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Old 11-09-2007, 06:54 AM
Uniqueuponhim Uniqueuponhim is offline
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Default Infinite bankroll paradox

Alright, I know this problem has absolutely no bearing whatsoever on real life, but I think it's an interesting problem so I'd like to ask what the "solution" is:

Say a man with an infinite bankroll comes up to me and bets me everything I own at 1:1 payout that if he flips two coins they will both turn up heads. Clearly this bet would be +50%EV for me to take. Now let's say that every time I win I have the option of making the bet again, but the man requires that I bet everything including previous winnings each time. Every time I take the bet, my EV is +50%, but if I keep taking the bet over and over again, eventually I'm going to lose.

Looking at it another way, the probability of me losing the bet at least once goes to 1 as the number of bets I make goes to infinity. Therefore I can set my total number of bets to an arbitrarily high number and my overall EV will increase exponentially with the number, but if I take the limit as the number goes to infinity, my EV goes to -100%.

So I have two questions: How do you reconcile this mathematically, and what is the correct (ie highest EV) action with respect to the man's bets? The answer can't be to keep betting infinitely (ie until you lose) because that makes your EV -100%. Yet you also can't set a finite number of bets that would be correct because it's always +EV to make at least one more bet. So what is the solution?
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  #2  
Old 11-09-2007, 08:34 AM
jason1990 jason1990 is offline
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Default Re: Infinite bankroll paradox

Everybody wants EV. Play a +EV game, and earn "Sklansky bucks." Got a decision to make? Turn off your brain and pick the one with the highest EV. EV, EV, EV. It is all about EV.

Really, EV is just an abstract mathematical concept. You cannot take Sklansky bucks to the bank. If your decision has an EV of 10 and mine an EV of 9, what do you have that I do not? An extra unit of EV. What are you going to do with it?

What people so easily forget (or perhaps never learned in the first place) is that the value of EV comes from the Law of Large Numbers (LLN), which states that if you repeat the same wager many times independently, then your winrate will be close to the EV of the wager. By itself, EV is just an abstract number. It takes the LLN to make a +EV game desirable.

In the situation you described, the LLN does not apply. You are not repeating the same wager, since the wagered amount keeps changing. And the wagers are not independent, since the wagered amount depends on previous results. So you cannot use the LLN to translate the +EV of each individual wager into a long term positive result. Clearly, you already realized this, but I wanted to point out that it is not a paradox. It only looks like a paradox if you forget that EV requires the LLN to be valuable.

[ QUOTE ]
what is the correct (ie highest EV) action with respect to the man's bets?

[/ QUOTE ]
You are assuming that "correct" = "highest EV". You are falling back on a general principle (always maximize EV) without thinking critically about whether or not it applies. You are not alone in this. Many people have it ingrained in their heads that the optimal choice in any situation is the one with the highest EV. They live by this motto without ever stopping to ask themselves why it is true. Stop and think. Highest EV is correct when that EV can be translated into long term profit. EV turns into long term profit by the LLN. The LLN depends on repeating the same wager many times independently. Are you not repeating the same wager many times independently? Then you cannot use the LLN. So you cannot translate your EV into long term profit in the usual way. So maybe the highest EV choice is not in your best interest.

At any step in this game, the highest EV choice is obviously to continue betting. As you pointed out, this is clearly not the best choice, by any reasonable definition of "best." There is no 100% objective answer to when you should quit. After every win, you will need to make a subjective evaluation to decide whether or not to continue.

You may be interested in reading about the concept of utility functions. The Kelly criterion is also connected to your question. In addition, you can read this post for some comments of mine which are related. Here is the relevant part:

[ QUOTE ]
Now, you also brought up EV and decisions. Probability theory (and science in general) cannot tell you what decisions you should make. It can only tell you the consequences of the various decisions you are considering. It is up to you to decide which consequences you prefer. Physics, for example, cannot tell us whether we should split an atom. It can only tell us what will happen if we do. The same is true of probability theory.

As far as EV is concerned, the relevant theorem is the Law of Large Numbers (LLN). The LLN, as you probably know, says that if you perform a sequence of independent and identically distributed wagers, then the overall average rate of change of your bankroll will converge to the EV of a single wager in that sequence. People often ignore the details of what the LLN says, and simply act as though it says you should always maximize your EV. But it is a mistake to ignore the details. In particular, it is mistake to ignore the fact that the LLN contains hypotheses which must be satisfied before it can be applied. In particular, the LLN requires not only that you have a sequence of wagers, but also that they are independent and identically distributed.

For example, suppose you are offered a 3:2 payout on the flip of a fair coin. What percentage of your bankroll should you wager on this bet? If you unthinkingly try to maximize your EV, then you will bet your entire bankroll. However, if you did this many times, you would eventually go broke. The LLN does not apply to a sequence of wagers in which you always bet a fixed proportion of your bankroll, because these wagers are not independent and identically distributed. The Kelly criterion will recommend a fraction smaller than 100% of your bankroll. If you accept Kelly's recommendation, then on a single wager, your raw EV will be smaller than it would be if you bet your whole roll. So Kelly, strictly speaking, is not recommending that you maximize your EV. If you study the Kelly criterion, you will find theorems that explicitly describe the long term consequences of following the Kelly system. The theorems do not say, "follow the Kelly system." They say, "if you follow the Kelly system, then here is what will happen." In some circumstances, the LLN is valid. In others, the Kelly theorems are valid. And in still others, neither might be valid and we may need to turn to something else in order to discover the consequences of our considered actions.

[/ QUOTE ]
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  #3  
Old 11-09-2007, 09:44 AM
pzhon pzhon is offline
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Default Re: Infinite bankroll paradox

[ QUOTE ]
I can set my total number of bets to an arbitrarily high number and my overall EV will increase exponentially with the number, but if I take the limit as the number goes to infinity, my EV goes to -100%.
...
How do you reconcile this mathematically


[/ QUOTE ]
f(y) is not always the limit of f(x) as x goes to y. This fails when f is discontinuous at y. A lot of reasonably nice functions are discontinuous at infinity, like sin(x). It's something you have to look for whenever you use infinity in a mathematical model. Discontinuity at a value is a hint that the value is outside the applicable context of the model.

[ QUOTE ]
what is the correct (ie highest EV) action with respect to the man's bets?

[/ QUOTE ]
First, E$ is not necessarily the same thing as E(Utility). You do not have to assume that the difference between being broke and having $1 billion is half of the difference between being broke and having $2 billion, so you do not have to find each toss better than not.

Second, there doesn't have to be an optimum. If the stranger says, "Tell me an integer, and I will give you that amount," there is no optimal strategy. If the stranger says, "Tell me a number strictly less than 1, and I will give you $1 with that probability," then there is no optimal strategy. There is neither a largest integer nor largest real number less than 1.
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  #4  
Old 11-09-2007, 10:04 AM
Uniqueuponhim Uniqueuponhim is offline
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Default Re: Infinite bankroll paradox

Your post does make a lot of sense, and answers a lot of questions, but I find myself in the dark about some things.

There is one assertion in particular that you've made which I'm unsure is correct and that is that the law of large numbers does not apply in this situation. The following is your definition of the LLN, which I completely agree with:

[ QUOTE ]
What people so easily forget (or perhaps never learned in the first place) is that the value of EV comes from the Law of Large Numbers (LLN), which states that if you repeat the same wager many times independently, then your winrate will be close to the EV of the wager. By itself, EV is just an abstract number. It takes the LLN to make a +EV game desirable.

[/ QUOTE ]

So in this situation, what we do is look at each individual situation and say "if I performed this exact bet a large number of times, how much would I end up with on average?" and you get the answer of 150% of what was bet. Therefore, on average, you make 50% profit every time you make the wager and each individual bet has an EV of +50%. There's no need to compare that bet to past or future ones in the hypothetical situation, just like in a poker game you don't compare your current hand with past or future ones to get the EV of different actions. Rather, you *simulate* the exact hand a large number of times and look at what your average profit or loss is for each decision. That doesn't mean that that exact hand actually has to happen a large number of times, you just pretend it does and look at the average result. Even looking at individual streets within a hand where the amount in the pot (and therefore the amount wagered) depends on past streets, you still just take that exact situation and iterate it a large number of times to get your average expectation value for various decisions, and both EV and LLN still apply. So how is this hypothetical situation really any different? The amount wagered in each bet may depend on the outcomes of previous bets, but each one can still be looked at individually and iterated a large number of times to get your average EV.

However, all of this does bring me to the idea of bankroll management, where you play the highest stakes that you can while retaining a negligible Risk of Ruin. As you mentioned, playing with your entire bankroll may have the highest EV in the short term, but in the long term you'll end up broke. Calculating the stakes of the game you should play seems to be a problem very similar to the one I mentioned above: How do you balance EV with RoR? I would imagine it would have something to do with the utility of an extra dollar decreases with the number of dollars you already had. But how do you actually perform the calculations?

I'm also curious about another problem which is very similar to the first, but also different on many levels:
Say that instead of wagering money on each game, you play the following game against a friend:

Neither of you can know what your opponent is doing or how many points they have until you both decide you're finished and compare points to see who won.
You each start with one point, and each round you can decide to either wager all of your points in the same manner as the first example or finish playing. If you lose a wager you're forced to finish playing. What is the optimal strategy to win this game?
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  #5  
Old 11-09-2007, 12:34 PM
Pokerfarian Pokerfarian is offline
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Default Re: Infinite bankroll paradox

From a real world point of view, the obvious point is that 2 billion is not twice as good as 1 billion.
From a mathematical point of view, you're wrong, your EV does not tend to 0 as (number of bets) tends to infinity.
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  #6  
Old 11-09-2007, 01:29 PM
pococurante pococurante is offline
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Default Re: Infinite bankroll paradox

The answer is 0. You should never put yourself in a situation where you're 25% to lose everything you own.
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  #7  
Old 11-10-2007, 05:39 AM
Syntec87 Syntec87 is offline
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Default Re: Infinite bankroll paradox

[ QUOTE ]
The answer is 0. You should never put yourself in a situation where you're 25% to lose everything you own.

[/ QUOTE ]

winnar! (Dad, Mom, is that one of you?)
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  #8  
Old 11-10-2007, 09:47 AM
R Gibert R Gibert is offline
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Default Re: Infinite bankroll paradox

Your conclusion that the EV is -100% is not right. What you did was separate out the calc of the frequency with which you are a winner, which is 0% and concluded that the EV was -100%. With more care, you would have realized that this approach actually results in an indeterminate form of 0*infinity e.g.

EV = lim ((3/4)*(2))^n - 1, as n goes to infinity
= (lim (3/4)^n)*(lim (2)^n) - 1, "
= 0 * infinity - 1, "
= 0 *** Not Kosher ***

Of course this is not correct. This approach does not work. Better is

EV -> ((3/4)*(2))^n - 1, as n -> infinity
-> lim 1.5^n - 1, as n -> infinity
-> infinity, as n -> infinity

So the conclusion is that the EV goes to infinity as n goes to infinity. The frequency calc portion you made is okay. This means that you wind up a winner overall, but you are a winner a vanishingly small percent of the time. A curious result, but correct.

Analogously, you can also devise a scenario where it is -EV, but you are nevertheless a winner virtually 100% of the time.

Reference:

http://en.wikipedia.org/wiki/Indeterminate_forms
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  #9  
Old 11-10-2007, 11:18 AM
R Gibert R Gibert is offline
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Default Re: Infinite bankroll paradox

This line:

= 0 *** Not Kosher ***

should have been:

= -1 *** Not Kosher ***

Which was(is?) your conclusion.
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  #10  
Old 11-10-2007, 03:36 PM
jason1990 jason1990 is offline
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Default Re: Infinite bankroll paradox

[ QUOTE ]
just like in a poker game you don't compare your current hand with past or future ones to get the EV of different actions. Rather, you *simulate* the exact hand a large number of times and look at what your average profit or loss is for each decision. That doesn't mean that that exact hand actually has to happen a large number of times

[/ QUOTE ]
Suppose you play 5/10 limit. On the turn, you have a flush draw and are faced with calling a $10 bet in a $60 pot. You calculate that you have +EV, so you call. If you continue to play 5/10 limit, then you are going to face this same situation many, many times in the future. It really will happen a large number of times. Your +EV really is going to turn into long term profit by the LLN.

Now suppose differently. Suppose you are very foolish. Every time you sit down to play poker, you play the highest stakes NL you can, and place your entire bankroll on the table. On the turn, you have a flush draw and are faced with calling all-in for your entire bankroll with 6 to 1 pot odds. If this situation repeats many times and you call every time, then you will eventually go broke. The EV of calling is still positive, but this +EV will not turn into long term profit by the LLN. So calling every time is no longer in your best interest.

[ QUOTE ]
Say that instead of wagering money on each game, you play the following game against a friend:

Neither of you can know what your opponent is doing or how many points they have until you both decide you're finished and compare points to see who won.
You each start with one point, and each round you can decide to either wager all of your points in the same manner as the first example or finish playing. If you lose a wager you're forced to finish playing. What is the optimal strategy to win this game?

[/ QUOTE ]
This is interesting. Let me add to some things to it, so that we may analyze it. At the end, the person with fewer points pays the person with more points $1, with no money changing hands on a tie. The game will be repeated many times, so that our objective is to maximize EV.

I will choose to play n flips; n can be 0, in which case I just stand with my 1 point. My friend will choose m flips. Here is one possibility. I will choose n = 1, 2, or 3, with probabilities 33/117, 4/117, and 80/117, respectively. If my friend chooses m = 0, then I win if I do not bust and I lose if I bust. So my conditional EV, given that I chose n, is

p^n - (1 - p^n) = 2p^n - 1,

where p = 3/4. Hence, my overall EV is

(33/117)*(1/2) + (4/117)*(1/8) + (80/117)*(-5/32) = 1/26.

If my friend chooses m = 1, 2, or 3, then a tedious calculation will show that my EV is 0. If my friend chooses m > 3, then I win if he busts and I do not, and I lose if he does not bust. So my conditional EV is

p^n(1 - p^m) - p^m
= p^n - p^m(1 + p^n)
>= p^n - p^4(1 + p^n)

Evaluating this gives

n = 1 --> 201/1024
n = 2 --> 279/4096
n = 3 --> -459/16384.

Hence, my overall EV is greater than or equal to

(33/117)*(201/1024) + (4/117)*(279/4096) + (80/117)*(-459/16384) = 513/13312.

So we see that if my friend chooses m = 0 or m > 3, then I have positive EV; and if he chooses m = 1, 2, or 3, I have 0 EV.

I believe I have a proof that this is the only equilibrium solution, but my calculations are very tedious. Perhaps there is a simpler analysis.
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