Why does this version of martingale fail please explain

I 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

Can you explain why this is a winning strategy?

Unless 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.

Quote: Dantheman

1,4,8,16,32 = 3 unit win each win, 61 unit loss each bust

Can someone explain why this isnt a winning strategy?

say 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

Quote: DJTeddyBear

Unless 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.

BUT What if, is the next question.

1, 3, 7, 15, 31. You win a unit a hand as long as you don't bust and you only lose 57 units when you bust.

So 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: Dantheman

I 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

Because you are playing a -EV game.

If it worked casinos would not let you do it.

Teams would simply over run casinos with monster bankrolls.

Quote: Dantheman

So 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?

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.

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