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Old 06-08-2012, 01:48 PM   #1
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Question regarding EV on a martingale variant

Hello.

Suppose we are offered 2-to-1 odds for a game. These odds correspond to a (fake) probability of 33% of our bet happening.
Since the house should have an edge, let's suppose that this edge corresponds to an (actual) probability of 23%, or something along the lines of 10-to-3.

Our bankroll is $150, we start with $10 and double up on every loss.
This means we can take up to 3 losses.

EV is calculated by the formula EV = Total Profit * (1 - q) - Total Loss * q, where q == the actual probability of losing every bet up to and including this. In simpler terms, q = (1 - 0.23)^(bet No#)

Bet #1: Bet: $10, (possible) Total Profit: $20, (possible) Total Loss: $10, EV = -3
Bet #2: Bet: $20, (possible) Total Profit: $30, (possible) Total Loss: $30, EV = -5.27
Bet #3: Bet: $40, (possible) Total Profit: $50, (possible) Total Loss: $70, EV = -4.08
Bet #4: Bet: $80, (possible) Total Profit: $90, (possible) Total Loss: $150, EV = +7.08

Now, this shows that every sequence of 4 bets produces a positive EV.

My question is this: Given that my calculations are correct (if not, please point out the errors), is this strategy sound? And if not, why is this?
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Old 06-08-2012, 02:23 PM   #2
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Re: Question regarding EV on a martingale variant

It seems you found the holy grail.
Please delete this thread!
This is how the whole forum makes their living and we do not want to have it out in the open.
Good luck.
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Old 06-08-2012, 02:47 PM   #3
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Re: Question regarding EV on a martingale variant

You can't combine -EV bets to make a +EV series, no matter what.

Even if its 2-1 or 100-1.
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Old 06-08-2012, 02:56 PM   #4
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Re: Question regarding EV on a martingale variant

Quote:
Originally Posted by acesover8s View Post
You can't combine -EV bets to make a +EV series, no matter what.

Even if its 2-1 or 100-1.
Thanks for your reply.

How come this does not show up in the calculations? Could you please point out where the mistake is in the math?
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Old 06-08-2012, 03:16 PM   #5
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Re: Question regarding EV on a martingale variant

Quote:
Originally Posted by panos View Post
In simpler terms, q = (1 - 0.23)^(bet No#)
So when you lose the first round karma then transfers over to the next I N D E P E N D E N T event adding %s to your hit probability
It would make a nice video game tho
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Old 06-08-2012, 03:27 PM   #6
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Re: Question regarding EV on a martingale variant

Quote:
Originally Posted by tukkk View Post
So when you lose the first round karma then transfers over to the next I N D E P E N D E N T event adding %s to your hit probability
It would make a nice video game tho
Isn't a prerequisite of the total loss (let's say in bet #3), that bet #3 and both bet #1 and bet #2 are unsuccessful?

Since they are independent events, isn't the probability of losing all 3 bets q=(1 - 0.23) ^ 3 ?
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Old 06-08-2012, 08:22 PM   #7
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Re: Question regarding EV on a martingale variant

Quote:
Originally Posted by panos View Post
Isn't a prerequisite of the total loss (let's say in bet #3), that bet #3 and both bet #1 and bet #2 are unsuccessful?

Since they are independent events, isn't the probability of losing all 3 bets q=(1 - 0.23) ^ 3 ?
that is the probability of losing all 3 bets in sum. but when you're placing bet 3, you've already lost the first two bets.
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Old 06-09-2012, 02:32 AM   #8
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Re: Question regarding EV on a martingale variant

brb printing money.
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Old 06-09-2012, 04:10 PM   #9
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Re: Question regarding EV on a martingale variant

Quote:
Originally Posted by wahoo3 View Post
that is the probability of losing all 3 bets in sum. but when you're placing bet 3, you've already lost the first two bets.
I finally got it. Thanks a lot!
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Old 06-12-2012, 07:53 PM   #10
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Re: Question regarding EV on a martingale variant

Glad the forum could do your homework for you.
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