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Blackjack and Fixed Fractional Position Sizing

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Milk Man

Joined
11 May 2005
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Hi guys,

I wanted some feedback about a fixed fractional position sizing idea i've been toying with. It is similar to my casino method when playing blackjack.

Basically when playing blackjack, every time I lose I increase my bet. E.g- I bet $10; if I win the bet remains at $10, if I lose then the bet increases to $15 then $20 then $30 and so on until I win. The premise being that each time a loss is incurred, a win is more likely next hand. This is because the odds of the losing streak continuing are less than it breaking. Just like when you flip a coin; the outcome is either heads or tails, the statistics of such experiments show that larger streaks are a lot less common as numbers increase. The odds increase dramatically each time up (13 of one side in a row is the highest reasonably possible). The theory has worked well a couple of times now.

Now it is my intention to apply the same logic to forex trading. Basically using normal fixed fractional position sizing except that the position size is never decreased. The logic being that when a loss is incurred the most likely outcome next time out is a winner. The system im using has a 60-70% win rate so this makes it all the better, I believe. I wouldnt use it with systems with larger numbers of losers though as the drawdowns would likely be horrendous.

Anyways, if you had the energy to read through all my prattle, can I have some feedback? C'mon, one of you academic types has to have a major in statistics.
 
Well no major but some systems experience.

I'm suprised your Blackjack method works as your returns at best can only be 1.5 x stake. at worse you can be risking massive amounts for only the initial stake price.Its only a matter of time that you blow up unless your Kerry Packer,last I heard he wasnt playing a great deal these days.

As for FOREX there would be no need to increase your position size as you are not limited to win % over risk.Unless you have done this.
IE you exit the trade at Y times Risk.

Infact the exact inverse I believe has been successfully employed by many.

That is decrease position sizing in losing trades (If you have consecutive losing trades generally its not moving consistently in the direction your attempting to trade) and Increase it during winning streaks (If you have consecutive winners then generally the instrument your'e trading is moving consistently in the direction your trading).

In my discretionary trading I will decrease position sizes during losing streaks--LIKE NOW.
 
Position sizing why 1.5times initial bet?
If you have two losses in a row say $10 and $15 you''ve lost $25, on third bet you bet $20 and you win but you are still down $5.

My friend plays a similar system but doubles his bet if he looses.

Now on the theory on winning streaks, I can see what your saying but each event whether a toss of a coin or spin on the roulette table is mutually exclusive to any event prior.
Even if by chance you had 10 reds in a row, the chances of the next spin being red is still the same ~49%.
I was at the casino last year and I think I saw 11 reds in a row, if you were playing that system you would need to bet $20480 to try to make a profit of $20, if started with a $20 bet
 
re:bvb's 11 in a row - once the table limits kick in, you cant get it back.

just do what i do - go every blue moon, win when you do, and be happy.

SEE ITS SO EASY

(tongue firmly in cheek)
 
Position sizing why 1.5times initial bet
Click to expand...

Maximum return on anyone hand--Blackjack,then even money then Stand off then loss.

Thats why card counters look for better than average decks so they get their win rate up and as they do hit the amount risked.

What Milk is describing isnt fixed Fractional position sizing it is Martingale position sizing.
What Im professing is Anti Martingale (in this case).
 
bvbfan said:
My friend plays a similar system but doubles his bet if he looses.
Click to expand...
That technique's as old as the hills, and while the basic procedure is sound and works in theory, in practice there are other conditions that can stop it working: namely available funds and betting limits.

If you double your bet with each loss, a run of losses can put a serious dent in your funds. Remember the story of the rice on the checker board? If your first bet is $1, your bet after 10 losses would be $1024, after 20 losses more than $1m (and also remember they're cumulative, so as you placed your bet for $1m, you'd have lost nearly that amount again in bets already). Chances are betting limits would have cut in before then too.

Now you might think that a run of 20 losses on something like black & red on a roulette table is highly unlikely, and indeed it is. However, you'd be unlikely to bet just $1 as well. If you put $100, then after 5 consecutive losses you'd be betting $3200, having lost $3100 already. Five consecutive losses is relatively common, and would you be prepared to risk another $3200 on an almost 50/50 bet by that time - if you lost, you'd be $6300 down and looking to bet $6400? That's a long way from your starting punt of $100.

Essentially you're trading a high probability of small wins against a small probability of a big loss. Casinos have deeper pockets than you, and the odds are in their favour, so if they're not too busy, they may well be happy to let you take that chance.

Cheers,
GP
 
Milk Man said:
13 of one side in a row is the highest reasonably possible
Click to expand...
Why do you say that?

13 of one side in a row has twice as much chance of happening as 14 in a row, and half as much chance as 12 in a row. What's so special about 13?

GP
 
GPs right you have to double up.

Lets take your idea of $10,then $15-,$20-,$25,-$30
and you then win on $35 so in 5 bets youve lost $100 and then on 6th bet youve returned $35---ooops

If you double.
10,20,40,80,160 total outlayed = 320
next bet 320 win 320---ooops.
 
Glad I asked. Blackjack system needs work then huh? I never fully crunched the numbers; I was just going by feel. Oh well, no harm done, im a few hunge richer anyways. See what happens when you get reward for doing the wrong thing?

GP- I think the whole 13 in-a-row works exponentially but im no rocket scientist.

Anyways, I guess what I propose is more anti-martingale not martingale like I was playing. Just to clarify; im not staking any more when I lose just not any less then previous risked.
 
Milk Man,

People have been working on systems for Blackjack ever since the game was invented, so it's highly unlikely that you'll ever find anything that actually works consistently . And as Tech keeps saying about share trading, it's consistent profits that are important (if you're going to do it regularly), not just having a few successes now and again.

Regarding you original premise that "each time a loss is incurred, a win is more likely next hand", you may want to take a look at this PDF file.

And for actually using the Martingale system in casinos, you might want to note the last line here.

Cheers,
GP
 
Infact the exact inverse I believe has been successfully employed by many.
Click to expand...

Yes, I use the exact inverse, and I have read alot about this, and play alot of blackjack. Was in Vegas only 10 days ago.

IMHO if a coin is tossed 1000 times and each time it comes up heads I'd be betting on heads next time. Not tails!!

Trends in all aspects of life occur, everyone knows it, people do get on winning streaks and losing streaks. Betting againt trends is unwise unless you get better odds for it (which occurs on the stockmarket, but not at a casino odds are fixed). If you lose 3 in a row leave the table, if you are winning bet more - simple as that.

However with Roulette, you can double your bet after a losing bet (on red or black 49/51 odds cause of 0) until you win, because the odds of you losing 8 in a row are very very small, if you had unlimited cash and did this on an unlimited table you could basically never lose. But there are table limits!! And losing 8 in a row means you can't bet again and you've lost alot of money! The odds of this are small of course, but if you keep doing it, it is bound to happen making this theory unwise.
 

Sorry but I TOTALLY disagree Ive seen 27 straight black and well over 8 times everytime I walk into a casino.If you went to Caesar's or MGM in Vagas
and told them you wanted the limit increased so you could play martingale they would usher you into a private table area fully serviced fasted than you could sign a cheque.
Odds can be less in the US many tables have 00

Dice (not Craps) is the only 50/50 game in town that I know of.

Least with the markets I can get better odds and a better edge.Well I "think" I can.
 

There are 18 red, 18 black and zero on a roullete wheel - therefore the mathematical probability of a red or black on any given spin is 18/37 = 48.6%

Regarding losing on 8 spins in a row betting on either black or red and not at all on zero, the probability is:

(18/37) x (18/37) x (18/37) x (18/37) x (18/37) x (18/37) x (18/37) x (18/37)

= 0.153%

Consequently the probability of picking red or black correctly 8 times in a row is 0.153% assuming an evenly balanced wheel and ball.

cheers

bullmarket
 
20 reds in a row were spun at a roulette table in the Alice Springs casino one night. The bloke doubled his bet till he hit the table limit and then bet the limit on each spin ($800.00 I think as it was 20 years ago). When the 21st spin came up black, he picked up his chips and left.
 
I haven't done the maths ( incapable is a better description) but I am told that martingale, and other staking plans do not remove the negative expectancy. The odds (and capacity risk) will get you sooner or later.

But I would love to do some statistical studies on distributions of runs of one colour.

One would expect to see normal distributions , but I suspect there would be a rather marked degree of kurtosis. The number of long runs observed seems higher than the pure mathematical probability.

Any statisticians care to comment?
 
LOL. Should have realised it was discovered previosly. I once thought up the same system for Keeno at the club (doubling up after each loss). Worked fine a few times, but darn it wouldn't you believe I lost 7 times in a row and came to the maximum bet limit! (which I didn't know existed). Fortunately I had a friend with me and we split the last bet. I won the last one, took my $10 reward (for a $600 final bet risk ) and vowed never to be so silly again. Then the stockmarket came along.....
 
bullmarket said:
(18/37) x (18/37) x (18/37) x (18/37) x (18/37) x (18/37) x (18/37) x (18/37)

= 0.153%
Click to expand...
For a losing bet you have to include the zero, so the chance of a loss is 19/37. Losing 8 times in a row has a probability of 0.484% (your answer should have been 0.314%).

Realist said:
because the odds of you losing 8 in a row are very very small
Click to expand...
Over the longer term, if you're a regular player, the odds are irrelevant. It's the average return that becomes important. With the house having the zero on the wheel, their long-term return is 1/37th, or about 2.7%. Therefore, on long-term average, for every $100 you bet, you'll lose $2.70. This means that if you start with a fixed amount and keep playing with it as long as you can (which means rebetting any winnings), you'll always eventually lose the lot.

The only certainty with gambling (on casino games) is that the more you play, the more you'll lose, as you'll gradually start matching the average return.

GP
 
wayneL said:
martingale, and other staking plans do not remove the negative expectancy
Click to expand...
That's correct. Martingale only works where the probability is exactly 50/50 - no house advantage.

Although you could perhaps compensate for that by slightly more than doubling each bet. However, you'd still be stuck with betting limits and potentially exhausting your funds in a run of losses.

GP
 
hello,

you need variable odds for each bet
i
e. 1/5, 1/21, 1/10, this makes it far more effective

and a constant item

thankyou
robots
 
Thanks GP - you are right

the probability of losing in my example is actually 19/37 = 51.3% and so my (18/37)s should have been (19/37).

What I calculated originally is the probability of picking red or black correctly 8 times in a row and I think I did one too many multiplications in my original example.

(18/37)^8 = 0.314%

(19/37)^8 = 0.484%

The method is correct but the numbers used were not

cheers

bullmarket
 
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