Seems fool proof to me with big enough bankroll to break even at least but I must be missing something. Put $100 (or $5, $10, ect.) on black. If it hits just walk away for the day up $100. If not keep doubling down on black until it hits. Walk away even for the day. Eventually its always going to hit black again be it 1 more spin or 10. Its probably something super easy Im missing just cant think of what it is. Thanks

Continue through your example during a long string of reds. Long strings can occur, and they occur more often than you may think.

If bankroll and table limits isn't an issue then even on the off day it goes red and green 20 times in a row it always going to hit black again. I guess Im trying to find the mathematical hole In the strat more then anything.

This is a trivial known problem but for every one that is not trolling (the old classic martingale) and is for real, there will be a first time of course this came as a thought.

How much money do you have to use for this? If you have 2^n-1 available money and you start with 1 unit of bet and then go to 2, 4 etc until you win, then what happens if you hit n times the opposite of what you are betting for?

You would have used after n bets 1+2+4+..2^(n-1)=2^n-1 ie all your money. This is why i chose the total to be that way so that the last bet would still be enough to recover everything lost up to that point+1 unit of profit.

So all it takes to run into trouble is to use all your money and then nobody will help you to bring the next bet to deliver a possible victory. What happens then? You have lost 2^n-1, that is what happens.

So if you have a chance to win that is p per bet then with a chance (1-p)^n you lose all your money in that trial meeting a deadly n-streak and with chance 1-(1-p)^n you will always walk away with +1.


As a result your expectation for this system is

(1-(1-p)^n)*1-(2^n-1)*(1-p)^n

For this to be larger than 0 ie the strategy to make sense as a plus EV effort (it may still be a bad way to play though even then as you will eventually learn/recognize)

you need to have (1-(1-p)^n)>(1-p)^n*(2^n-1)

First of all realize that if p<1/2 no amount of betting (in a game that you win 1 with probability p and lose 1 with probability 1-p) will ever make playing that game positive in return and no sum of any such bets however structured will lead to an average result that is larger than 0. This is because the average of the sum (of any sequence of bets) is the sum of the averages and the average each time for any bet is negative so the average of the sum which is the sum of the individual averages will be negative too no matter how you construct the sum.

You see whatever x you bet the expectation will be E=x*p-(1-p)*x=x*(2*p-1) and since p<1/2 => 2p-1<0 so E<0. An elaborate sequence of bets wont change the average result.


What if p=1/2

In this case

(1-(1-p)^n)*1-(2^n-1)*(1-p)^n becomes 1-1/2^n-1/2^n*(2^n-1)=1-1/2^n-1+1/2^n=0

This is expected of course because any bet you make in that game has expectation 0 since its 50-50.

Now things can get better if p>1/2 but you do not have this here.

Even then this is not the best way to use your money when p>1/2 as you will tend to be bankrupt typically even then unless p or n or both are very large.

What would p need to be to have a nonzero probability to never lose? This may be true because although there is always a chance to have a streak that kills everything, your position also grows every time you win so you will eventually get big enough that the chance to deplete everything is much smaller than what it was when you started.


For example if p>1/2 one can claim that

To double your bankroll from K=2^n-1 to say 2*(2^n-1)=2^(n+1)-2 you need to survive 2^n-1 attempts of this strategy (https://en.wikipedia.org/wiki/Martin...betting_system) )

The chance to survive them is P(n)=(1-(1-p)^n)^(2^n-1)

As a result the chance to survive an infinite sequence of doublings (ie never die) when you start with 2^n-1 bankroll is


S= survival probability

S=P(n)*P(n+1)*P(n+2)....

Log S= Sum[(2^k-1)*log(1-(1-p)^k), {k,n, Infinity}]

Given that (1-p)^k is typically very small<<1 if we pick 2^n-1 the starting bankroll big enough

We can say log(1-(1-p)^k)~-(1-p)^k

So LogS~Sum[(2^k-1)*(-(1-p)^k), {k,n, Infinity}]=

Sum[-(2*(1-p))^k+(1-p)^k,{k,n,Infinity}] =

= -(((1 - p)^n (1 - 2 p + 2^n p))/(p (-1 + 2 p))) (When p>1/2)


So survival probability S of infinite doublings is

S~e^(-(((1 - p)^n (1 - 2 p + 2^n p))/(p (-1 + 2 p)))

So eg for p=0.55 and n=10

S~e^(-3.486)~0.03 or 3%

But if p=0.55 and n=20

S~e^(-1.21577)~29.6%

If p=0.55 and n=40 (trillionaire)

S~e^(-0.1478)~86%

See even then with such edge its kind of risky and much better ways exist to play in those cases that not only grow you faster but can take the ruin probability to much smaller levels even with real life min bets restrictions in place. As always see this https://en.wikipedia.org/wiki/Kelly_criterion


PS: Hopefully i didnt make any stupid math error in these last parts. Will revisit if so.

As others have posted, this strategy makes you $100 almost every time, and you lose everything you have infrequently.

All those hundos are not enough to make up for the big inevitable loss

I can probably modify a little bit that derivation for the survival probability bound when p>1/2 (plus i need to call it a bound to be technically correct) to be more consistent and accurate with what i did by demanding we go from 2^n-1 bankroll to 2^(n+1)-1 bankroll that actually requires 2^n bets attempts that are successful.

Under such move the chance to survive that near doubling (its a bit better than doubling in a trivial difference) is

larger than (1-(1-p)^n)^(2^n)

I need to call that larger and not the exact survival probability because one can see easily that after you have won a few times you are no longer at 2^n-1 bankroll so technically even if you lost n bets in a row you are still able to recover by using the remaining profits you had up to that point and betting it all, then winning and betting it all again etc as many times as needed to either lose it or reach a recovery level that resets the martingale. That correction is not significant for the beginning but near the middle say of the way to doubling the bankroll you have reached levels that the correction is significant as its only 2 wins away for a recovery hence a decent probability boost correction.

Essentially i should have done this instead;

I should have tried to bound the survival probability between 2 terms (until a more thorough exact calculation can be produced that is the exact or very accurate at least with minor corrections missing).

What can be said is that the chance to survive a process of moving up 2^n from 2^n-1 to 2^(n+1)-1 position is P(n) such that ;

(1-(1-p)^n)^(2^n)<P(n)<(1-(1-p)^(n+1))^(2^n)

So the chance to survive infinite number of doublings (ie to never lose everything ) is

S=P(n)*P(n+1)*P(n+2)*....

and it is bounded by ;


R1(n)*R1(n+1)*....<S<R2(n)*R2(n+1)*...

Where R1(n)=(1-(1-p)^n)^(2^n) and R2(n)=(1-(1-p)^(n+1))^(2^n)

So logS now with the usual approximation for log(1-(1-p)^n)~-(1-p)^n and log(1-(1-p)^(n+1))~-(1-p)^(n+1) is bounded by

Sum[2^k*(-(1-p)^k), {k,n,Infinity}]<logS<Sum[2^k*(-(1-p)^(k+1)), {k,n,Infinity}]

or (p>1/2)


-((2^n*(1-p)^n)/(2p-1))<logS<-((2^n*(1-p)^(1+n))/(2p-1))

or

e^(-((2^n*(1-p)^n)/(2p-1)))<S<e^(-((2^n*(1-p)^(1+n))/(2p-1)))

for example to see the difference from above

for p=0.55 and n=10

or 3%<S<20.8%

for p=0.55 and n=20

29.6%<S<57.9%

for p=0.55 and n=40 (trillionaire)

86.3%<S<93.6%

A simple way to figure out that the strategy is not profitable is noting that for a series of bets to have positive EV, at least one of the bets must have positive EV which no roulette bet ever has.

The other day at my weekly poker game, one of the players started to tell a story about how he did this kind of betting at blackjack. The way he started, I knew he was going to tell how he ended up losing lots of money.

So I said "that is a good way to lose a lot of money". He then sort of argued with me, telling me that the dealer has to bust some time, and that this should be a good way to guarantee you win. Then he proceeded to finish the story about how he lost 12 in a row (or some big number) and lost a whole lot of money.

I think just about everyone who has ever entertained notions of hitting it big at gambling has thought about some variation on this type of betting system. Most have either realized it is a very dangerous game to play and avoided it. Some have probably used it successfully, winning small amounts - just as they would have if they bet in any other fashion. And some have gone broke trying.

If you are interested in a fool proof method to avoid losing at roulette, search for Mike Caro's Roulette System #1.

I'm honestly curious what it feels like to play this system. Like say your bankroll is big enough to start with $10 bets and go 12 bets deep, doubling each time. That means your 12th bet is gonna be $40,960. What does it feel like to place a $40,960 bet, where, if you win, you'll have a $10 profit on the series?

Nobody knows, because they always hit the table maximum while trying to bet $5,120.

I thought high stakes blackjack was a thing that existed?

But if you start off at $10 you will need to switch tables once your bet gets above that table's limit. Throws everything off. New dealer, new deck. Resets everything.

How it feels? It feels like how the f did i get myself here? I would never be betting 50k in a single hand. Who am i? Is this the only way i can win now?

Screw this. Losing 50% of your money is not the end of the world. The end of the world is to not care anymore to be rational. So it's never too late to care to realize that the real victory is to put an end to the madness. Only very few can do that! Because if you win one more time you will be back again here eventually one day. If you win you wont stop. And if you dont want to stop then ask yourself what is better a 51% chance to never get the chance to stop losing everything vs 49% chance to stop at double the position (if its some 49-51 game say) or a 100% chance to stop at 50% the position?

Plus if we speak of blackjack the win probability is not even very close to 50%. Eg see here http://wizardofodds.com/games/blackjack/appendix/4/

My point being if your objective at that moment is to maximize the win probability, not the EV of the game - any win is ok, then blackjack is probably a bad choice of a game for the last bet.

I mean you can for example place bets in a variety of number choices in roulette that nearly take you to recovery but happen close to or over 50% of the time, a lot better than a single BJ hand or a black/red bet in roulette.

The reason is that in BJ the strict loss probability per hand in some familiar online formats say (meant as either loss right away or loss after a series of pushes before any win) is (see above link) about 48.04%+8.71%*48.04%+8.71%^2*48.04+....=48.04%*(1/(1-8.71%))=52.6%

That may be slightly off also because you no longer split or double as this is your last bet but i dont think the correction from the deviation in strategy affects substantially the rough loss probability estimate (as much as it does the EV i mean) (first of all double or splits are a minority of hands that require specific combination of events and the deviation in how played is not massively affecting the loss probability then).


Funny how in the roulette (red/black bet) the probability to lose in the all in bet is 20/38 or 52.6% (in US roulette) so close enough although the EV of the game is worse.



I wonder what a mad player that is forced to play martingale can do in Blackjack, where counting is possible, to maximize the final all in win probability by delaying when it is made and doing it only when its super advantageous in terms of win probability (not EV).

That is a good question to wonder about.

I assume you do the crazy stunt and when you reach the final bet you continue (assuming its money you dont mind losing i mean or forced to risk by some constraint) only when its correct to do so, otherwise you start small bets for a while until the count gets ideal and then strike the last time.

Does anyone know if there tables out there for the win probability as function of the count?

I mean if you have to do the bad thing at least be good at it lol.

I read somewhere that longest streak of black/red was 39x. Now think about that "rice and chessboard" story

btw. same number was 6 in a row

Doesn't matter which table or dealer. Every spin is new, it has no history

If there are 15 bil people that lived past 4 centuries and and 5 bil of them played roulette (far less but whatever) and they did so for 1000 times each in their lives (some probably only a few others thousands but whatever say 1000 avg for all) and each time they did there were say another 5 people around betting we would have had at most 5*10^9*1000/5~10^12 ball spin events so far.

Which means the chance to have had 39 of the same color at some point is reasonable as say 2*0.474^39~1 in 2 *10^12.

simplify for me

Why walk away after you win $100? If your system is sound why not grind it out?

My question exactly, however I thought you started with Red?

everytime ive made my mind up no matter what to press my winnings ive gone to the table and busted a hundo the first hand . lol. ive given up on dreams accepted the slow inevitable grind.

This seriously made me laugh. I agree!

All this jazz below.....is this legit or a level? I honestly don't know? The old saying, it's all Greek to me, well Greek makes more sense than that to me.lol

Wow you are the boss! As I see it there's no way winnings can be sustained in the long run. Odds are against you. Yeah you can mess with it at much much lower stakes compared to your bankroll, just for entertainment and research sake, or so you could win yourself a free pizza + beer, but I would never try to make a profit at it, just pick another gambling game that you have a chance of having an edge, like sports betting, or poker. The house wins @ casino.

Negative expectation games are not beatable, so unless you can get an edge due to wheel bias, etc., it is pretty much a non-starter (Martingale is a non starter). At some point, reality sets in. It reminds me of a famous physicist who also made some money in the markets. He said that when he was young and his father stated that they might run out of gas, the young physicist/investor stated that it wasn't possible to run out of gas, since whatever amount his dad currently had in the tank could be divided by half, and you could just keep having half a tank less, and so you could never really run out, since you could pretty much divide by half close to an infinite number of times.

Eventually, the reality of table limits sets in, and the unGodly bad streak will happen. It isn't a bad strategy to have fun with, as long as you don't go the distance. People do it for very small amounts, but bail out at maybe a bad streak of 5 against.

So, you would bet $5 to start. If you lost that you then bet $10. If you are willing to keep doing that you might get on a good streak, as long as you promised yourself that you would never go past 5 or 6 against. In this case, you are basically playing until you lose 5 or 6 in a row, and you would stand a good chance to get on a decent run. But inevitably, you will run up against the streak and busts the bankroll.