sheet values not equalling the expected probabilities of the sheet values.

Although I got a bit lost trying to follow the explanations.


Rob

Maybe there is a way to compare the strategies SWITCH vs STAY even though the EV's are infinite.

Suppose we choose two values M and V. M is the number of game trials to play. V is some positive value that the players hope to attain. Then we compute the probability that player SWITCH will gain at least value V in M trials and we compute a similar probability for player STAY. It seems to me that these two probabilities will be equal for all values of M and V. So wouldn't that say the the two strategies are totally equal?

I think it has to do with the fact that we are taking a divergent infinite sum and grouping terms in two different ways. When you group by N you get EV=2/9N for each N. When you group by T you get EV=0 for each T. But this kind of effect is common with divergent infinite sums.

T = 0: N = 1 or N = 3: ev(switch from 1) = 2 ev(switch from 3) = -2 these cancel
T = 1: N = 3 or N = 9: ev(switch from 3) = 6 ev(switch from 9) = -6 these cancel
T = 2: N = 9 or N = 27: ev(switch from 9) = 18 ev(switch from 27) = -18 these cancel
etc.
they all cancel so we get a sum of zeros

but if you take the same series and combine the terms that have like N's you get...
ev(switch from 3) for T = 0 and T = 1 ---> you get a positive number
ev(switch from 9) for T = 1 and T = 2 ---> you get a positive number
etc.

If I try your way, I am comparing apples and oranges. I am making the fairest comparison. I'm not using an MT algorithm, I am using an enumerative technique that eliminates variance.You are proposing I compare the 3rd and 6th row of the enumeration while ignoring the 4th and 5th row. Why do you think that is an improvement?It wasn't diverging. The Monte Carlo version, I ran for 10,000,000 iterations several times. Sometimes the result showed a largish net EV that was negative and sometimes it was positive. The result was getting dominated by a small number of trials with very large N.I already looked at your post. Both your methods ignore the N = 1 case.

Look at these 2 steps in your method 1:
If you choose them at random with a frequency of 50%, it does not matter if you switch. For example, to generate that 50% frequency, you must flip a coin and assign:

"show larger slip value" to heads
"show smaller slip value" to tails

Always switching is equivalent to the reassignment:

"show larger slip value" to tails
"show smaller slip value" to heads

and always staying. Yes?

Since there is no reason to prefer one method of assignment over another, you know that switching vs staying are equivalent. No need to run your MT sim. You already know the answer with zero variance.

Your method 2 has a built in selection bias.

Yes, because 9+9+1+1+1+1 (22) > 3+3+3+3+3+3 (18). This should be "obvious" ... 1/3 of the time you are getting 3x as much, so for the gamble to be even the other 2/3 of the time you'd need to get 0.

This is not the same problem as OP, where (apart from when you get 1) it's a strict 50/50 (the difference would be if the distribution of flips was predetermined, and not random each time ... then you could act on the relative frequencies and what has happened in the past -- like counting cards in BJ).