A Less Obvious Martingale Fallacy - Page 6

PairTheBoard · #51 07-17-2005, 03:47 PM

So the expected number of coin flips to win a dollar betting a dollar at a time is infinite. But the expected number of coin flips to win a dollar using a Martingale looks like:

1/2 + 2/4 + 3/8 + 4/16 + 5/32+ ... =
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... +
000 + 1/4 + 1/8 + 1/16 + 1/32 + .... +
000000000 + 1/8 + 1/16 + 1/32 + .... +
000000000000000 + 1/16 + 1/32 + .... +
0000000000000000000000 + 1/32 + .... +
.
.
.
= 1 + 1/2 + 1/4 + ... = 2

Can that be right? You mean that using a Martingale reduces the expected value for the number of coin flips to win a dollar from infinite to 2?

PairTheBoard

PairTheBoard · #52 07-17-2005, 03:52 PM

I think you should try putting yourself in the Casino's shoes and saying to yourself, I'm going to judge how I'm doing by looking at the totals when the Martingale gambler is way behind after a long string of losses.

PairTheBoard

SheetWise · #53 07-17-2005, 03:53 PM

I think the problem in this thread concerns a probability of zero. There is an assumption that if an events probability has any value, that given an infinite number of trials it will eventually occur. I disagree.

[ QUOTE ]
Corollary I: The sum of the probabilities of any event and its complement is unity. That is P = 1.

The complement of an event is, of course, the nonoccurrence of that event. Characterizing an impossible event, we can state;

Corollary II: The probability of an impossible event is zero, or P = 0.

It is worth noting that the converse of Corollary II is not true. Given that the probability of some event equals zero, it does not follow that the event is impossible (there is a zero probability of selecting at random a pre-specified point on a line).

Theory of Gambling and Statistical Logic, Richard Epstein, page 15

[/ QUOTE ]

There are many events that I will easily assign a probability of zero, even though I know the probability is a number above zero.

I believe the probability of playing golf on a full length 18 hole PGA course, and shooting a score of 18 is impossible. I would without thought use this value in any equation.

I believe the probability of having a single selection in the PowerBall lottery, and selecting the winning combination in four consecutive draws is zero.

In the first example, the probability is obviously not zero -- but it's a perfectly good number to use. In the second example I can even assign a numeric value to it. Just because something can happen -- such as a run of 1000 heads or tails from a fair coin -- does not mean it ever will.

The theoretical concept of infinity is not really necessary for this problem, a definition of a theoretical zero is, then you can calculate some very large, yet finite, numbers.

drudman · #54 07-17-2005, 04:12 PM

As long as each series can complete (and I think, by definition, it must) you will always win in the long run.
Isn't this correct? I can't see how it couldn't be.

PairTheBoard · #55 07-17-2005, 04:12 PM

Saying that it "will eventually occur" is a somewhat loose use of language. What can be said more precisely is that if the probabilty of the event occuring in 1 trial is strictly positive, Then the limit of the probability that it occurs in N trials goes to 1 as N goes to infinity.

Here's an example you might like. Let the random variable X be uniformly distributed on the interval [0,1] - ie. X is equally likely to be in subintervals of the same length. Let x be the outcome of a Trial for the random variable X. ie. x is a specific number in the interval [0,1]. Then x has occurred as a result of the Trial but prior to the trial the probabilty for x occurring was 0. The event {x} had zero probabilty yet it happened anyway.

PairTheBoard

Dov · #56 07-17-2005, 04:34 PM

[ QUOTE ]
think you should try putting yourself in the Casino's shoes and saying to yourself, I'm going to judge how I'm doing by looking at the totals when the Martingale gambler is way behind after a long string of losses.

[/ QUOTE ]

That's what I was doing before when I asked why casinos bother with table limits.

The answer I got was that the variance will kill them.

I know this is correct, but I also know that the gambler MUST win if he is allowed to play each Martingale series to completion.

Dov · #57 07-17-2005, 04:39 PM

It's funny, but this reminds me of the other thread where they were discussing traversing a distance in steps that are each half as large as the next.

When someone finally says, "I don't get it, just take your first step twice!"

That's what the martingale does. Takes the first step as often as necessary to get to the other side.

PairTheBoard · #58 07-17-2005, 04:43 PM

[ QUOTE ]
[ QUOTE ]
think you should try putting yourself in the Casino's shoes and saying to yourself, I'm going to judge how I'm doing by looking at the totals when the Martingale gambler is way behind after a long string of losses.

[/ QUOTE ]

That's what I was doing before when I asked why casinos bother with table limits.

The answer I got was that the variance will kill them.

I know this is correct, but I also know that the gambler MUST win if he is allowed to play each Martingale series to completion.

[/ QUOTE ]

So the gambler must "win" AND the Casino must "win". There will be times when the gambler is ahead and there will be times when the Casino is ahead. So how do you measure who is doing better?

PairTheBoard

SheetWise · #59 07-17-2005, 04:47 PM

It's wrong to look at the outcome of a trial (series of trials) and then assigning it a probablity. What you are calculating now is the probability that it will re-occur.. And that probability may well be zero.

There are many things we can observe -- then assign a probability -- and conclude that they will never happen again. The leap in logic is to assign a probability of zero to an event you just witnessed.

SheetWise

Dov · #60 07-17-2005, 04:52 PM

[ QUOTE ]
So the gambler must "win" AND the Casino must "win". There will be times when the gambler is ahead and there will be times when the Casino is ahead. So how do you measure who is doing better?

[/ QUOTE ]

The casino will NOT win if the gambler is allowed to always complete the series. It's impossible.

If there is a precondition that the gambler is always allowed to complete the series, then the casino must account for its reverse implied odds. They are actually paying out more than they are winning provided that the gambler can always complete his series.

Paradoxically, the only way the casino can win here is to put an end to his play or stop the martingale.