A Less Obvious Martingale Fallacy

[mosta]

I shouldn't have sounded like I was criticizing every post in the thread. it's certain posters in particular that always impress me in a certain way...

[mosta]

is your claim that EV is < 0? skimming the thread I didn't notice any mathematical statements from you. I'd LOVE to see you try to _prove_ this (ie, not with analogies and word pictures). do you do any proofs on near 200 score IQ tests?? do you know what a proof is??

[Dov]

Thanks Mosta,

This is a good thread.

I wish you had linked it a couple of days ago.

[PairTheBoard]

[ QUOTE ]
here's an eg. I post at the end just to cite/paste an email from a mathematician friend. some of the discussion gets highly techincal, but if you focus you can keep track of the central issues:

archive thread

edit: you can make various statments that intuitively sound equivalent (eg: 1. P of eventual succes = 1 (yes) 2. EV > 0 (no--it is undefined!)), but when you do some math instead of talking out your a$$, you get some results, some surprising, some confusing, some not.

[/ QUOTE ]

Here is the last post in that archive thread. As it happens, the post is by mosta:

mosta --
"after consulting with my resident expert, I'm going to vote--Henke 1, bigpooch 0, on split decision: I think it's incorrect that EV=1, but no one says why saying that roulette trick works has to mean that EV>0. good bout all around!.

comments quoted/ lifted from personal email:

Q: Does there exist a sequence of random variables Y_n such that
i) Y_n is almost surely eventually 1, and
ii) E(Y_n) is negative for all n?

["Almost surely" is a technical term meaning "with probability 1".]

The answer is yes, and winnings from the doubling strategy (p individual win
< 1/2) is an example. The first post in the string proved i) and Henke's
first post proved ii). [Well, both of them close enough for government work,
anyway -- they're right in spirit at least].


----------


bigpooch started off on the wrong foot when he tried to prove i) implies
E(Y_n) -> 1 as n gets large:
"Then one can see that
E(F(tau)) = 1 or lim (n --> infinity) E(F(n)) --> 1. "
This is just false --- he's confusing the process at the stopping time
F(tau) [about which he has made a true statement] with the unstopped process
F(n).

-------


There were several side issues, but nobody asked the real question. Why is
ii) so bad for gamblers? Note Henke's first comment:
"You only proved that the probability of at least one win approaches one as
the number of trials approaches infinity. However, to state that the
roulette trick works, I think you would have to prove that it's +EV which
isn't the same thing."
He's absolutely right "[it] isn't the same thing", but why should "roulette
trick works" mean "+EV"? Unfortunately, all subsequent debate is focused on
attacking the part he was right about.

------


Side issues: Henke's distinction between being zero and approaching zero is
not meaningful, but he is right when he says you need further assumptions to
make sense of 0*infinity. bigpooch was pretty careless to let the 0*infinity
genie out of the bottle.

------


Haha, I just read some more of the thread, where bigpooch makes this
seemingly ironclad claim:

E = +$1 x P(tau=1)
+$1 x P(tau=2)
+$1 x P(tau=3)
+...

and then one post later accuses Henke of the mistake _he_ just made:

"Okay I understand your mistake in thinking: going to the
infinite case is not necessarily the same as taking the
case for finite n and taking the limit as n approaches
infinity."

Elaboration: If Y_n is our net change in wealth after playing the nth game,
then bigpooch has implicitly assumed that there is a limiting random variable
Y = lim Y_n as n gets large and that E(Y) = lim E(Y_n). [You can check (as
Henke did) that E(Y_n) is getting large and negative with n (if p<1/2), so
this equality of expectations is definitely violated.] If bigpooch were able
to make sense of the Y = lim Y_n statement (he would say "pointwise a.e."),
then indeed Y would be the constant 1, but he would not be able to obtain
the limit in a way necessary to justify the E(Y) = lim E(Y_n) statement (he
would have to say "in L^1")."
=============================
=============================



It seems to me that drudman's assertion that he will just pile up win after win as each martingale series ends, is countered by the simple observation that the expected value of his net gain/loss after n spins, En, is negative and the limit of En as n goes to infinity is negative infinity. It looks like mosta is saying this is true Even if the Martingaler stops after one win. I was not sure about this, although I was convinced it was true in the MMMMM model where the Martingaler plays on forever.

I also think that En can be seen at work in the model of a properly expanding pool of Martingalers at play.

I think in order to really find satisfaction with this thing we need more statements about the probabilty space of infinite sequences of wins-losses when coupled with the varying bet amounts. People want to know what happens "in the end" which involves the infinite sequences in this case. I understand identifying the sequences with binary expansions of real numbers and using the lesbegue measure on the space, although I'm not exactly sure why that should be the obviously correct way to do it. Maybe that yields nothing of particualar interest though, because the sequences almost surely cannot be identified with specific dollar results. Or can they?



PairTheBoard

[MMMMMM]

[ QUOTE ]
is your claim that EV is < 0? skimming the thread I didn't notice any mathematical statements from you. I'd LOVE to see you try to _prove_ this (ie, not with analogies and word pictures). do you do any proofs on near 200 score IQ tests?? do you know what a proof is??

[/ QUOTE ]

Are you addressing me?

Of course the gambler's EV is less than zero on a roulette wheel, using the Martingale or any other system.

EV = total volume bet * advantage (or disadvantage). In this case the advantage for the gambler is negative on every spin (positive for the house), so no matter how many spins he might make (>1) his EV is <0. The house edge on a roulette wheel is fixed somewhere around 5.25% as I recall. So his EV would be -5.25% * total amounts bet.

Also, I never claimed an IQ near 200.

I do sense a hostile tone in your post. Why?

[MMMMMM]

So, Mosta, how does this point that impress you:

As the projected chance of eventual success doubles with each projected future trial, the amount that must be wagered at that point doubles also. So if you are going to use limits (or any other principle) to assert that the chance the gambler will lose a future bet is zero, you must also use the same principle to assert that the gambler has to be able to wager his entire infinite bankroll on one spin.

You can't claim that halving forever is equivalent to zero, unless you agree that doubling forever is equivalent to infinity.

[MMMMMM]

It is also a mathematical fact that NO betting system can overcome the house advantage when utilized against an independent-trials-based game with a built-in house advantage on every trial. The house extracts an EV of the house edge times the volume bet.

I will not attempt to PROVE it, but any mathematician will tell you that that is so.

It should also be obvious that it is so: because the EV for the gambler is negative on every single roll.

Adding a string of negative numbers cannot produce a positive number. Not a formal proof, but pretty basic, and clearly true.

[MMMMMM]

All gambling games have an EV of: Advantage * TotalBet.

The gambler has a disadvantage against the roulette wheel ON EVERY SINGLE WAGER.

You can't sum a string of negative expectations and obtain a positive expectation. You can't sum a string of negative numbers and obtain a positive result.

Hence those who think an attempted Martingale series has an EV of 1 are wrong.

This has nothing to do with bankroll limits. It has everything to do with the fact that no amount of negatives, summed, produces a positive.

Very simple when looked at in this way.

[Dov]

[ QUOTE ]
All gambling games have an EV of: Advantage * TotalBet.

[/ QUOTE ]

Neither of these numbers are negative. Therefore this will not be a negative number, ever. In Martin's case, (Advantage * Bet) - Bet will be negative, but that's not how we calculate EV as you pointed out.

[ QUOTE ]
Hence those who think an attempted Martingale series has an EV of 1 are wrong.

This has nothing to do with bankroll limits. It has everything to do with the fact that no amount of negatives, summed, produces a positive.

Very simple when looked at in this way.

[/ QUOTE ]

I like the way you tried to simplify it, but I don't think it works. You are not, in fact, adding negative numbers, you are adding numbers that are smaller than your initial wager until the win (I know, I'm agreeing with you) which is greater than the sum of the losses by 1. The net effect is that you are adding 1/n trials where n is the number of trials in the series.

That gives me an idea. How's this for a rudimentary proof?

Since the house advantage is fixed, it is impossible for it to increase or decrease on any given spin of the wheel.

Since the Martingaler's 'advantage' per spin = 1/n where n is the number of trials to complete a series, his advantage fluctuates with each series.

If the Martingaler's advantage changes, then the house edge cannot be fixed.


Example:

Lose - Win

If he wagers $1 on the first bet and loses, then wagers $2 on the second bet and wins, then Martin will be up $1 for 2 spins or $.50 / spin.

Lose - Lose - Win

-$1, -$2, +$4 = +$1 over 3 spins = .33 / spin

It feels like there's something wrong with this, but I'm not sure what it is.

I should go back to sleep.

Any good?

[MMMMMM]

Hmm, Dov, not sure if I made myself clear or if I followed your point entirely, but quickly:

The house's advantage is fixed on every spin of the roulette wheel. So too is the Martingaler's disadvantage.

If the house has an positive expectation of 5.25% on every spin, then the Martingaler has a negative expectation of 5.25% on every spin. Neither party's advantage or disadvantage on each spin fluctuates according to the bet sizing or patterns or anything else. It is simply fixed for each trial.

A string of negatives cannot be summed into a positive.