A Less Obvious Martingale Fallacy - Page 11

MMMMMM · #**101** 07-18-2005, 05:00 PM

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The thing is that I can choose to stop after I win my series

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No you can't. In this game the Martingaler plays forever,

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Ok, but isn't the whole point of the theory that it is broken into series and you can stop after a win? I'm not sure it is supposed to be valid if you aren't given that ability.

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Even if you plan to stop, you must realize that you are gambling that you will eventually win a bet. You aren't absolutely 100% sure you will, though. And the penalty for not winning a bet increases geometrically after each losing spin.

Even if you have lost 15 bets in a row, or 15,000 bets in a row, the odds of your winning the *next* spin are still less than 50%. So while you may look into the future, saying "I have to win a bet *eventually*", the wheel doesn't know that, and each new spin is a new chance because the wheel has no memory. Meanwhile your losses are multiplying geometrically.

It may seem like it is too farfetched to imagine that you might NEVER win another spin. But if you had essentially unlimited players each playing on independent wheels, there could be one player who just loses every single bet he makes. This is parallel to the chances of your sitting down and just losing every bet--it's an infinitesimally small chance, but given unlimited time, you could fall into that slot. And the finacial penalty would be unlimited, too.

For every winning player or winning sequence, there is a theoretical counterpart which is losing the same amount or more. So theoretically the system isn't going to produce a profit even with no restrictions on table limits or bankroll.

MMMMMM · #**102** 07-18-2005, 05:09 PM

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Whoa, you're saying that in an infinite series of rolls, he will necessarily have an infinitely long losing streak?

I'm pretty sure that's not true.

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In an infinite series of rolls, which is akin to an infinite number of players playing independently, there could well be one player who just never wins. That doesn't mean his losing streak will necessarily continue to extend to infinity. It just means that no matter where you are on the timeline when you ask him, that's the result he will be having at that time.

At least, that's the way I see it.

bobman0330 · #**103** 07-18-2005, 05:15 PM

I don't think you understand MMMMMM... If you strip the problem of all meaning and make a lot of shaky assumptions, you can WIN ONE UNIT!!

The casino shouldn't be too concerned though, because in any real-time setting, they can put all the money the Martingaler is currently down, keep it in a money-market account, and make a ton of money in interest.

Dov · #**104** 07-18-2005, 05:28 PM

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I don't think you understand MMMMMM... If you strip the problem of all meaning and make a lot of shaky assumptions, you can WIN ONE UNIT!!

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LOL, I know you're just being sarcastic, but...

What if we make each unit infinite too? Then would it be worth winning JUST ONE?

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The casino shouldn't be too concerned though, because in any real-time setting, they can put all the money the Martingaler is currently down, keep it in a money-market account, and make a ton of money in interest.

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Until they lose a ton of money in early withdrawal fees when he wins a series earlier than they expected.

pzhon · #**105** 07-18-2005, 05:34 PM

Here is a proof I think is more intuitive, with more details in white:

Let p(n) be the probability you have never been ahead before the nth wager.

The expected first time to be ahead is the sum of the p(n).
<font color="white">In the usual sum for the expected value, change i to the sum from n=1 to i of 1. Then change the order of summation.</font>
p(2k-1)=p(2k)= (2k C k)/2^(2k) ~ 1/sqrt(pi k)
<font color="white">Use the reflection principle to write the number of lattice paths to (a,b) staying above the diagonal as a difference between binomial coefficients. The sum telescopes. Then use Stirling's approximation for n!.</font>
The sum of c/sqrt(n) diverges, so the expected number of turns before your first win is infinite.

MMMMMM · #**106** 07-18-2005, 05:37 PM

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Whoa, you're saying that in an infinite series of rolls, he will necessarily have an infinitely long losing streak?

I'm pretty sure that's not true.

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In an infinite series of rolls, which is akin to an infinite number of players playing independently, there could well be one player who just never wins. That doesn't mean his losing streak will necessarily continue to extend to infinity. It just means that no matter where you are on the timeline when you ask him, that's the result he will be having at that time.

At least, that's the way I see it.

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Let me be clear about something here, because I realize this could quickly get a lot more complicated. I'm not saying it's guaranteed that at any time you ask, there will be someone who has never won a bet in their current series. I'm saying it could well happen. And I'm saying that no matter when or whom you ask, you can't be absolutely sure it won't happen in the future.

SomethingClever · #**107** 07-18-2005, 05:38 PM

Bottom line... if you can always bet 2x after a loss, you can choose to be a winner as long as you stop on a win.

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If you were truly to play this game eternally, your chips would slowly go down, on average, due to the house edge.

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The "average" amount of your bankroll is meaningless, though, when you can recoup all your losses with a single bet.

Sure, if you play eternally, you will experience bigger and bigger losses, on average, but if you quit on a win, you have won.

Edit: Oh, and the probability of being a losing player under the system with an infinite bankroll, infinite betting limits and an infinite amount of time appears to be infinity - 1. Do you see why? [img]/images/graemlins/tongue.gif[/img]

hoyaboy1 · #**108** 07-18-2005, 05:44 PM

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For every winning player or winning sequence, there is a theoretical counterpart which is losing the same amount or more. So theoretically the system isn't going to produce a profit even with no restrictions on table limits or bankroll.

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For this to work, the odds of never winning x potential losses would have to be greater than the odds of eventually winning x 1 unit, right? Is this necessarily true?

Edit - I'm sure this can be figured out, but I haven't done much stats stuff in a while and forget how

bobman0330 · #**109** 07-18-2005, 05:46 PM

Here's the killer proof:

Eventually, the house will have so many of your checks stacked up on their racks that their mutual gravitation will overwhelm the electrostatic and other quantum mechanical forces that hold atoms apart, causing the checks to collapse into a black hole. Special relativity and quantum mechanics as we understand them today state that it is impossible to retrieve any matter from a black hole. Ergo, your chips will be lost forever.

Grivan · #**110** 07-18-2005, 05:50 PM

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Bottom line... if you can always bet 2x after a loss, you can choose to be a winner as long as you stop on a win.

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You cannot have it both ways. You can't have the game go on for infinite and have the ability to stop on a win. Since if you stop it is no longer an infinitly long game. So stopping at any point is not valid. What people are doing is coming up with a way to measure who is winning, since you cannot reasonably use a single point in time to do this.