Modified Martingale and Baccarat system Questions

Hi Wizard, I came across this link:
/ask-the-wizard/baccarat

Where one of your fans sent you a question:

Q. I have a thick, and I mean thick, friend who is intoxicated with having won a fair amount betting Player only in baccarat. He plays $10 units and does the dumb play that 9 straight Banker wins won't happen. He escalates to a risk of $3,980 by going 10 10 30 60 120 250 500 1000 2000. How can I find solid mathematical evidence to try to convince him to stop?

A. This is a close variation of the Martingale betting system, in which the player doubles after every loss. Usually, the Martingale player will win but occasionally he will have more consecutive losses than he can handle and suffer a major loss. Assuming your friend is betting on the player, the probability that any given bet will begin a streak of nine losses in a row is (2153464/(2153464+2212744))9 =~ .001727, or 1 in 579, assuming ties are ignored. There is more information available about the folly of the Martingale in my section on betting systems. However, the more ridiculous a belief is the more tenaciously it tends to be held. It usually takes a big loss to possibly convince a believer in any particular betting systems to stop.

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So what I was wondering is what I'm missing in this situation, since it seems profitable if 9 losses occurs 1/579.

This means, over millions of hands, you will win 578 units before losing $3980 (as used in the posters question).

Isnt this profitable at a bare minimum of $5780, but likely much more since during legs 6-8, the profit is 2 units. I.e. profit would be somewhere around the $7K mark on average?

Im sure im missing something obvious because it can't be that simple but yea, please advise!

Thanks!

From first glance, what you appear to be "missing" is, if you lose your first bet but win your second, which happens slightly less than 1/4 of the time, then you break even.

Here's a breakdown of what you would make when you win after 0, 1, ..., 8 losses, and when you lose all 9 bets:

Bet	Prob	Profit	EV
10	0.493211500688927	10	4.93211500688927
10	0.249953916277104	0	0
30	0.126673770126999	10	1.26673770126999
60	0.0641968098647375	10	0.641968098647375
120	0.0325342049319086	10	0.325342049319086
250	0.0164879608937209	20	0.329759217874417
500	0.00835590895802845	20	0.167118179160569
1000	0.00423467856121919	20	0.0846935712243838
2000	0.00214608639310505	20	0.0429217278621009
9 losses	0.00220516330425065	-3980	-8.7765499509176

The sum of the EVs = -0.985894398670407.

deleted because Don's chart fully answered the question

Oh snap! Duh i cant believe i missed that.

So does -0.98 mean that this strategy carries a ~-1% loss over long term?

Also does every Martingale add up to -.98?

Quote: waznboi03

So does -0.98 mean that this strategy carries a ~-1% loss over long term?

1% loss of what? Your expected loss is 9.8% of your base bet of 10.

This has gotten a bit of debate here. Technically, it's 9.8% of your base bet, but if you are willing to commit to following this through and losing the whole 3980 each time, then you can consider the HE as about 100% x 1 / 3980, or 0.4%. Personally, I don't consider HE with multiple-bet systems - just EV.

Quote: waznboi03

Also does every Martingale add up to -.98?
link to original post

No. A straight 9-step baccarat player bet Martingale (10-20-40-80-160-320-640-1280-2560) has two possible outcomes; +10, with probability 452.48 / 453.48, and -5110, with probability 1 / 453.48, so the EV is -1.29.

BTW, your 1 / 579 number is the probability of nine consecutive player wins, not nine consecutive banker wins.

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a modified 9 step marty such as this would "seem" to better than the OP's because from the 3rd step on there would be a 2 unit ($20 instead of $10) win (if there is a win at any point) with $10 less total risk than the traditional marty

note the jump to 40 in the 3rd bet after the 2nd bet which with a win will cause a break even on that sequence allowing the bettor to win 2 units ($20 instead of $10 with a win on any bet from the 3rd bet onwards

but the seeming must be just an illusion because any system betting only Player has to have the same - EV in percentage terms

.
10 10 40 80 160 320 640 1280 2560


I like this much better than the traditional marty and will play if for fun on the Wizard's free bacc game


.

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I've modified the modified 9 step marty even more to make it more fun - the same total risk - yes, I know it can't be a winning system and it's unprofessional - but it's fun - to me anyway

here it is:

10 20 20 80 160 320 640 1280 2560

note the third step won't get the bettor ahead - he'll be down 10 - so if he wins that one he goes back to the 2nd step to try to get ahead

if he loses at the 3rd step - then a win on steps 4,5,6,7, 8, and 9 will cause a win of 30 instead of a win of 10 on the traditional marty which is doubling every time

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