Why does this version of martingale fail please explain
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I worked out that if your first bet is 1 unit and you increase your second bet to anything more than 2 units (for example 4) then continue to double for say upto 3 more attempts (5 step) until winning or busting, it is approx 3 times more profitable and only costs approx twice as much as a 5 step standard martingale on a bust.
Eg
1,2,4,8,16 = 1 unit win each win, 31 unit loss each bust
1,4,8,16,32 = 3 unit win each win, 61 unit loss each bust
Can someone explain why this isnt a winning strategy?
Not interested in smart alec comments cheers
Big deal.
Martingales don't work because, sooner or later, you'll hit a string of bad luck where the next bet is either greater than the table limit, or exceeds your bankroll.
say you bet on a fair coin flip at Sally's CasinoQuote: Dantheman1,4,8,16,32 = 3 unit win each win, 61 unit loss each bust
Can someone explain why this isnt a winning strategy?
your first bet is on A side (A and Z are the bets)
you WIN!
you won 1 cuz ur 1st bet was only 1
not 3
that happens 50% of the time
now
you must calculate your average bet
for the game you wants to play in a casino
and you will C the light
have fun and win!
Sally
BUT What if, is the next question.Quote: DJTeddyBearUnless I missed something, all you did was skip one step. As a result, when (if) you finally win, you win three units instead of one.
Big deal.
Martingales don't work because, sooner or later, you'll hit a string of bad luck where the next bet is either greater than the table limit, or exceeds your bankroll.
1) win after a win = 1
2) win after a loss of less then 5 in a row = 3
3) lose 5 bets in a row = -61
I know losing 5 bets or more in a row is common but im only interested in multiples of 5 eg 5,10,15,20
as for example 6 losses in a row then a win would mean i bust once (-61) and win after a loss once (+3)
How do I find out the probability of losing exactly 5 times in a row (-61), 10 times in a row -122), 15 in a row (-183)
and so on?
Because you are playing a -EV game.Quote: DanthemanI was playing around with different variations of martingale recently.
I worked out that if your first bet is 1 unit and you increase your second bet to anything more than 2 units (for example 4) then continue to double for say upto 3 more attempts (5 step) until winning or busting, it is approx 3 times more profitable and only costs approx twice as much as a 5 step standard martingale on a bust.
Eg
1,2,4,8,16 = 1 unit win each win, 31 unit loss each bust
1,4,8,16,32 = 3 unit win each win, 61 unit loss each bust
Can someone explain why this isnt a winning strategy?
Not interested in smart alec comments cheers
If it worked casinos would not let you do it.
Teams would simply over run casinos with monster bankrolls.
How many examples of 10 or more do you want? Go though a casino and look at digital table game tally boards like roulette black/red even/odd You will have no problem seeing 10 or more.Quote: DanthemanSo theres basically 3 outcomes
1) win after a win = 1
2) win after a loss of less then 5 in a row = 3
3) lose 5 bets in a row = -61
I know losing 5 bets or more in a row is common but im only interested in multiples of 5 eg 5,10,15,20
as for example 6 losses in a row then a win would mean i bust once (-61) and win after a loss once (+3)
How do I find out the probability of losing exactly 5 times in a row (-61), 10 times in a row -122), 15 in a row (-183)
and so on?
Quote: mustangsallysay you bet on a fair coin flip at Sally's Casino
your first bet is on A side (A and Z are the bets)
you WIN!
you won 1 cuz ur 1st bet was only 1
not 3
that happens 50% of the time
now
you must calculate your average bet
for the game you wants to play in a casino
and you will C the light
have fun and win!
Sally
Would the avg win be approx 2?
Half the time win 1 unit half the time win 3 units (1+3)/2
1: 50%
2: 25%
3: 12.5%
4: 6.25%
Lose 5 in a row: 6.25%
Payouts at each round are: 1,3,3,3,-61
EV of 1: 0.5
EV of 2: 0.75
EV of 3: 0.375
EV of 4: 0.1875
EV of lose 5 in a row: -3.8125
Expected loss of this system is a loss of 2.75 per "round", which is playing until you win OR lose 5 times.
The expected gain or loss of flat-betting is 0. The more "levels" you play martingaling, the lower the chance of hazard, and the bigger the hazard.
EV of 1: 0.5
EV of 2: 0.25
EV of 3: 0.125
EV of 4: 0.0625
EV of lose 5 in a row: -1.9375
Expected loss of this system is a loss of 1.05 per "round", which is playing until you win OR lose 5 times.
Quote: DanthemanI was playing around with different variations of martingale recently.
I worked out that if your first bet is 1 unit and you increase your second bet to anything more than 2 units (for example 4) then continue to double for say upto 3 more attempts (5 step) until winning or busting, it is approx 3 times more profitable and only costs approx twice as much as a 5 step standard martingale on a bust.
Eg
1,2,4,8,16 = 1 unit win each win, 31 unit loss each bust
1,4,8,16,32 = 3 unit win each win, 61 unit loss each bust
Can someone explain why this isnt a winning strategy?
Not interested in smart alec comments cheers
To answer your question directly, yes, you triple your wins if you win. Yes, you double your hazard if you lose. Triple is more than double, right? ... well... at first that makes sense, but it's really your brain fooling you. Tripling 1 is 3, so a magnitude of 2 change. Doubling ~30 is 60, a magnitude change of 30. Doubling the big loss is way worse than winning 2 more a few times. Cheers.
Quote: surrender88sTo answer your question directly, yes, you triple your wins if you win. Yes, you double your hazard if you lose. Triple is more than double, right? ... well... at first that makes sense, but it's really your brain fooling you. Tripling 1 is 3, so a magnitude of 2 change. Doubling ~30 is 60, a magnitude change of 30. Doubling the big loss is way worse than winning 2 more a few times. Cheers.
Thats what is confusing me-
If it is still going to lose with the same frequency as the standard martingale, doesnt the extra amount in wins before that happens mean it is better?
Quote: DanthemanThats what is confusing me-
If it is still going to lose with the same frequency as the standard martingale, doesnt the extra amount in wins before that happens mean it is better?
Like every other martingale variant, it works great until you can't place a big bet.
When you can't place a big bet - either due to table limit or bankroll constraint - you've lost a big chunk of money, and that is quite hard to recover from, even winning 3 units at a time.
These people spend hours a day (daily) in a casino with some losing system. They treat it like it's a job. They've figured out how much $$$ they think they should be making per roll/shooter/hour/day whatever. But they THINK they're gonna be winners.
With that being said, do what you want with your money -- the casino will be glad to take it from you.
Quote: DanthemanThats what is confusing me-
If it is still going to lose with the same frequency as the standard martingale, doesnt the extra amount in wins before that happens mean it is better?
Actually, if "lose" means "reach the point where you can't make the next bet necessary," then losing will happen more often if you bet more because the amounts you bet after each number of losses are larger as well.
Compare the examples, and assume your bankroll is 1500:
"Normal" Martingale: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024; after 10 losses, you are behind 2047
"1-4" Martingale: 1, 4, 8, 16, 32, 64. 128. 256, 512, 1024; after only 9 losses, you are behind 2045
And don't even think about going to the ATM and "temporarily borrowing" the money for your next mortgage payment, under the assumption that "I'll win it back eventually."
Remember the words of Oscar Madison (that's Quincy Oscar Madison, not the new Chandler one): "There's no such thing as a sure thing; that's why they call it gambling." ("Pro wrestling is fixed, right? There's no possible way Vince McMahon lets Brock Lesnar beat Undertaker at WrestleMania...")
noQuote: DanthemanWould the avg win be approx 2?
Half the time win 1 unit half the time win 3 units (1+3)/2
you lose 1/32 of the time so the avg win is a weighted average
cuz u win 31/32
I gets 61/31 or 1.967741935 (rounded) for the avg win
(I bet 3.4 cents one wants to see the math on that one)
so you lose 1 time in 32 tries
-61
and win 1.967741935 31 times
all on average of course
1.967741935 * 31 = 61
a wash (ding! i am worth millions and still have to do my own laundry)
just like a regular Marty 5 step
you lose 1 time in 32 tries
-31
and win 1 (avg win = 1) 31 times
all on average of course
hey, echo
but good luck finding a 50/50 game in a real casino
the games are not fair there (by definition that is)
Sally
