Beating the odds

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rogerthat

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Nov 1, 2013
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After years of trying different methods I have concluded you simply cannot beat the odds on a 50/50 chance game.


Rules:

You can start your wager at .01 the max is 300.00
The payout is 2 to 1. So for every .01 unit you bet you win .02 back.
Coin flip style,


You can wager in any amount up to 300.00

You start out with 500.00

Is there a way to statistically beat this?


I was thinking starting bet at .03, if lose double up and +.01 and increases by .01 per turn and stopping at 6 bets in.

so bets would be, .03, .08, .19, .42, .89, 1.84

chances of losing 6 in a row is 1.5%
accumulation of loss is 3.45 per 100 so 3.45 x 1.5 = 5.175

based off of the above formula the average gain per bet is .06 times 100 means 6.00 per hhundred does that mean you can beat the system?

help please
 
Well, this game is 50/50. This is not roulette, as roulette you only have a 48% chance of winning if that's european roulette, american roulette is 47%.

The question I'm asking is at 50/50, can someone verify my formula, does it work statically?
 
Martingale

50/50% Martingale betting systems are no better than even odds in the long run.
You are insightful in your attempt.
Starting slow minimizes the risk of ruin. You increase your bet, only as your bankroll grows. That's good -- you will play a LOT longer than otherwise. But ruin you will, if there is a limit.

I used to play red and black (at the same time), doubling the color that loses.
Some one will try to help by explaining (correctly) that I will surely lose the first bet.
Now, they're hooked. I explain my system.
I make a (large) side bet that I can buy drinks for both of us from my winnings.

I always won that bet. I put my winnings from the wheel into escrow (not drinks) against the day that I must pay it all back (+vigorish) if I get very unlucky. I get to spend the "sucker's" money (less the vigorish)
 
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Bob Brown MSEE said:
50/50% Martingale betting systems are no better than even odds in the long run.
You are insightful in your attempt.
Starting slow minimizes the risk of ruin. You increase your bet, only as your bankroll grows. That's good -- you will play a LOT longer than otherwise. But ruin you will, if there is a limit.

I used to play red and black (at the same time), doubling the color that loses.
Some one will try to help by explaining (correctly) that I will surely lose the first bet.
Now, they're hooked. I explain my system.
I make a (large) side bet that I can buy drinks for both of us from my winnings.

I always won that bet. I put my winnings from the wheel into escrow (not drinks) against the day that I must pay it all back (+vigorish) if I get very unlucky. I get to spend the "sucker's" money (less the vigorish)
Click to expand...


Is there any way to statically come come ahead? If you can bet in any increments, with a a range of 30k?

Units start at .01 and can go up to 300.00, is there any way to set up a betting sequence to come ahead?
 
rogerthat said:
After years of trying different methods I have concluded you simply cannot beat the odds on a 50/50 chance game.


Rules:

You can start your wager at .01 the max is 300.00
The payout is 2 to 1. So for every .01 unit you bet you win .02 back.
Coin flip style,


You can wager in any amount up to 300.00

You start out with 500.00

Is there a way to statistically beat this?


I was thinking starting bet at .03, if lose double up and +.01 and increases by .01 per turn and stopping at 6 bets in.

so bets would be, .03, .08, .19, .42, .89, 1.84
Click to expand...
I don't understand this. If you start at .03 and, losing, "double up", that would be .06. If you "+.01" that would bring it to .07, not .08. If, betting .08, you lose again "double up", to .16 and "+01" that would be .17" or do you "increase by .01" the amount added? I think you mean that you start betting at .03 and, each time you lose, you double the previous bet, then add .01+ .01=.02, .01+ .02= .03, .01+ .04, etc.

Anyway, betting .03, .08, .19, .42, .89, 1.84, then going back to .03 when you lose you can
a) Win on the first turn. You win .03 and the probability of that is 5.
b) Win on the second turn. You win .08- .03= .05 and the probability of that is .25.
c) Win on the third turn. You win .19- (.03+ .08)= .08 and the probability of that is .125.
d) Win on the fourth turn. You win .42- (.03+ .08+ .19)= .12 and the probability of that is .0.0625.
e) Win on the fifth turn. You win .89- (.03+ .09+ .19+ .42)= .16 and the probability of that is 0.03125.
f) Win on the sixth turn. You win 1.84- (.03+ .09+ .19+ .42+ .89)= .22 and the probability of that is .0.015625
So far you have an expected value of .03(.5)+ .05(.25)+ .08(.125)+ .12(.0625)+ .16(.03125)+ .22(.015625)= 0.053575.
But if you lose 6 times in a row which, as you say has probability .015625, you will have lost .03+ .09+ .19+ .42+ .89= 1.62 (was this what you forgot?) so that there will be a true expected value of .053575- .162= -0.108425. That is, in each round of 6, you will lose an average of .11.

chances of losing 6 in a row is 1.5%
accumulation of loss is 3.45 per 100 so 3.45 x 1.5 = 5.175

based off of the above formula the average gain per bet is .06 times 100 means 6.00 per hhundred does that mean you can beat the system?

help please
Click to expand...
 
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