I did both of those things in measure theory.

I will see if I can get a copy to read sometime.

What were your guys arguments for why a series of +ev bets, whilst betting too much leads to negative ev after an infinite number of bets?

Take the simplest example: you go to an infinitely-bankrolled casino with no max wager. You find a +EV game and decide to play until you're infinitely rich. You start with $100. You're impatient, so you go all-in every wager. Each wager is independent.

Each wager is +EV, but the EV of your overall wager "Infinity or bust" is -100 because P(bust) = 1. You have a 100% chance of losing $100 (which is half the answer to your question) and a 0% chance of winning infinity (0%, but it is "possible" that you'll never lose a wager, so it's not the same 0% as rolling a 7 on a 6-sided die).

Infinity times zero is undefined, so that usual EV equation (for a finite series of wagers) doesn't apply here. The actual EV equation wouldn't have nonsense terms like 0*inf. Maybe RR can show the math, which will fully answer your question and also perhaps elucidate why the 0*inf term amounts to 0 (unless that's a wrong way of thinking about it altogether).

If you can actually make an infinite series of bets, you will end up making money. The catch is that you cannot make an infinite series of bets. This is why we talk about risk of ruin; if you run out of money you can never bet again. I could go through some math showing that as you bet more of your bankroll your probability of ruin approaches 1.

That's the easy part which doesn't require measure theory. The harder part (assuming it's not just a matter of definitions) is showing why the 1/infinity times you gain infinite wealth don't offset the "almost certainty" of losing $100.

It may have already been explained in one of the threads I alluded to, which I could dig up.

Yes, it is the part I was wanting to avoid doing unless someone really wants to see it. This weekend I was boxing up books I don't use. I debated and then kept out my real analysis/ measure theory books.

After a large number of bets at 1.1:1, wining 50%, betting 100% of roll each time.

your EV would be

prob(win all)*(how much you win)

so

EV=0.5^n*2.1^n
= 1.05^n
This clearly diverges as n grows

if you choose to bet p% of your roll each time, your EV becomes

EV=0.5^n*(1+p*1.1)^n
=(0.5+p*0.55)^n
for positive p this clearly diverges as n grows


Looks to me that any infinite sequence of +EV bets will always lead to +ev overall return.

Can anyone tell me why im wrong and the EV actually goes to 0?

The expected value will got to infinity, but the probability of losing all your money goes to 1. This is why Lebesgue integration exists; Riemann integration is inadequate to deal with this sort of problem. My wife starts home dialysis in the morning, so right now I don't have the time to go through a formal proof, but I will work something up later in the week if nobody else does first.

Lim n~inf 1.1^n, diverges, lebesgue integration isn't required.

You are assuming an infinite bankroll when you do that. The reason this discussion exists is we don't have an infinite bankroll.

I'm not assuming an infinite roll, I'm not sure if it even matters, your betting 100% each time, you either lose it all or you win.

If you have an infinite roll, why would you ever bet?

Maybe you misunderstood the problem. The problem isn't whether or not making a sum of +EV bets is +EV. The problem is maximizing the bankroll. If you lose all your money you lose the ability to make +EV bets in the future.

To put this another way, as n (number of bets) goes to infinity, the expected value goes to infinity (assuming it is a +EV bet); however, as n goes to infinity the probability of surviving to collect the EV goes to 0. I might be able to show that there is an optimal (maximum) bet size that maximizes wealth without Lebesgue, but I am not sure because at first glance there is going to be a 0 times infinity term in the bankroll calculation.

Here it is. I am glad I didn't sit down and try to do it off the top of my head (I don't think I would have managed).

http://projecteuclid.org/download/pd...msp/1200512159

Wait but does that show why it becomes -EV at infinite time (and greed)? I skimmed it and it only seems to be talking about why Kelly is the best. I thought we already established that Kelly is super, and had moved on to a purely academic question about the EV.

Well that is the thing, it isn't that EV becomes negative; the probability of having a bankroll different from 0 approaches 1. I will read that paper in detail later (today is the first day of doing dialysis at home, so a lot going on here).

I know you know this already, but suppose we have a chance to flip a fair coin and get paid 10:1. 50% pf the time we will be broke. If we take 2 tosses 75% of the time we will go broke. Each toss adds to the expected value, but each toss increases the chance of being broke. Even though each toss adds to the "expected value" the probability of going broke approaches 1.

And kind of the funny thing with flipping "infinite times" is that the game only stops when you bust out. So, you might get to have theoretical infinite wealth, but it's all on the table, you never get to cash it out and spend it.

Good point!

Im getting a bit lost on what we are actually discussing.

i)kelly is good, over double kelly is bad for growth (discussed)

ii) an infinite series of +ev bets is/isnt +ev (discussing?????)


an extension to ii) would be.

If we have a very large number of gamblers (appraoching infinity) all betting with an edge and all betting independently of one another. How does the population of gamblers maximise their wealth in the long run? betting Kelly or everyone betting 100% every bet?

I think (but don't know) that betting Kelly gets there because they make more bets. Everyone betting 100% may put less money in action.

For any finite number, everybody betting 100% all the time maximizes EV. The recursive proof is obvious once you think about it that way.

(a completed infinity of bets is a nonsense concept, so I'm not going to waste any time on it)

I think that only holds when there is a finite number of bets, and you know what that number is. I haven't given this too much thought, but an infinitely repeated game is used to model a finite game with an uncertain ending point, so I suspect that this needs to be treated as an infinity of bets, or at least one where there is some positive probability of a future betting round.

Fyp to be precise. Over double causes negative growth, but any amount over Kelly also results in a sacrifice of growth compared to exactly Kelly.

Fyp again. There is no debate when it comes to bets that aren't double Kelly.

Pretend it's one rich person making simultaneous wagers, and use Kelly for that situation. The answer is a little different (smaller) than simply everone making their own individual Kelly bet. They'd have to coordinate with one another after each completed wager to calculate each person's next bet size.

With a finite number of bets, you use the geometric mean maximization formula provided by Ralph Vince in his books, which is more aggressive than Kelly (but as time approaches infinity, the formula approaches Kelly). If there's only to be one wager ever, then the formula says to go all-in.

Not only is betting everything every time positive EV for a finite number of bets, it also produces the highest EV of all betting schemes (even though you are usually busted out well before the sequence ends.)

Doesn't the sequence end when you bust?