Not to mention that if you have $127000 to throw around, is winning another grand really that big of a deal?
Most roulette systems lead to small, regular wins but can also lead to very big losses. The Reverse Labouchere system does it backwards and thus leads to small, regular losses but sometimes leads to very big wins. Since it requires you to either be the Rain Man or have access to notebook, pen and calculator at all times (guess if those are allowed anywhere near a roulette table) it’s not exactly practical, and you need to really put the time in if you want to get anywhere.
When you get right down to it, barring mechanical bias, there’s just no way to get around the fact that the game is rigged in the house’s favour.
That may not be true: if he sold it to a pawnshop, he could buy it all back for a reasonable amount of “interest”. He could actually make a good killing on this racket.
The Martingale method results a very small probability of a very large catastrophe.
For example, with 10 flips of a coin, doubling each bet after a loss, walking in with $1,024: You can handle up to 10 bad flips. The odds of 10 bad flips in a row is 1 out of 1,024 (1/2 to the 10th power). But if, god forbid, there are 10 bad flips in a row, you have lost all $1,024.
So with one coin flip, you have a 1-in-2 chance of winning 1 dollar, and a 1-in-2 chance of losing one dollar. Using the Martinglae method on 10 flips, you have a 1,023-in-1,024 chance of winning 1 dollar, and a 1-in-1,024 chance of losing 1,024 dollars. In other words, a very small chance of a very large catastrophe.
A friend of mine did this with the field bet in craps. After a year, he was up tens of thousands of dollars. After two years, he was in the hole, based on two disastarous outings. That’s Martingale for ya.
P.S. Odds above are slightly off, losing 10 in row is a loss of $1,023, not $1,024. I kept it a bit simpler to illustrate the point.
Interestingly, one of the themes in Fooled by Randomness is that the author, as an investment firm, invests to take advantage of the catastrophies figuring that he can weather the steady losses for that huge payout. The model broker OTOH invests to minimize risk and thereby will eventually take it on the chin when all the bad outcomes stack up against her.
He didn’t detail his methods, however, and I’ve always wondered if he had been foold by randomness as well…
that’s the part i don’t understand, isn’t it 624 times more likely i’ll win every 1k?
i’ll repeat that i understand changing the way you bet does not change the expectation or the odds. i also understand the question the OP asked, being that after 1000 heads, the next throw will still be 50/50.
what i do not understand is why can’t i take out 127k on the gamble that a streak of 10 does not happen more than once or twice in 625 games?
thanks for sharing muttrox. so i guess the answer is by pitting our bank vs the casino’s, we lose because their pocket is way deeper than ours, and is better able to weather catastrophes without breaking a sweat?
I didn’t see where your 624 number came from, so I can’t respond to that.
You can certiainly go in with 127K, and the odds are vastly in your favor that you’ll win a few thousand easily. However, should you be on the bad side of those odds, you’ve lost vastly more. Those two “vastlys” exactly balance out. Your going in with 127K does not in any way change your odds or average expected gain. If I understand your question correctly, it may be 624 times more likely you’ll win the $1000, but if you lose, you lose exactly 624 times as much.
So the Martingale method is no worse than just “normal” betting, but it’s no better either, in terms of your expectations over the long run. The bank factor doesn’t much enter into it – the casino wins because there are no 50% bets, every bet has an edge for the house. (Please please let’s not bring card counting or craps odds bets and such into this thread). Doesn’t matter how you bet, the odds are against you, and the more action you lay, the more the odds are working against you. Also note that in the real world of the casino, the Martingale method is not limited by your bank role, it is limited by the table maximum. My friend started with a $5 chip. He may have had $10,000 in his pocket, but the table maximum was $2,000 on any bet, so he had to stop a couple bets shy of his personal limit.
Yes, it is much more likely you’ll win any $1K bet. (However, if you don’t win you are much more screwed in exact proportion.)
If you are a gambling type (like my friend is), then you can certainly play the strategy based on the hope that the very small chance of a very big loss does not occur. Nothing wrong with that, it’s as good or bad as any other strategy that varies betting amounts, since they’re all basically equal. Just don’t go in with the illusion that your long term prospects are any better using this method.
It seems to me that he’s saying that not only is there no weighting towards sixes coming up again, there is also no weighting towards sixes not coming up again, other than the normal 5/6 odds. He’s approaching it from a different direction but he’s still saying the same thing.
muttrox, i got 625 from rounded down from the roulette example above (20/38)[sup]10[/sup].
let’s use your example instead. 10 bad flips in a row is 1 out of 1,024. 11 bad flips in a row is 61/125000 or very roughly 1 out of 2,084. so using $1024 and waiting for 1 bad flip to happen first before jumping in, do i get to bet with a 1 in 2084 chance of losing that $1024?
it all sound very fallous to me, but i’m not sure how to pinpoint it and kill that notion for good.
Hmmm. Did you mean a streak of 7 instead of 10? Or an amount other than $127K? Because the Martingale strategy applied to a run of 10 bets could result in a loss of $1023K (over a million). Maybe I misunderstand your proposal.
The expected rate of return can be thorny to compute in general, but in this case I think it condenses down to: (1–P[sub]L[/sub])A[sub]W[/sub] – P[sub]L[/sub]A[sub]L[/sub], where P means “probability of”, A means “amount of money for”, and the W and L subscripts mean “win” and “loss” respectively.
For Hampshire’s proposed scheme (post #45), P[sub]L[/sub] = (20/38)[sup]7[/sup], A[sub]W[/sub] = $1K, and A[sub]L[/sub] = $127K. The expected rate of return is therefore negative $431.97. That is, if Hampshire were to try his strategy many many times, I would expect him to lose money at an average rate of $431.97 per attempt. For each particular attempt of course, he either wins exactly $1K or loses exactly $127K.
For betting on a run of 10, the expected rate of return worsens to –$670.18. (P[sub]L[/sub] = (20/38)[sup]10[/sup], A[sub]L[/sub] = $1023K. A[sub]W[/sub] doesn’t change.) So again, a net loss for you and a net win for the casino — on average.
Let’s throw out roulette. No reason to make this more complicated than 1/2 raised to various powers.
The odds of 11 bad flips in a row is (1/2)^11, or 1 in 2,048. Which you notice is exactly half the odds of getting 10 bad flips in a row, as we would expect. (That’s the point of this whole thread, that 11th flip has the same odds no matter what happened the first 10 flips.)
The odds of 11 bad flips in a row is 1 in 2,048. But the odds of 11 bad flips in a row given that one bad flip has already occurred is not 1 in 2,048, it is 1 in 1,024. Also, the odds of 11 bad flips in a row given that one good flip has already occurred is not 1 in 2,048, it is 1 in 1,024. (I am sparing you the math – I feel sure someone in this long thread has posted the formula for conditional probabilities.)
In other words, it is the exact same as the probability of 10 bad flips in a row. It doesn’t matter what the first flip did, the next 10 are just as likely or unlikely to have any particular sequence. The odds of getting 10 bad flips in a row is 1 in 1,024 regardless of whether the flip before was bad, good, on it’s edge, a quarter or a nickel, flipped on the 3rd Wednedsay in April, Bush gets impeached, or Cecil Adams is revealed to the world as being Madonna’s penname. None of those have anything to do with the odds of the next 10 flips being bad.
The fact that sixes have been thrown twice in succesion isn’t sufficient cause for anything. It’s irrelevant. Poe seems to be arguing that it is relevant.
Anyhow, there are some interesting correlations between the Gambler’s Fallacy and the Monty Hall puzzle:
-Suppose I flip a coin 20 times in a row and record the results. I then show you the results of 12 of those flips, all of which are heads. We then pick one of the unshown results at random. What is the probability that it’s heads? What if we don’t pick it at random? What if I specify ahead of time which one it will be?
First problem: You lose your 127K 7 out of 100 times; you’d have to win on a 93/100 bet 127 times in a row in order to win that money back. Heck, even doing it 35 times in a row is less than a 10% chance. It’s like Russian Roulette- your chance of surviving a single spin of the chamber is quite high (83%). Of course, if you lose, the consequences are drastic.
Second problem- in most casinos, there’s a betting limit. In theory, one can always win at any even payoff game (Red/Black in Roulette; Blackjack) just by doubling the money bet after each loss- you’ll always recoup the original bet in winnings, even if it takes you 6 or 7 losses to succeed. But when there’s a limit in place, then the strategy fails- how can you bet $32K on loss #6 when there’s a 20K limit? And on a table where you can bet as low as 1K, chances are that the limit is 20-50K. Drop your last two chances to win the original bet back, and now you have a much higher chance of losing it all to the casino.
Let’s play a game, we find a quarter lying on a table. I flip it once - heads. SO now we bet what will be next, I get heads, you get tails - $1 to the winner from the loser. A second flip - heads again. We repeat the bet - assuming we are both risk neutral - and it comes up heads again. And a fourth and a fifith time and each time it comes up heads.
At how many tosses coming up heads do you stop thinking it is a fair coin and stop betting? 10? 50? 100? 1000?
ALso interesting, someone with less “stistical acumen” then us probably belives after maybe a couple or so tosses of heads that a tails is “due” and so their expectations probably go from 50/50 heads/tails to something similar to 45/55. but as more and more heads come up they will move back to 50/50 and then evenutally reverse their probabilities. After enough tosses they may even come to the conclusion that it is 100/0.
For me, 2*log[sub]2/sub, which is slightly less than 7. The chance of getting 7 heads in a row from a fair coin is less than .01. That’s a highly significant result.
probably belives after maybe a couple or so tosses of heads that a tails is “due” and so their expectations probably go from 50/50 heads/tails to something similar to 45/55. but as more and more heads come up they will move back to 50/50 and then evenutally reverse their probabilities. After enough tosses they may even come to the conclusion that it is 100/0.