Sklansky, tortoise, hare.....why?

[Johnny 99]

Sklansky writes: [ QUOTE ]
your excess winning or losing swings do not increase proportionately to the number of hours you play. Rather, they increase proportionally to the square root of the number of hours played.


[/ QUOTE ]

I'm curious how this conclusion is reached (and I have no cause to doubt its validity). Sure one's win rate will start to chip away at their bad luck until they are no longer losing, but how could we know that it increases proportionally to the square root of the number of hours played?

Is it the result of empirical studies? Is it derived from self evident principles? Does it just "sound about right"?

I have similar questions as to how statistical concepts are derived and perhaps the answer to this will help.

Thanks,

[wagon30]

I believe Sklansky is referring to the fact that when you add n independent identically distributed random variables the variance of the final variable is sqrt(n)*Var(X)

That being said, I don't understand the article. Variance is not the most you can win or lose in a session. The most you could win or lose in a session has doesn't have that much explanatory power. Obviously, when he says,"almost certainly be between -50,000 and 50,000." He means getting a result of -50,000 is a rare event, but certainly possible.

Obviously though, if 49.99% of the time he wins $1 and 49.99% he loses $1, and the remaining times are equally divided between him winning or losing $500, then the range of "likely" events after 100 sessions can be narrowed considerably.

[wagon30]

IMO they would have been better off publishing BruceZ's posts in this thread.

[wagon30]

I'm wrong. The sum of variances of iid variables is n^2var(x), so I don't know what he is referring to.

[PairTheBoard]

[ QUOTE ]
I'm wrong. The sum of variances of iid variables is n^2var(x), so I don't know what he is referring to.

[/ QUOTE ]

You are wrong on both counts.

If X1,X2,...Xn are independent random variables and
Y = X1+X2+...+Xn then
Var(Y) = Var(X1)+Var(X2)+...+Var(Xn)

The important thing is that they are independent. They need not be identically distributed. However, if the Xi are iid as X then the above simplifies to:

Var(Y)=n*Var(X)

But it's your standard deviation that determines "your excess winning or losing swings" as Sklansky puts it. Taking the square root of both sides above you get:

sigma(Y) = sqrt(n)sigma(X)

where sigma is the standard deviation of the random variable.

Your expected win increases in proportion to the number of hours played while your standard deviation increases in proportion to the square root of the number of hours played. What this means is that as your hours increase, excessive swings decrease as a proportion of your expected win. That's what causes your actual win rate to "converge" to your expected win rate over the long run.

PairTheBoard

[wagon30]

Ok, I see that your right. The variance of the sum is indeed the sum of the variances (I always forget such facts and try to derive them). But I if he is talking about standard deviation, why does he call it the most you can win or lose in a game. I think that is misleading.

Also, I can't figure out why this didn't work.

Y=X+X = 2X Var(Y)=4Var(X)
or more formally
Var(Y)=E[Y^2]-E[Y]^2
=E[(2X)^2]-E[2X]^2
=4Var(X)

I know the right way to do it, but I don't know why this wrong way is wrong. If someone could please advise.

[PairTheBoard]

[ QUOTE ]
if he is talking about standard deviation, why does he call it the most you can win or lose in a game. I think that is misleading.


[/ QUOTE ]

Where does he use the words, "most you can win or lose in a game"? According to the OP's quote he says, "your excess winning or losing swings". If he had used your words he would just be wrong. But "excess winning or losing swings" is another thing. The word "excess" is a little vague, but it's not too hard to see he is talking about the tails of a normal distribution and for those it's the standard deviation that most directly applies.

[ QUOTE ]
Also, I can't figure out why this didn't work.

Y=X+X = 2X Var(Y)=4Var(X)
or more formally
Var(Y)=E[Y^2]-E[Y]^2
=E[(2X)^2]-E[2X]^2
=4Var(X)


[/ QUOTE ]

The error here is that hour1 and hour2 must be represented by two distinct random variables, X1 and X2. While they are identically distributed, as say the random variable X, they are not the same random variable. It is not the case that X1+X2 = 2X. For example, X1+X2 :=: 2X1. Adding the result of the first hour's play to the second hour's play is not the same as multiplying the first hour's play by 2.

PairTheBoard

[wagon30]

Ok, should have been obvious thanks.

Here is Sklansky's 3rd paragraph:

Say you are a break even player, playing in a game where the most you would win or lose in a six hour session is $500. After a hundred such sessions (600 hours) it would be very wrong to conclude that your results would range from minus $50,000 to plus $50,000. Rather they would almost certainly range from negative $5,000 to positive $5,000. You multiply the one session possible swing by the square root of 100.

[PairTheBoard]

[ QUOTE ]
Ok, should have been obvious thanks.

Here is Sklansky's 3rd paragraph:

Say you are a break even player, playing in a game where the most you would win or lose in a six hour session is $500. After a hundred such sessions (600 hours) it would be very wrong to conclude that your results would range from minus $50,000 to plus $50,000. Rather they would almost certainly range from negative $5,000 to positive $5,000. You multiply the one session possible swing by the square root of 100.

[/ QUOTE ]

Ok. He's being a little loose with his language here, trying to make the point in a simple way I think. Technically, if the absoulte maximum you possibly could win or lose in a session were $500 then the absoulte maximum you could win or lose in 100 such sessions is $50,000. Technically that would be your "range". But it wouldn't be your likely range. And he makes it clear that's what he's talking about with the words, "almost certainly range".

PairTheBoard