- #1
salzrah
- 81
- 0
Hi, I have been learning/experimenting with the Martingale betting system recently. I have read a lot about how no "system" works for betting in casinos. However, I want to either prove or disprove the validity of the system by looking at its expected value/payout. I will be using the game of roulette as an example, assuming I have a 50% of getting black or red. Here is how the payout will work-
If the ball lands on black, I win $5
If the ball lands on red 7 times in a row, I lose $155
What is the expected payout of me playing roulette with this system?
winning payout = $5 * (1/2) ;have a 50% chance of getting black
losing payout = -$155 * (1/2)^7 ;have a 50% chance of getting red, 7 times
net payout = winning payout - losing payout
=$1.29
So according to this, if I use this system then there will be a positive payout. What do you guys think?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
If you want to how i got the winning/losing payouts then here is the explanation for that-
The way the system works is that you first bet your initial, let's say it's $5. If you lose, you double your previous bet, and if you lose again you double up again, etc. So if I start at $5 and keep losing, my bets will go like this - $5,$10,$20,$40,$80...
Using this system, you will earn $5 every time you win. To see this, assume I lose $5,$10,$20 but on the $40 bet I win. I will have lost $35 from before but gained $40, i.e. $5 profit.
***Now, according to this if I have enough money I can eventually earn $5 as long as I have a crap load of money. This is why tables have maximum bets, but many cheap casinos don't.
Moving on, if i lose 5 times in a row, I will lose 5+10+20+40+80 = $155. Let's say I "restart" after I lose 5 times. Also, because I am so clever, I will bet only until I see two reds. Therefore, I can technically "lose" 7 times in a row and still lose only $155.
All in all, one can see that this ^ is where the losing payout comes from, and that the winning payout is me simply winning $5 on black.
If the ball lands on black, I win $5
If the ball lands on red 7 times in a row, I lose $155
What is the expected payout of me playing roulette with this system?
winning payout = $5 * (1/2) ;have a 50% chance of getting black
losing payout = -$155 * (1/2)^7 ;have a 50% chance of getting red, 7 times
net payout = winning payout - losing payout
=$1.29
So according to this, if I use this system then there will be a positive payout. What do you guys think?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
If you want to how i got the winning/losing payouts then here is the explanation for that-
The way the system works is that you first bet your initial, let's say it's $5. If you lose, you double your previous bet, and if you lose again you double up again, etc. So if I start at $5 and keep losing, my bets will go like this - $5,$10,$20,$40,$80...
Using this system, you will earn $5 every time you win. To see this, assume I lose $5,$10,$20 but on the $40 bet I win. I will have lost $35 from before but gained $40, i.e. $5 profit.
***Now, according to this if I have enough money I can eventually earn $5 as long as I have a crap load of money. This is why tables have maximum bets, but many cheap casinos don't.
Moving on, if i lose 5 times in a row, I will lose 5+10+20+40+80 = $155. Let's say I "restart" after I lose 5 times. Also, because I am so clever, I will bet only until I see two reds. Therefore, I can technically "lose" 7 times in a row and still lose only $155.
All in all, one can see that this ^ is where the losing payout comes from, and that the winning payout is me simply winning $5 on black.