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## Martingale doubling chances

Hello all. I recently calculated the chances of doubling 1000\$ using Martingale method. I did this with two methods: 1) Analytical 2) Numerical

First some definitions:

Roll: a single roll, you either win or lose, chances are 50% win and 50% lose.
5R method: in this method gambler believes he can't lose 5 times in a row, so he bets (amount he has)/((2^5)-1), that is (amount he has)/31.
Round: For 5R method round ends either when gambler wins before he runs out of money or loses all 5 rolls and has no more money to bet.

1) Analytical: In this scenario the gambler believes that there is no way he is going to lose 5 times in a row, chooses 5R method, so he starts with 1000\$ and bets (1000/31)\$, and each time he loses, he bets double previous amount (2000/31)\$. I calculated that to double 1000\$ (win 1000\$) he must complete 22 rounds with wins. Now, the chances that he is going to lose 5 times in a row and leave the casino bankrupt are (1/2)^5, that is 1/32. So the chances he is going to go bankrupt before he completes 22 rounds and doubles his money are 22/32, that is 68.75%.

Result: 68.75% that he will lose 1000\$ he had at the beginning, 31.25% that he will win 1000\$

2) Numerical method: I used turbo pascal and wrote a simulation for this scenario, will not in detail.

Result: 50% he loses 1000\$ he had, 50% he wins 1000 and leaves casino with 2000\$ in his pocket.

Question: I really think that I must have made a mistake in my analytical calculation. But I check it over and over and can't find what is wrong with it. Can you guys help me? If you have any clarification please tell me.

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Hello ?

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Originally Posted by roboticmhd
Hello ?
Your description of the gambler's strategy isn't clear. For example, suppose he wins on the first toss and loses on the next two tosses. What is the size of his bet on the fourth toss?

4. Originally Posted by roboticmhd
I calculated that to double 1000\$ (win 1000\$) he must complete 22 rounds with wins.

5. This tactic often results in big wins and big losses, and how that happens is proportional to how much you start with. Ultimately, if you hang out long enough, you always lose it all. Adding a goal of \$2000 is likely skewing things. What happens when you start with \$100 and want \$200?

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Ok to clear, here how the gambler applies his tactics. In this example I will assume that he doesn't lose 5 times in a row.

A round is when he bets 1/31 of his money and doubles everytime he loses a roll.
For example if gambler has 31\$:

Round 1:

Roll 1: bet 1\$, win (now has 32\$). Round is over.

Round 2:

Roll 1: bet 1.03 \$, loss (now he has 30.97\$)
Roll 2: bet 2.06 \$, loss (now he has 28.91\$)
Roll 3: bet 4.12 \$, win (now he has 33.03\$) Round is over.

Round 3:

Roll 1: bet 1.06\$ .....
.........

And so, on

Everytime he stars a new round, he first bets 1/31 of existing money, and if he loses the roll he doubles the bet for next roll.

As long as he doesn't lose 5 times in a row, after 22 rounds he doubles his the very first initial money (Now he has 62\$).

If he ever loses 5 times in a row then he loses all money, since 1+2+4+8+16=31.

So my question is, I seem to have done analytical analysis correct but I believe that it is not. Must be mistake somewhere, because pascal simulation, in which man visits casino 10000 times with 1000\$ in his pocket reveals that he doubles only about 5000 times, the rest he loses.

7. I don't get that, he wins only 1 stake each round (and loses the lot after five losses apparently.)
so after 22 rounds he has 22/31 of his original? not double.
the twin problems are you only win the stake and you hit the house limit by doubling.

8. OK I get it you count rounds and not plays but you assume a random number to get the size of each round?

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Originally Posted by tashirosgt
Your description of the gambler's strategy isn't clear. For example, suppose he wins on the first toss and loses on the next two tosses. What is the size of his bet on the fourth toss?
I agree that the description could be clearer, but the gambler is engaged in a doubling-up strategy that results in a profit of X/31 (where X is initial wealth) with probability 31/32, and a loss of X (so all of the money) with probability 1/32. The strategy is to bet X/31 the first time. If a win, then the gambler has achieved the goal of winning X/31. If a loss, then he has X*30/31, and bets 2*X/31. If a win, his total wealth is 32*X/31, and he has achieved the goal of winning X/31. If a loss, he now has X*28/31, and bets 4*X/31. A win means the goal of winning X/31 is achieved. A loss means he now has X*24/31, and bets 8*X/31. A win achieves the goal of winning X/31. A loss means he now has X*16/31, and he bets it all. A win achieves the goal of gaining X/31, and a loss makes him bankrupt.

So the gambler believes (mistakenly) that it is impossible to lose five times in a row, and that this strategy therefore guarantees a profit of X/31.

So the idea is, the gambler repeats the whole thing 22 times, increasing his wealth by a factor of 1/31 each time. So if he never loses five times in a row, he ends up with (32/31)^22, which is slightly more than a doubling, a factor of about 2.0107.

roboticmhd - the error is in the analytic approach. The probability of increasing by a factor of 1/31, 22 times in a row, is not 22/32; it is (31/32)^22. This is about 0.4973, which is (surprise surprise!) the reciprocal of 2.0107.

So the simulation is correct, the analytic approach is done incorrectly. The gambler has a 0.4973 chance of increasing wealth by a factor of 2.0107, and a 0.5027 chance of going broke. You are probably getting 50/50 from the simulations because you need a really big number of simulations to be able to distinguish between 50/50 and 49.73/50.27 accurately.

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Originally Posted by profloater
I don't get that, he wins only 1 stake each round (and loses the lot after five losses apparently.)
so after 22 rounds he has 22/31 of his original? not double.
the twin problems are you only win the stake and you hit the house limit by doubling.
Each time he resets the initial wealth. So the first time, he ends up with X*32/31, where X is what he started with (unless he loses five times in a row). The second time, he starts betting 1/31 of X*32/31, not of X. So if he successfully executes the strategy twice, his wealth is now X*(32/31)^2. 22 times does result in a doubling, within about 1.07%.

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No, assume he has 31\$, he wins 1\$ first round. For the second round he starts with 1/31 of 32\$ and when round is complete he has 33.03\$, third round he starts 1/31 of 33.03\$, and so on, so he only need 22 rounds, instead of 31, do double his money, as long as he does not lose 5 times in a row. And no matter how many rolls he plays in a round he always wins what he bet the first roll of a round.

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for each round newbalance=oldbalance+(1/31)oldbalance, if you keep doing this you will see that after 22 round he doubles his money. this of course assuming he does not lose 5 times in a row.

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You can check by doing this, if you have casio calculator:

Write 31+0, answer is 31, now write ANS+ANS/31, and keep pressing "equal to" button until you get 62, count while doing so. result is after you press 22 times you get 62. (actually you get 62.33)

14. OK I get the simple part. the chance of five losses is 1/32 in a 50/50 game. to have enough to keep betting: 1+2+4+8 =15 so next bet is 16.
that's how you get the 31 start stake. But the error i think is in thinking the 1/32 failure chance happens only after 32 tries. You use the binomial:
nCr. p^r.Q^(n-r)
to find the chance of 0 fails in 22 rounds nCr =1
r =0
p ^0 = 1 also
q is 1-p = 31/32 ^ 22
my calculator gives 0.497 for that so in 22 rounds you have a 50% chance of losing by having five losses in fact if correct just over 50%
50.3% chance of losing. So in the long term you lose.

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You know that if the game is 50/50, no matter what betting strategy you use, your expectation value is zero gain and zero loss. So any strategy you use that is expected to lose your entire stake half the time is also expected to double your stake the other half of the time. So if your goal is to double your money, then just bet once, all in. Why waste time? That's also the best strategy when there are house odds against you, as you will want to minimize the number of times you play against those odds. The bottom line is, there is no "good" strategy for doubling your money, the only way to use strategy is to suffer a small chance of losing everything in order to get a large chance of gaining just a little, or to suffer a large chance of losing everything to get a small chance of really cashing in big.

16. At most casino odds, the optimal gambling time is 49 minutes.

This is what leads to the phenomena of casino workers who play on their lunch breaks seeming to make money consistently, quitting to gamble full time, and quickly going broke.

I've employed this casually and it seems to work better than random chance.

But I've also known very successful gamblers who laugh at this.

17. If it all boils down to strictly chance, but The House is going to keep a percentage off the top, then I know I'm going to lose and hence won't bother. So, I (effectively) win by staying away.

Makes me feel better for being lazy.

18. Originally Posted by Ken G
You know that if the game is 50/50, no matter what betting strategy you use, your expectation value is zero gain and zero loss. So any strategy you use that is expected to lose your entire stake half the time is also expected to double your stake the other half of the time. So if your goal is to double your money, then just bet once, all in. Why waste time? That's also the best strategy when there are house odds against you, as you will want to minimize the number of times you play against those odds. The bottom line is, there is no "good" strategy for doubling your money, the only way to use strategy is to suffer a small chance of losing everything in order to get a large chance of gaining just a little, or to suffer a large chance of losing everything to get a small chance of really cashing in big.
The point is wasting time. The issue with this method is that it appears to be successful when you go after a relatively short term goal. You want to appear to be beating the odds while people are watching. It is gambling for the sake of gambling. Winning and losing doesn't matter, but being seen playing is the point. This is unlikely to break you over the short term, is easy to understand the (flawed) idea, and seems to be a good idea until you do the math or lose your shirt. There is a very, very small window where this tactic does provide better than normal odds at allowing to win back money, which isn't likely breaking even.

19. the danger of a method is that you might get the impression you are winning and that your formula actually works. This would encourage you to keep going and that is how you will always, always lose in the end. If you take KenG advice and put the lot on red, then you must walk away if you win or lose. The only way to make money with a formula is to invent a fairly complex version and sell it to vulnerable people who do not believe the brutal odds. Games with skill of course are very different, poker players win because they have good psychological skills and also know the odds.

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Originally Posted by profloater
But the error i think is in thinking the 1/32 failure chance happens only after 32 tries.
I'm not sure we are interpreting the thinking behind the error the same way, but in any event, you have identified the source of the error, as per my post #9.

Originally Posted by profloater
You use the binomial:
nCr. p^r.Q^(n-r)
to find the chance of 0 fails in 22 rounds nCr =1
r =0
p ^0 = 1 also
q is 1-p = 31/32 ^ 22
my calculator gives 0.497
Yes, this agrees from the 0.4973 from post #9.

Originally Posted by profloater
for that so in 22 rounds you have a 50% chance of losing by having five losses in fact if correct just over 50%
50.3% chance of losing. So in the long term you lose.
I'm not sure what the long term is here. If I understand the game here, it has a well-defined end, and will never last more than 110 rounds, so there is no long term.

But, if the gambler is not happy with doubling, and just keeps playing beyond 22 rounds, forever, then he is going to go broke with probability one (which contradicts Ken G's statement that the expected gain of every strategy is zero).

However, as I understand the game, the gambles has a probability of slightly less than 50% of increasing his money by a factor of slightly more than two, and a probability of slightly more than 50% of losing it all. It is pretty close to the same outcome as just betting all of your money once (a strategy Ken G suggests). Neither winning nor losing on average, but coming out even on average.

Originally Posted by Ken G
You know that if the game is 50/50, no matter what betting strategy you use, your expectation value is zero gain and zero loss.
This is actually not true without some technical restrictions on the strategies that can be used. If we can use strategies that are unbounded in the number of gambles that can take place, and unbounded in the amount that can be bet on each gamble, then the expected gain need not be zero. If the game must come to an end after some maximum number of tries, or if the casino only accepts bets of a certain size, or if we impose some other restrictions on the strategies that can be used, then the expected gain will be zero.

Originally Posted by Ken G
So any strategy you use that is expected to lose your entire stake half the time is also expected to double your stake the other half of the time.
Subject to the technical restrictions above, yes. However, I think the OP was puzzled by the fact that s/he was getting some results that appeared to contradict this. The error was identified by me in post #9, and by profloater in post #14.

Originally Posted by Ken G
So if your goal is to double your money, then just bet once, all in. Why waste time?
Fair enough, but I think the original post indicates that the gambler in this game is suffering from some delusional thinking.

Originally Posted by BigDon
At most casino odds, the optimal gambling time is 49 minutes.

This is what leads to the phenomena of casino workers who play on their lunch breaks seeming to make money consistently, quitting to gamble full time, and quickly going broke.

I've employed this casually and it seems to work better than random chance.

But I've also known very successful gamblers who laugh at this.
I think not just very successful gamblers. But OK, I'll bite, where this particular gem of wisdom come from?

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If we show due respect to probability theory, we can't say what definitely will or will not happen to a player who uses a given strategy. We can only talk about "expected" values or assign a probability to certain outcomes. I agree that the urge to translate probability theory into a theory a theory of actuality is irresistible. It's nice to have moral guidelines that say "If you do ...such and such...then you will (definitely) be sorry."

People who advocate the "ensemble" viewpoint of physical probability try to overcome the hurdle of "probable" vs "actual" by saying the physical probability of an event is its frequency of its occurrence in a population of actual events. However, the population involved seems impossible to define.

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The impossibility of a true betting "strategy" is the "Markov" character of gambling-- if you don't count cards or some such thing, then every result is uncorrelated with everything previous. So you can never alter your expected winnings in the net, you can only alter the distribution of winning probabilities over the plus and minus. You don't really need a special strategy to achieve a large probability of small winnings along with a small probability of losing your entire stake, that's just naturally what people do if they are not gambling a long time and use the "quit while you're ahead" style of thinking. If they instead gamble a long time, with a "I'm on a roll" (non-Markov) illusion, they will generally lose their entire stake. If they use the attitude "I can't end up behind, just one more chance is all I need to get back up again", then they will go past their entire stake and into the realm of gambling disorder.
Last edited by Ken G; 2017-Dec-16 at 04:18 PM.

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The best system to win at Las Vegas is to buy stock in a corporation that owns a casino and call your dividends your winnings.

24. Yes. I learned early on the rule that the dice (wheel, whatever) have no memory. The fact that you've had bad rolls recently doesn't mean that you're due for some good ones.

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Right, and ironically, people think they are "due" if the rolls have been bad, and they are "on a roll" if the rolls have been good! That kind of uninformed optimism really sets them up to lose their shirts, given that the house odds are against them.

26. I remember that the baseball announcers I used to listen to had that kind of logic.

"Watch out! Duke is oh-for-four today, so he's due to smash one!"
...
"They don't want to face Rocky today: three hits, including a homer ... he's hot!"

Although, these cases are likely to be streaky.

27. Originally Posted by BigDon
At most casino odds, the optimal gambling time is 49 minutes.

This is what leads to the phenomena of casino workers who play on their lunch breaks seeming to make money consistently, quitting to gamble full time, and quickly going broke.

I've employed this casually and it seems to work better than random chance.

But I've also known very successful gamblers who laugh at this.
I'm a very successful gambler. The secret is to never gamble in your entire life.
(Full disclosure: In my Sixth Year Studies class in Statistics and Probability, we did run a poker school in the lunchbreak - we all used pocket calculators, and we were all very good at the necessary sums, but I won overall, hands down, because I was the only one bluffing. The nerd beats the geek, every time.)

Grant Hutchison

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Thank you Ioga for clear answers to my question. I understood ! I knew I was wrong, can't believe the answer is so simple. I knew I was wrong because if the gambler was to play 100 rounds that gives, if to use my analytical method, 100/32 probability of losing which is grater that 1 ! (greater than 100%). This can not be true, since you can't have probability higher than 100%.

Actually I came across this whole gambling thing from here [link removed] (sorry if such links are not allowed here)

This is great site, it gives small amount of bitcoin every hour for free ! Also there is Multiply BTC section where you can play Hi-Lo game, like a casino, and earn a lot of money if you are lucky. I was trying to find a strategy to win, but later I realized it is impossible to increase chances.

The site is also provaby fair, which means that they can prove that the games are fair.

I just can;t believe that highest win on that site is 4500 BTC which is like 90 million \$ right now.
Last edited by Swift; 2017-Dec-18 at 07:02 PM. Reason: link removed

29. I have a friend who uses a gambling site that uses red and black and has the double zero to bet on too. If I got this right, he believes you can influence the number of zeroes by betting all the time on both red and black. I think he believes the other players also influence the frequency and the odds on zero are, I think, 32:1. and thus unpopular. He tells me he wins regularly on that method until he gets bored because its a series of small wins. In other words he believes there are more than 1:32 zeroes appearing. If the casino assumes most people use red and black, they may deliberately insert more zeroes or maybe my friend is right that it depends on the number of red/black bets.

so he bets say 1 on zero and 32 on red and 32 on black (at 2:1) He enjoys it but it seems, well, boring to me to win about 1 every half hour.
But as was said earlier it passes the time and he feels satisfied in beating the casino.

30. I am fairly good at games of chance, but I would never play with a machine interface and rarely think casinos are a good way to spend my time. The whole thing is time vs. money vs. odds. Any extended period of time means the odds will force you to give money to the machine/casino. Having a strategy is really only going to help you in the short term (don't risk what you don't have) and at the end of the day, it will allow you to look graceful while removing your shirt. I figure going to the casino is almost exactly like going out to dinner and movie and would never spend more than that. The experience is the payoff. You (probably) aren't going to leave with more than what you had.

I do like playing games against people, informally. I don't have a problem with someone I know walking off with my money. My wife lets me gamble like this on the condition that I give the money back. She has a hard time with numbers, so I generally give away whatever I brought to the table, too. It really isn't fair, but it is often fun.

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