I just thought of this while contemplating a gambling problem. Perhaps it has been thought of by others. Or perhaps its an obvious example of routine paradoxes of the infinite and thus doesn't deserve to be named after me. But maybe not.

There are two indistinguishable coins in a hat. One produces heads 51% of the time. The other 48%.

You choose a coin and flip it up to a million times. Every flip you bet one dollar on heads.

If you blindly flipped it all million times your EV would be negative 10K since you will wind up either about 20K winner or 40K loser. But if you are good with statistics you will give up when the evidence that its the bad coin is strong but will very rarely quit when you picked the good coin. So your actual EV will be somewhat less than PLUS 10K as you will win 20K almost half the time and lose far less the other times.

The more flips you are allowed the higher your EV. (Exactly how many flips you need promised to you before your EV is positive is a nice problem for a computer programmer.) If you were allowed a googol flips your EV would be just about 10 to the 98th dollars. And since there is that much waiting for you, you would need awful strong evidence before giving up. 30 tails to start wouldn't be enough.

But what if there is no limit to your allowable flips? With no limit to your profit if you determine its the good coin, there is no limit to how discouraging the evidence has to be before you give up. So you should never give up. But that means that your EV in this spot would be negative infinity instead of positive infinity.

Don't be greedy David, you already have royalties from your Skylansky Bucks rolling in nicely to Skylansky Towers.

If you reduce infinity to a perpetually expanding finite set...

Build an automatic coin flipper, then give the blueprints to someone and induce the same dilemma in them.

Walk away.

There is a long history of these types of problems (where just looking at EV gives a crazy result), I think the St Petersburgh paradox might be the oldest. They end up not being difficult math problems, but more about economics. You define some utility function and maximize that, which is a much better guide than just maximizing EV in terms of how real people should behave.

There are two indistinguishable hats. One hat has a piece of paper that reads "You win infinity dollars" and the other hat has a piece of paper that says "You lose infinity dollars."

How much would you be willing to pay to play this game?

That's actually cute, but.. any strategy for the game can be declared in advance as pairs of (a,b) a>=0 where if you have flipped b or more heads in a flips, you continue, otherwise you stop. There's obviously no way to derive that (0,0), (1,0), (2,0) ... (aka always flipping) is a good idea because it isn't. What your argument is doing is effectively saying that for any N, there's a +ev strategy that contains (0,0) up to (N,0), which is true. It just doesn't follow that (0,0) ... (inf,0) IS a +ev strategy. Your argument that it appears to be +EV consists of arguing about the EV of something that isn't a game strategy, which is meaningless (your argument that it isn't +EV is correct).

I don't consider the St Petersburgh Paradox analogous. If there is an analogy it is more like how those infinite sequence of numbers add up to -1/12 and stuff like that.

There is no paradox here for the following reason; (this is different than St. Petersburg paradox)

We can show there is a strategy that has positive expectation if we do not do the play forever no matter what thing from the start but later under some conditions met only.

Our starting profile if we started with the stop never approach will lead after a number of steps n to an expectation that goes like;

1/2*(0.51-0.49)*n+1/2*(0.48-0.52)*n=-0.01n.

All we need is to show now that there is a strategy that is superior to the never stop one for some big enough n. The never stop one has -0.01n. Lets create one that has kn expectation with k>0 in leading terms for some n and higher.

Imagine we had a strategy that says play until you endure a certain loss of N or you reach a positivity condition that is not available at the beginning in the never stop play mode. By that i mean if we had a profile that was eg 70% 0.51 and 30% 0.48 the EV would be (0.7*0.02-0.3*0.04)n=0.002n. So if we set as our goal to reach such a state we would very often reach that state of conviction with the good coin and rarely (can be made astronomically low for some big enough N) with the bad coin.

So you could introduce a stop under some condition strategy as a way to turn this into a plus EV game. Or you can avoid the loss depth and instead suggest a test trial phase (probably an inferior method though) for some flips N (not the losses now, the total flips).

We can proceed next to show an example of such strategy of course by finding a particular N that realizes it. Only a very nice profile (like 70-30 etc) will deserve a play indefinitely decision approach.

Ask yourself for example how often after 1 mil runs you have a profile for 70-30 if you started with 50-50? There will be some number that this is true which is smaller than 50% of course. The approach might be play 1mil flips and then play indefinitely only in those cases that you have reached (Bayesian updating) a 70-30 or better profile now and stop in the others (in fact there is an optimal number that this is true say call it p*0.02-(1-p)*0.4>0 or p>2/3). So give the system a testing phase to clean up the profile from 50-50 to eg ~67-33. Then play indefinitely only the 67-33 and better cases always. Now sure some of them will turn bad (a real real very low chance but on avg the strategy is positive now over all cases).

So basically find a testing phase long enough , ie with enough number of flips that has a nonzero population which has at least now at the end of testing an updated profile of 67-33. For example you cant have such confidence with only 10 or 100 flips etc. But for some big enough N it will have a nonzero population that clears it.

That is not the best approach for a finite number of available flips. But it is a positive expectation strategy for n steps that is big enough and the play indefinitely always is a negative expectation strategy as function of n in leading terms unless it starts from such profile.


So if a strategy exists that does better than the never stop from the beginning, it must make no sense to never stop because that other strategy is better.

Just to show something for illustration purposes;

Ask yourself what profile do coins that result in 550+ heads have after 1000 flips?

If p =0.51 the chance to get to 550+ in 1000 is
Sum[1000!/k!/(1000 - k)!*0.51^k*0.49^(1000 - k), {k, 550, 1000}]=0.00619383

If p=0.48 the chance to get to 550+ in 1000 is;

Sum[1000!/k!/(1000 - k)!*0.48^k*0.52^(1000 - k), {k, 550, 1000}]=5.46076*10^-6

so it may start at 50-50 but among those that have 550/1000 or better the profile is

0.00619383 to 5.46076*10^-6 or 99.9119% is good coins.

As you see we have a severely biased population profile now at the 550+ level.


As an idea from this one can imagine continuing indefinitely only if a run so far for a given coin that started as 50-50 offers a 2/3-1/3 profile probability now. I mean if one took all the coins that had such profile at some point and decided to continue indefinitely only with them we have created a positive expectation strategy for large enough n. We do not have such EV available if we used that never stop approach from a negative starting profile.

So we could start with a coin and either hit a loss limit or clear the 2/3 probability to be a 51% coin whichever happens first. Then continue only with the coins that have reached that probability. There will be a large enough number that always clears that in say 1000 flips or earlier. If we set the barrier at 530/1000 10.8658% of the 51% clear it and only 0.0868852% of the 48 ones clear it.

If we set it at 500/1000 74.675% of the good ones clear it and only 10.8581% of the bad ones.

So its fairly easy to hit such numbers and those that do definitely have a positive game now if it was played without ever stopping. Of course even this can be improved.


I suppose if one played a stop loss N strategy with all coins ie keep flipping until N losses is reached (for some big enough N) then stop, this creates a highly positive strategy for any starting profile eventually. The good coins have a tiny probability to ruin if you select big enough N. The bad ones will always ruin eventually. You can use ruin formula results to study this better ie to see what kind of filtering you obtain with such stop loss level N (what % of good ones you lose basically. You also of course lose all the bad ones). Fix N so that you have a positive expectation now if you start with 50-50. Then maximize that too over all population. If you set N big enough eg 100 you will lose almost none of the good ones practically if you start with 51-49 for them.

The ruin probability if you have p>q=1-p for a coin is;

((1-p)/p)^N if your stop loss is N.

So for N=100 and p =0.51 you get 1.8% ruin. You lose all the bad ones but only 1.8% of the good ones.


For big enough n this strategy will have EV that goes like 0.5*0.98*0.02n~0.0098n vs -0.01n for the never stop nomatter what one.

[QUOTE=masque de Z;48813014
So if a strategy exists that does better than the never stop from the beginning, it must make no sense to never stop because that other strategy is better.[/QUOTE]

Any strategy that just may have you giving up after n flips has a lower EV than a strategy that has you possibly giving up only after a much much much larger finite number of flips. As long as an unending number of flips is offered to the brave. Are you disputing that?

This is a conceptually simple though computationally complex game.

There are two indistinguishable hats. One hat has a piece of paper that reads "You win $1" and the other hat has a piece of paper that says "You lose $10,000,000." Once you pick a hat, you can draw the slip of paper out as many times as you want and the corresponding action is taken.

Here's your N draw strategy: Pick one hat at random and draw the slip of paper. If you lost money, stop drawing. If you won money, draw N-1 more times. Your EV is -$5,000,000 + 0.5*N. As N goes to infinity, so does your EV.

And it doesn't actually matter that there are two hats. Let's say that there are a billion hats and only one of them has the winner paper in it. You still have that one large win to offset any amount of loss you might sustain. It all comes down to the fact that there is a situation in which you can win infinite money and that you can identify stop-losses when you're in a losing situation. All that the extra probability element does is make it so that you have to do some calculations to figure out whether to play on or stop, because your knowledge of the distributions will require you to flip the coin many times rather than simply knowing right away.

So the only interesting thing is identifying when you're in the winning condition, not that you should be willing to pay infinite money to play this game.

I am not disputing that. Surely if i set N not to 100 but at 1000 the ruin probability will collapse to ridiculous levels like 10^-18. It will do so at a loss though near term because we allow the bad ones to do a lot worse damage. In the end we recover all that damage from the good ones we saved of course. But only if we stop at some point. If we dont stop we dig a bigger and bigger well for ourselves basically at the -0.01n rate. We just postpone the recovery but if we never put an end to it we never put an end to the accumulation of losses and the beginning of the recovery process.

Basically a proposed strategy after a while will be going like 0.0098n (tending to 0.01n in leading terms if you make more room) and the other will be going as -0.01n.

Any stop strategy at some given N that is big enough will always do better than the never stop strategy of all of them to any level n.

Only if you stop you realize the superiority of continuing further to save more good ones from bad luck ruin (and only after a great while eventually, because the allowed bad ones have created a big loss, that time needed to recover and lead in EV may be very big already but of course will be reached in the limit).

I think the following strategy will work to create an infinite expectation for the original problem.

Stop condition: Your net winnings is equal to -$1

You will never lose more than $1 with this strategy. But since there is a way to win infinite money (when you draw the winning coin and you start off lucky), you will be able to drag that EV up as high as you want.

It's more like Martingale than Petersburg.

Right, this is a positive expectation increasing with n strategy. But allowing more room creates an even better result. If you can afford losses near term you can create as close as to 0.01n rate of profits as you wish eventually, instead of relying on the never losing lucky ones that are basically just 1-(0.49/0.51)^1=4% of the good coins.

So your method establishes a 0.04*0.5*0.02n=0.0004n winning rate by setting the bar at -1 only.

The other limits can go as close to 0.01n as they want but only if they have a finite stop level this becomes possible, otherwise its -0.01n for the nonstop ones until they give up lol.

So is the slightly more-in-spirit "play the strategy for n=bazillion, then keep flipping after that as long as it appears that the next flip is +ev". That gets almost all the available equity up to any point in time (ratio of ev of that after n>>bazillion vs ev of perfect strategy for that n will be pretty close to 1).

Right. So set the stop loss at $2 instead of $1. Or $3. Or $3,000,000,000. It doesn't really matter. As long as you have some finite stop loss, your EV goes to infinity. Any massaging that you do beyond this is all just an optimization problem, which isn't of interest to me. For example, if you were once up $2, then maybe you allow yourself to fall down to -$2 instead of just -$1. Or maybe your stop loss starts at $2, but if you got above a certain amount, then your stop loss is -$1, so that you maximize when you got lucky with a bad coin to start.

Whatever. It's all just on the fringe at this point.

No.

I think this is right. Some Stopping strategies will give you a higher probability of achieving infinite winnings with the Good Coin than others. If you choose that probability, p, that you desire then there may be an optimum strategy for p which minimizes your expected loss for when you stop under it. Let's say that strategy is S(p). You can achieve any p < 1. But if you denote by X[S(p)] the random variable giving your loss under S(p) when you stop (0 if you don't stop), then EV(X) ---> -infinity as p ---> 1.

The other thing of interest is what your EV is under S(p) after N flips. If we denote this by EV[S(p),N] then what we should see is as p gets closer to 1, EV[S(p),N] will stay negative for larger and larger N. If we denote

M = Max(N such that EV[S(p),N]) < 0)

then we should get

M ---> infinity as p --->1.

In other words, as your strategies get your probability of infinite winnings with the Good Coin closer to 1, you stay -EV longer and longer. So is it really a discontinuity to imagine those strategies converging to a "Never Stop" strategy where you always have a -EV?


PairTheBoard

Well that seems ludicrous. St Petersburgh is only a "paradox" if you consider maximizing EV as all that matters, so when you have an incredibly small chance of infinite gains you have to lose any finite amount, which is the same as your problem.

They are analogous in that sense. But I wasn't bringing the concept of worrying about dirty money into it. That is beneath me.

In what possible way is this analogous to Ramanujan Summing or Zeta Function Regularization? I still have no clue how you could possible compare the two, other than gross ignorance of the previous.

And of course money doesn't matter....it's an optimization problem with some sort of utility function. Money just happens to be the most interesting/relevant thing to build utility functions around.

The discontinuity is in the logic. "Stop" and "Never stop" are a binary logic, so that any jump from one to the other is discontinuous. You cannot continuously go from "Stop" to "Never stop" because "Almost never stop" doesn't exist.

But this is no more profound than simple limits. What's the limit of x^2 as x goes to infinity? For any finite x, the value of x^2 is finite. But in the limit, it makes the jump from "finite" to "infinite" without there ever being an "almost infinite" intermediate condition.

With an infinite amount of flip...
An uninteresting question is; lets say your coin is 48% to win. Is your stock worth infinite, or nothing?

Hey! Your stock's worth is all that matters in our social world.
I mean value is derived somewhere...

hmmm... marketing!