Wizard is wrong on his Martingale analysis

I disagree that it all averages out.

Lets say you have a 1/2 chance of winning any bet (coin flip), then the chance of losing two times in a row is 1 in 4. That would be the case with steady odds. But lets take blackjack -- say one bet you have a 1/4 chance of winning, and the next bet you have a 3/4 chance of winning (exaggerated, but just for example). The average chance of winning overall is the same -- 1/2 as the coin flip. However, the chance of losing two in a row is now 3/16, which is less than 1/4 with constant odds. So when you have variable odds on each bet, you are less likely to have a losing streak then with a constant odds game. It doesn't matter whether or not you know when you have the 1/4 chance or the 3/4 chance, the very fact that it is variable is in your favor when calculating chance of streaks. That's the math.

Once again, with constant betting it makes no difference -- an average is an average. But when you are interested in streaks (Martingale), you are less likely to have a streak with a variable odds game than constant odds, even if it averages the same overall chance of winning. Now of course it is less likely to have winning streaks as well, but you don't care about that in a progressive betting system -- you just care about not having any streaks at all.

So if the variability is high enough, you might be able to make up for the house advantage for all practical purposes when playing Martingale? Take an extreme example, where half the time the odds are 1/10 to win and another time 9/10. The average is 1/2. But the chances of losing 5 in a row that might kill Martingale are not 1 in 64 but actually 27 in 1,000,000,000. Not that you can find a game like that... but you get my point. The question is, what is the most variable odds game? Is there one? That's where a system based on streaks can be used best...

Quote: andraaklly

I disagree that it all averages out.

Lets say you have a 1/2 chance of winning any bet (coin flip), then the chance of losing two times in a row is 1 in 4. That would be the case with steady odds. But lets take blackjack -- say one bet you have a 1/4 chance of winning, and the next bet you have a 3/4 chance of winning (exaggerated, but just for example). The average chance of winning overall is the same -- 1/2 as the coin flip. However, the chance of losing two in a row is now 3/16, which is less than 1/4 with constant odds. So when you have variable odds on each bet, you are less likely to have a losing streak then with a constant odds game. It doesn't matter whether or not you know when you have the 1/4 chance or the 3/4 chance, the very fact that it is variable is in your favor when calculating chance of streaks. That's the math.

Only if you assume a 1/4 game is immediately followed by a 3/4 game. It isn't, and if you knew it was you'd min-bet the first hand and Kelly bet the second. No Martingale required.

A better example would be to model as follows :

When you play a game, you have 50% chance of playing a game where you win 25% of the time and 50% chance of playing a game where you win 75% of the time. You have no idea a priori which of the two games you are about to play.

What's the chance of losing two in a row in this case?

Chance of winning a single game :

1/2 * 1/4 (playing the bad game)
PLUS
1/2 * 3/4 (playing the good game)

=> 1/8 + 3/8 = 4/8 = 50%.
=> Two in a row = 1 in 4.

The EV is zero. If you assume these are NOT independent trials, then you need to start looking at Bayesian probabilities. Blackjack is non-independent, which is why you count, but in that case instead of martingaling you're better to bet with the cards rather than betting due to a previous loss.

I have no idea where you get your 27 in 1 billion number from. You'll have to show your working as I think you are making assumptions on the games being dependent.

In fact, I don't see if it matters if some of the games are more or less fair... the pattern of results will still look like coin flips.

Actually the chance of getting a long losing streak is greater in blackjack than in other games (e.g. Roulette, Craps). Somebody posted an interesting analysis of it: http://wizardofvegas.com/forum/gambling/blackjack/2456-streaks-in-blackjack-some-actual-data/ For example, the dealer will get three times as many 10 win streaks as the player will. Those are $5120 losses in your progression system, not counting any double downs or splits. You make up this sizable disadvantage in split and double down situations, but you will still lose a great majority of the hands in the game.
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I agree with the person who said baccarat is the best game for Martingaling. But nobody here will agree with your proposition that Martingaling will create a positive expectation in a negative expectation game, and trying to prove that point will be like trying to break a rock with your head.

Quote: teddys

I think most people would disagree, that it's not a worthwhile thread. It's a quick and easy disproof. And you just proved your own point -- you can't turn lead into gold!

Please speak for yourself, or show us all all the "quick and easy disproofs" involved. (Bet you can't even begin, lol.)

The only thing for sure so far... mkl654321 has backed off a pretty-stupid (for such a renowned so-called teacher) remark about "turning lead into gold". (And apparently you have carried it through to this topic, itself, double lol.)

Quote: Garnabby

Please speak for yourself, or show us all all the "quick and easy disproofs" involved. (Bet you can't even begin, lol.)

The only thing for sure so far... mkl654321 has backed off a pretty-stupid (for such a renowned so-called teacher) remark about "turning lead into gold". (And apparently you have carried it through to this topic, itself, double lol.)

Yo, Garnabby, dude. I'm not renowned, but I AM a teacher. And if you think that it's stupid to say that you can't turn lead into gold--well, I have news for you--you can't. (And before you bring up the nuclear reactor process that someone else mentioned, that isn't turning lead into gold, it's breaking down gold into its constituent atomic particles and rearranging them into lead, or vice versa--but in each case, some material is lost/discarded.)

I guess I have to presume that you don't get the point of the analogy--that in the same way a knowledgeable chemist would not bother to investigate the intricacies of a claim to be able to (chemically) turn lead into gold, a knowledgable gambler should not bother to examine in detail a claim for an effective Martingale system.

And far from "backing off"--I REITERATE that statement.

Quote: Garnabby

Please speak for yourself, or show us all all the "quick and easy disproofs" involved. (Bet you can't even begin, lol.)

Well, lol, there is a quick and easy, lol, disproof, lol, that can be easily stated, lol, thusly: it is impossible to add a series of negative numbers and end up with a positive number, lol, as a result.

Lol.

Quote: mkl654321

Well, lol, there is a quick and easy, lol, disproof, lol, that can be easily stated, lol, thusly: it is impossible to add a series of negative numbers and end up with a positive number, lol, as a result.

Lol.

Quote: mkl654321

Now, you're starting to babble.

And by the way, there's no such word as "there're" (it is indeed used quite a bit, but so are words like "anyways").

Quote: Garnabby

Nothing

Thanks for contributing.

Quote: andraaklly

I disagree that it all averages out.

http://wizardofvegas.com/forum/gambling/blackjack/2456-streaks-in-blackjack-some-actual-data/

There is some data here on streaks in blackjack. Look at dealer streaks for lets say 5. If your hypothesis is true, then the number of dealer streaks that end at 5 should be more than the sum of dealer streaks of 6, 7, 8, etc. More generally, streaks of n would be greater than the sum of all n+i streaks where i goes from 1 to infinity. This notion would imply that the dealer losing 5 and then giving a way a winner is more likely than the dealer continuing to win. If you add up the numbers here, you will find that that the number of streaks of lesses >5 are higher than the total number of 5 loss streaks.

It is now up to you to either refute this evidence through finding potential flaws of calculation or present your own statistical analysis that refutes these findings.

Quote: andraaklly

I disagree that it all averages out.
[snip]
So if the variability is high enough, you might be able to make up for the house advantage for all practical purposes when playing Martingale?

No, because increasing the variance (variability) of a series of bets doesn't change the mean (house edge). Consider two extremes:
A) You take $100,000 and make a single blackjack bet. The house edge (in dollars) is approximately $500. Variance is through the roof.
B) You take the same $100,000 and flat-bet $1 for months in a row until you've bet $100,000 in action. The house edge in dollars is still approximately $500, but variance is much, much lower.

If you play a Martingale with $100,000 in handle, your variance will be somewhere between A and B. It can't be greater than A, and it's certainly higher than B. The question you need to answer is "is the house edge still $500?" If it's $500 at both ends of the variance scale, then intuitively does it make sense that it would change somewhere in the middle?