What Math is Missing from this Martingale Strategy?

[lothariorowe]

So obviously no betting strategy will ever work in the long run, I know this, so this is driving me crazy. What is missing from the following remedial formula?

double your bet on losses, and return to the minimum bet after a win, typical.

start with a bankroll big enough to allow, say, 6 losses in a row before total exhaustion of bankroll.

therefore,

x = minimum bet

63x = starting bank roll

for the sake of simplicity, assume playing perfect strategy, odds are winning 1 of every 2 hands.

therefore, odds of losing 6 hands in a row are 1/64, or 6 consecutive losses for every 384 hands.

with a minimum bet of x and odds of winning 1 of 2 hands, there are average earnings of x/2 per hand.

with the odds of losing your starting bankroll (63x) every 384 hands, that's an average loss per hand of 63x/384

therefore,

average gain per hand - average loss per hand = net gain per hand

therefore,

x/2 - 63x/384 = ~x/3

or

average earnings of 1/3 the minimum bet per hand over the long run.

Then, to protect from long strings of losses, you always remove your winnings by reducing your bet to x after losing your 6th hand in a row. I forgot how that works into the math.

So obviously there's a major flaw in this, but what is it?

Is it the fact that losing 6 consecutive hands = a total loss of 63x whereas winning 6 consecutive hands = a total gain of 6x? Isn't that accounted for in the above equation though?

 

[FLASH1296]

Your math is seriously flawed.

In Blackjack you will win 42.9% of your hands and push 8.9%.
The remainder are LOSSES.

Ignoring pushes you then have a ratio of 42.9 wins to 48.2 losses.

Some simple division will lead you tothis:

Out of your non-pushed hands you will win 47% of your hands and lose 53%

The difference there - of 6% - is HUGE.

Incidentally, even if you had a game that actually offered 50% wins your progression will impoverish you, as it has done to so many people, for quite a few centuries.

 

[lothariorowe]

That 6% does make a huge difference, but replacing x/2 in the equation with 47x/100 yields:

47x/100 - 63x/384 = ~3x/10, or, average gains of 30% of the minimum bet per hand.

So considering a win 47% of the time, you should double your initial bank roll every 134 hands.

Also considering the above criteria, you should lose your initial bank roll every 384 hands.

After incorporating the rule of never betting more than your initial bank roll (returning to a bet of x if and when losing 6 consecutive hands), since you are doubling your entire bankroll at a quicker pace than you are losing your entire bankroll, it would seem that you would have a net gain in the long run.

Obviously in the real world however, and contrary to the above math, this type of strategy will of course impoverish anyone who uses it in the long run - so there must be a major part of the equation unaccounted for in the above math even after considering the 6% gap. The math should provide an average negative return per hand...

 

[FLASH1296]

The Law of Large Numbers needs to be ignored for anyone to imagine that a progression can yield profits.

Not only will you court disaster - you MUST lose over time. There is no two ways about it.

 

[MangoJ]

for the sake of simplicity, assume playing perfect strategy, odds are winning 1 of every 2 hands.

therefore, odds of losing 6 hands in a row are 1/64, or 6 consecutive losses for every 384 hands.

It is okay to approximate BJ with a coin flip game just for argumentation why your system is flawed.

probability of losing 6 hands in a row is indeed 1/2^6 = 1/64.
Your flaw is: This will happen every 64 hands on average, and not every 384 hands as you somehow assume.

 

[lothariorowe]

It is okay to approximate BJ with a coin flip game just for argumentation why your system is flawed.

probability of losing 6 hands in a row is indeed 1/2^6 = 1/64.
Your flaw is: This will happen every 64 hands on average, and not every 384 hands as you somehow assume.

That's not true because the odds are representing 6 hands per unit.

therefore the odds of 1/64 really = 1 (string of 6 hands) / 64 (strings of 6 hands).

Considering actual losses occur 53% of the time, the more accurate odds of 6 consecutive losses would be 1/45, or, 6 consecutive losses per 270 hands.

 

[MangoJ]

That's not true because the odds are representing 6 hands per unit.

therefore the odds of 1/64 really = 1 (string of 6 hands) / 64 (strings of 6 hands).

Considering actual losses occur 53% of the time, the more accurate odds of 6 consecutive losses would be 1/45, or, 6 consecutive losses per 270 hands.

I'm not following you. Are we discussing blackjack or coin flips ?
You stated "odds are winning 1 of every 2 hands" as approximation before, so I would like to discuss fair coin flips.

You asked for your flaw in calculation about your martingale system. You knew already it was flawed but couldn't quite find the error.

So again here is your error:
(A) Flipping 6 heads in a row on one try (of 6 flips) has probability of 1/64. Agree/Disagree ?

(B) Having a continuous flow of flips, at any specific instant of time, probability of just finishing 6 heads in a row is 1/64. Agree/Disagree ?

(C) Having a continuous flow of flips, average distant between 6 heads in a row is 64 flips. Agree/Disagree ?

A and B is certainly equivalent. If you're looking for 6 heads in a row at the very instant of (B), you just need to look at the last 6 flips. Discard the rest of the continuous flow. Then you are at scenario (A)

If you have N flips you are likely to find M=N/64 times your sequence. Average distant between completion of your sequence then is N/M = 64. (C)

 

[lothariorowe]

I'm not following you. Are we discussing blackjack or coin flips ?
You stated "odds are winning 1 of every 2 hands" as approximation before, so I would like to discuss fair coin flips.

You asked for your flaw in calculation about your martingale system. You knew already it was flawed but couldn't quite find the error.

So again here is your error:
(A) Flipping 6 heads in a row on one try (of 6 flips) has probability of 1/64. Agree/Disagree ?

(B) Having a continuous flow of flips, at any specific instant of time, probability of just finishing 6 heads in a row is 1/64. Agree/Disagree ?

(C) Having a continuous flow of flips, average distant between 6 heads in a row is 64 flips. Agree/Disagree ?

A and B is certainly equivalent. If you're looking for 6 heads in a row at the very instant of (B), you just need to look at the last 6 flips. Discard the rest of the continuous flow. Then you are at scenario (A)

If you have N flips you are likely to find M=N/64 times your sequence. Average distant between completion of your sequence then is N/M = 64. (C)

Ok, so you are basically saying that for every 64 hands, there are 64 different 6 card series made (provided no beginning or end)? That makes sense. I was isolating every 6 card series instead of viewing them continuously...

 

[tthree]

If you did your math right in a game of equal chance to win or lose with even money payoff your result must be zero!!!!!!!!!!!!!!!

Your odds of failure are 1/64. If there are no pushes your odds of success is 1 - 1/64 = 63/64.

(1 unit)*63/64 - (63 units)*1/64 = 63/64 - 63/64 = 0

Your mistake is 63x/64 is your loss per cycle. The figure of 63x/384 is used improperly in your formulation(x/2 is not average win per hand as you are not flat betting).

 

[Midwestern]

with the odds of losing your starting bankroll (63x) every 384 hands, that's an average loss per hand of 63x/384


So obviously there's a major flaw in this, but what is it?

there's your flaw buddy.

you're assuming that because you have a 1/64 chance of the event LLLLLL
that it will happen once out of every 384.
WHICH IS A VAST UNDERESTIMATION OF THE PROBABABILITY!

sure, if you only play 6 hands total, your chance on going LLLLLL is 1/64.
but in a run of 384 hands.. you will have a large distribution of wins and losses which can be represented as
WLWLWLLLLWWLW... etc...

and the chance of a string of LLLLLL occuring in a string of 384 is much larger.
there is nothing preventing a 6-loss string of LLLLLL happening more than once in a 384 hand run...

 

[tthree]

I had a hard time with the words to describe your errors. You are not including all your losses that ended in a successful progression as hands lost. The odds of winning a hand is 1/2 but you are not flat betting so to do a formulation expressed per hand you need to figure the percentage that you would make each bet amount.

Your approach makes an extremely easy problem into a mathematical nightmare where you must consider to many things. Simplify to an approach where the odds of success or failure are your concerns.

Doing this for an even money payoff and 50% chance of winning MUST give a result of zero. If it doesnt you made a mistake. Its an exercise in futility.

If you choose to play this system it doesnt give you any advantage over the house. What it does is exponentially increase your volatility. Your expected loss is a function of your AVERAGE bet.

So you will loss (average bet times house edge times 100) every 100 bets. Martingale your average bet would be larger depending on how many times you are willing to double your bet. Each time you double it raises your average bet raising your expected loss rate every 100 hands.

The affect of the system is you will very gradually win until you lose big. The sum of these two events will equal the expected loss for your average bet over time. If you lose your progression a little less frequently than expectation you may show a small profit for a while but if you lose a little more frequently you will show HUGE losses. With the expected number of failures, you lose proportionate to your average bet in the long run. Remember the casino has the advantage.

In blackjack the real house edge on any hand depends on the cards yet to be played. You may be making your largest bet were you are at a very large disadvantage. Not a good plan.

 

[tthree]

average gain per hand - average loss per hand = net gain per hand

therefore,

x/2 - 63x/384 = ~x/3

or

average earnings of 1/3 the minimum bet per hand over the long run.

Net gain per hand = average gain per hand - average loss per hand
Gain per hand = bet size *1/2 (in a coin flip)
Now we multiply this by bet size PER HAND.... How many times did I double my bet?


To do a per hand approach(as you chose) when not flat betting you have to break each bet amount into its own net gain. Then you have to multiply each by the frequency it is expected to be bet. Then you add these products together.

Let us assume for convenience you will win your bet 49.5% and lose 50.5% of the time and are paid even money.

NET GAIN: = average gain per hand - average loss per hand

for all 1 unit bets = (1 unit)*.495 - (1 unit)*.505 = (1 unit)*(.495 - .505) = -0.01 unit
for all 2 unit bets =(2 units)*(.495 - .505) = -.02 units
for all 4 unit bets = (4 units)*(-.01) = -.04 units
for all 8 unit bets = -.08 units
for all 16 unit bets = -0.16 units
for all 32 unit bets = -0.32 units

You will make the 1 unit bet far more frequently than your 32 unit.

Let the frequency of making all bets of n units = fn
ie frequency of 1 unit bet = f1

Let the net gain for all bets of n unit bet = gn
ie net gain for 2 unit bet = g2

The overall net gain is the sum of the product each bet amounts individual net gain and its frequency.

Overall net gain = g1*f1 + g2*f2 + g4*f4 + g8*f8 + g16*f16 + g32*f32

Enter net gain figures:
Overall net gain = -.01*f1 + -.02*f2 + -.04*f4 + -.08*f8 + -.16*f16 + -.32*f32

Enter net gain frequencies:
f1 = .515625 , f2 = .25, f4 = .125, f8 = .0625, f16 = .03125, f32 = .015625



OVERALL NET GAIN = -.0051625 + -.005 + -.005 + -.005 + -.005 + -.005 = -.0301625 units

As you can see when you flat bet you start with a -0.01 net gain per hand.
EACH TIME YOU DOUBLE YOUR BET YOU ADD ANOTHER 50% OF YOUR STARTING NET GAIN TO YOUR INITIAL NET GAIN OF -0.01 UNITS.


Net gain:
Flat betting -0.01, 2 bet martingale -0.0101625, 3 bet martingale -0.0151625, ... 6 martingale bets -0.0301625 units/hand


When you choose a martingale your general path has a steeper per hand decline than flat betting. With each additional bet in the sequence your path becomes steeper in exchange for a more winding path. So you lose money faster and are much, much more likely to deviate away from expected loss.

The short term results trend above the path very slowly; however, the corrections jump below the path. Your actual results are just as likely to deviate wildly in EITHER direction. Either slowly up or precipitously down. Think about how far behind you will be after a run of unsuccessful attempts that will eventually happen. EACH DOUBLE BET makes your expectation that MUCH WORSE and HOW FAR BELOW THIS EXPECTATION YOU ARE LIKELY TO DEVIATE BECOMES GREATER!!!!!!!!!!!