Roulette Odds Using Martingale
November 4th, 2015 at 8:51:58 PM
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Hi,
What is the probability that you could double your bankroll in European Roulette and in American Roulette using the martingale betting strategy while only being able to double your bet 10 times?
Thank You,
Drew
What is the probability that you could double your bankroll in European Roulette and in American Roulette using the martingale betting strategy while only being able to double your bet 10 times?
Thank You,
Drew
November 5th, 2015 at 12:45:06 AM
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is your session bankroll enough so that if you lose your first 10 bets that youll end with $0?
and youre talkimg about playing an even money wager, yes?
and youre talkimg about playing an even money wager, yes?
November 5th, 2015 at 1:20:36 AM
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Someone can give you better more accurate information but this might help some. IIRC. You have more than a 83% of winning about 14 units if you play for about an Hr on an average European roulette table . 17% of the time you'll lose 91 units.Quote: dtanthonyHi,
What is the probability that you could double your bankroll in European Roulette and in American Roulette using the martingale betting strategy while only being able to double your bet 10 times?
Thank You,
Drew
The longer you play the worst your odds of being ahead is
Play for 7 or 8 hrs and that 83+% drops about in half.
This is all from memory and I have no simulation or math to back it up.
♪♪Now you swear and kick and beg us That you're not a gamblin' man Then you find you're back in Vegas With a handle in your hand♪♪ Your black cards can make you money So you hide them when you're able In the land of casinos and money You must put them on the table♪♪ You go back Jack do it again roulette wheels turinin' 'round and 'round♪♪ You go back Jack do it again♪♪
November 5th, 2015 at 2:22:30 AM
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Let's say you have a starting bankroll so that you have just enough to survive a 10-streak loss and would be unable to bet on the 11'th round. If your starting bet is $1, then you need:
1
2
4
8
16
32
64
128
256
512
+_____
1023 units (or dollars).
For american roulette, the probability of LOSING 10 hands in a row is:
(20/38) ^ 10 = 0.00163103766
Or 0.163...%
Probability of winning any given hand in a 10-hand cycle (I believe) is 1 - 0.00163103766 = 0.99836896233
The probability to win a cycle 1023 times in a row is:
0.99836896233 ^ 1023 = 0.18826330207 or about 18.8%.
For Canadian Roulette I'd do (19/37) ^ 10 = 0.0012750252 = chance of losing 10 in a row.
Chance to win 1023 in a row is:
1 - 0.0012750252 = 0.9987249748
0.9987249748 ^ 1023 = 0.27112290764 or about 27% success rate.
Of course, my math may be off. Someone might wanna check that out.
1
2
4
8
16
32
64
128
256
512
+_____
1023 units (or dollars).
For american roulette, the probability of LOSING 10 hands in a row is:
(20/38) ^ 10 = 0.00163103766
Or 0.163...%
Probability of winning any given hand in a 10-hand cycle (I believe) is 1 - 0.00163103766 = 0.99836896233
The probability to win a cycle 1023 times in a row is:
0.99836896233 ^ 1023 = 0.18826330207 or about 18.8%.
For Canadian Roulette I'd do (19/37) ^ 10 = 0.0012750252 = chance of losing 10 in a row.
Chance to win 1023 in a row is:
1 - 0.0012750252 = 0.9987249748
0.9987249748 ^ 1023 = 0.27112290764 or about 27% success rate.
Of course, my math may be off. Someone might wanna check that out.
November 5th, 2015 at 7:11:13 AM
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Making the assumption that, if you have 10 losses in a row, you will continue playing, resetting your bet to the minimum and betting what's left of your bankroll if you don't have enough to double your previous bet, I ran a simulation of 1 million runs of four different conditions:
If it is a double-zero wheel and your bankroll is 1023 times your starting bet, you double it before losing it all 23.3% of the time.
If your initial bankroll is 2046 times your starting bet, you double it 9.3%.
If it is a single-zero wheel (and assuming that you lose the entire bet if it lands on 0), you double your bankroll 35.4% with a bankroll of 1023 initial bets and 24.0% with a bankroll of 2046 initial bets.
Yes, the larger the bankroll, the less likely you are to double it. This is because you need more winning bets to double it, and the more bets you make, the more money you expose to the house edge.
If you are wondering what the probability is of doubling your bankroll before losing 10 in a row, RS's numbers are correct - 18.83% on a double-zero wheel, and 27.11% on a single-zero one.
If it is a double-zero wheel and your bankroll is 1023 times your starting bet, you double it before losing it all 23.3% of the time.
If your initial bankroll is 2046 times your starting bet, you double it 9.3%.
If it is a single-zero wheel (and assuming that you lose the entire bet if it lands on 0), you double your bankroll 35.4% with a bankroll of 1023 initial bets and 24.0% with a bankroll of 2046 initial bets.
Yes, the larger the bankroll, the less likely you are to double it. This is because you need more winning bets to double it, and the more bets you make, the more money you expose to the house edge.
If you are wondering what the probability is of doubling your bankroll before losing 10 in a row, RS's numbers are correct - 18.83% on a double-zero wheel, and 27.11% on a single-zero one.
November 6th, 2015 at 2:07:00 PM
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You'd have a much better chance of doubling if you bet your whole bankroll at once which will succeed with probability 18/37 ≈ 48.65%. However, there is a strategy which can double your bankroll with a higher probability than this. That is to bet on a number. You always bet enough to win on one bet, and if that's not possible, then you bet your whole bankroll. So on the first bet you would bet 1/35 of your bankroll. This method has a lower total average bet which exposes less of your bankroll to the house edge, giving you a higher EV and hence a higher probability of success. You can double up with probability 49.0%. But there's a catch. In order for this to be better than betting your whole bankroll on a color, you have to place your bets with sufficient precision, which means that you have to buy in for a sufficient number of chips that represent your whole bankroll. Unfortunately, depending on your bankroll, this can require a lot of chips.
For example, if you want to double a $100 bankroll, buying in for 100 chips so that each chip is worth a dollar will only win at a rate of about 45%, much worse than betting your whole bankroll on a color. If you buy in for 1000 chips so each is worth 10 cents, this becomes about 48.63%, almost the same as betting it all on a color. But if you buy in for 10000 chips so each chip is worth a penny, then you will double your bankroll with a probability of about 49.0%. Clearly large bankrolls might require an impractical number of chips, in which case it will become better to bet on a color. The point at which betting a number starts to become better is when you have more than about 1233 chips per 100 dollar bankroll. The theoretical maximum if you had infinitely many chips is about 49.03%.
For American roulette, you don't need nearly as many chips to make betting on a number better than the 18/38 ≈ 47.37% chance you get by betting it all on a color. It starts to become better when you have more than just 541 chips per 100 dollars of bankroll. 2000 chips per 100 dollars will give a 47.9% rate. The theoretical max for infinitely many chips is about 48.1%.
These numbers come from an R function I wrote that computes the exact success probability for reaching a goal for any game with a fixed payout using this strategy. You input the starting bankroll, goal bankroll, payout odds, win probability, and chip value. For example, if I have a $100 bankroll, my goal is a $200 bankroll, betting a number with a 35:1 payout for European roulette with win probability 1/37, and each chip is worth a dollar, I would enter
> parlay(100,200,35,1/37,1)
which returns a success probability of 0.450183. But if I make each chip worth 1 cent, then we have
> parlay(100,200,35,1/37,.01)
which returns 0.48986.
For example, if you want to double a $100 bankroll, buying in for 100 chips so that each chip is worth a dollar will only win at a rate of about 45%, much worse than betting your whole bankroll on a color. If you buy in for 1000 chips so each is worth 10 cents, this becomes about 48.63%, almost the same as betting it all on a color. But if you buy in for 10000 chips so each chip is worth a penny, then you will double your bankroll with a probability of about 49.0%. Clearly large bankrolls might require an impractical number of chips, in which case it will become better to bet on a color. The point at which betting a number starts to become better is when you have more than about 1233 chips per 100 dollar bankroll. The theoretical maximum if you had infinitely many chips is about 49.03%.
For American roulette, you don't need nearly as many chips to make betting on a number better than the 18/38 ≈ 47.37% chance you get by betting it all on a color. It starts to become better when you have more than just 541 chips per 100 dollars of bankroll. 2000 chips per 100 dollars will give a 47.9% rate. The theoretical max for infinitely many chips is about 48.1%.
These numbers come from an R function I wrote that computes the exact success probability for reaching a goal for any game with a fixed payout using this strategy. You input the starting bankroll, goal bankroll, payout odds, win probability, and chip value. For example, if I have a $100 bankroll, my goal is a $200 bankroll, betting a number with a 35:1 payout for European roulette with win probability 1/37, and each chip is worth a dollar, I would enter
> parlay(100,200,35,1/37,1)
which returns a success probability of 0.450183. But if I make each chip worth 1 cent, then we have
> parlay(100,200,35,1/37,.01)
which returns 0.48986.
######################################################################
# Function parlay(br, goal, odds, p, chip=1)
#
# Computes probability of parlaying bankroll of size br to
# size goal with given payout odds, probability of a win p,
# and size of a chip which is a power of 10 like .1, 1, 10, etc.
######################################################################
parlay = function(br, goal, odds, p, chip=1) {
prob = 0
p_fail = 1
count = 0
while ( (br >= chip) & (br < goal) & (count < 1000) ) {
# Compute probabity of completing parlay up
consec = 0
while (br + br*odds <= goal) {
br = br + br*odds
consec = consec + 1
}
p_parlay = p^consec
# Compute probabilty of reaching goal
bets = 0
if (br < goal) {
while ( br >= chip*ceiling((goal-br)/odds/chip) ) {
br = br - chip*ceiling((goal-br)/odds/chip)
bets = bets + 1
}
}
p_goal = ifelse(bets, 1-(1-p)^bets, 1)
delta = p_fail*p_parlay*p_goal
prob = prob + delta
p_fail = p_fail*p_parlay*(1-p)^bets
count = count + 1
}
prob # Probability of reaching goal
November 8th, 2015 at 1:11:09 PM
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But 51.35% of the time they feel sick to their stomach until they forget about itQuote: BruceZYou'd have a much better chance of doubling if you bet your whole bankroll at once which will succeed with probability 18/37 ≈ 48.65%. ]
48.65% of the time they may be happy for a while but they will just end up losing it.
Also there is no entertainment value.
If you're desperately trying to double your BR I say you should find a sports pick from a known winning sports bettor and bet it all on that.
You can enjoy the game and avoid the humiliation and shock of losing everything in seconds.
♪♪Now you swear and kick and beg us That you're not a gamblin' man Then you find you're back in Vegas With a handle in your hand♪♪ Your black cards can make you money So you hide them when you're able In the land of casinos and money You must put them on the table♪♪ You go back Jack do it again roulette wheels turinin' 'round and 'round♪♪ You go back Jack do it again♪♪
