I have been wondering about the following question for a while. Consider the martingale strategy for roulette, where you bet on a color, such as red. If you put \$1 on red and win. You walk away, or bet another on \$1. If you lose, you double your bet and put it on red. That way if you win, you win back your previous investment, plus the \$1 payout from the original wager. If you lose, you double your bet and try again. Obviously with a finite amount of money this strategy will eventually bankrupt you.

Now consider the augmented strategy, where after a certain number of losses, you accept the loss and rather than keep doubling your bet you start over at \$1. For example, after 4 losses in a row, you would have \$15 Invested ;1+2+4+8. Now rather than keep doubling you start the system over.

So the question becomes are you able to hit on 15 reds before you hit on 4 blacks in a row. Runs of black less than 4 are obviously allowed.

In roulette there are 38 equally likely outcomes, 18 red, 18 black, 2 green.

  • 2
    $\begingroup$ What is the question exactly? $\endgroup$ – Peter Foreman Jul 6 at 23:13
  • $\begingroup$ In a game of roulette, what will happen first, 15 reds interspersed with runs of less than 4 blacks. Or a run of 4 blacks. $\endgroup$ – Jerome Turner Jul 6 at 23:23
  • $\begingroup$ The question doesn’t seem to be at all related to the strategy. It sounds like you can play the game forever so all possibilities will occur. $\endgroup$ – Alex R. Jul 6 at 23:25
  • $\begingroup$ If you stopped the game after one of the two outcomes happened. Either 15 reds or a run of 4 blacks. Which is expected to happen first. Sorry about the clumsy wording. $\endgroup$ – Jerome Turner Jul 6 at 23:33

Each bet is a losing proposition. No sum of negative numbers is positive. The strategy as a whole is losing.

We can compute the expected value of one run of the strategy, which is playing until you get either one red or four blacks in a row. On each spin you win with probability $\frac {18}{38}$ and lose with probability $\frac {20}{38}$. You lose $4$ in a row with probability $\left(\frac{20}{38}\right)^4$ and lose $15$ in that case. In all other cases you win $1$. The expected value is then $$1\cdot\left(1-\left(\frac{20}{38}\right)^4\right)-15\left(\frac{20}{38}\right)^4=-\frac{29679}{130321}\approx -0.2277$$ So the expected value of each trial is about $-23\%$ of the initial bet. On each run you have $1-\left(\frac{20}{38}\right)^4\approx 92.3\%$ chance of winning, but because the losses are so much larger than the wins the expected value is negative. The chance of winning $15$ times in a row is $\left(1-\left(\frac{20}{38}\right)^4\right)^{15}\approx 0.302$


As to the question in the comments of which will happen first, four straight blacks or $15$ total reds, we can model this as a finite-state absorbing Markov chain. There are $60$ non-absorbing states of the form, $(r,b),\,0\leq r<15,\, 0\leq b<4$, where $r$ is the total number of reds, and $b$ is the current number of consecutive blacks.

import numpy as np

# non-absorbing state is (r, b), 0 <= r <= 14, 0 <= b <= 3
# numbered as 15*b + r
# r is total reds, b is current blacks
# state 60 is 4 straight blacks
#state 61 is 15 total reds

P = np.zeros((62, 62))
for r in range(15):
    for b in range(4):
        s = 15*b + r  # state (r, b)
        s1 = r+1 if r <14 else 61   # state(r+1, 0)
        s2 = 15*(b+1)+r if b < 3 else 60 # state(r,b+1)
        P[s,s] = 1/19   # roll green
        P[s,s1] = 9/19    # roll red
        P[s,s2] = 9/19   # roll black
P[60,60] = 1
P[61,61] = 1
Q = P
for n in range(100):
start = np.zeros(62)
start[0] = 1
S = start@Q
print('4 straight blacks', S[60]) 
print('15 reds', S[61])
print('total', S[60]+S[61])


4 straight blacks 0.6201875941847539
15 reds 0.3798124058152452
total 0.9999999999999991

so if I haven't made a mistake, you lose $62\%$ of the time with this strategy.


Let’s start with the question in the exact form you asked it.

When you start playing, you can partition the possible outcomes into two events:

$Q.$ Red, possibly possibly preceded by one or more greens which in turn might be preceded or interleaved with sequences of one to three blacks.

$R.$ Four blacks, possibly preceded by one or more greens which in turn might be preceded or interleaved with sequences of one to three blacks.

Work out $P(Q)$ (either directly or by finding $P(R)$ and subtracting it from $1$), and work out the probability of 15 consecutive successful Bernoulli trials with probability $P(Q).$

I find this uninteresting, however, because it ignores all the losses you got when the ball fell in a green slot.

I would rather look instead at these two events:

$A.$ Red, possibly preceded by up to three non-reds.

$B.$ Four non-reds in a row.

Now $P(B)=\frac{10000}{130321} \approx 0.0767336$ and therefore $P(A) =\frac{120321}{130321} \approx 0.9232664.$ We need $A$ to occur $15$ times in a row. The probability of that is $$ (P(A))^{15} =\left(\frac{120321}{130321} \right)^{15} \approx 0.301929 .$$ In other words, you are probably going to encounter four consecutive losses before you accumulate $15$ wins.

To put it another way, for every $1/P(B)=13.0321$ trials of $A$ vs. $B,$ on average $B$ will occur once. The rest of the time you’ll get $A.$ Since you win $1$ when $A$ occurs but lose $15$ when $B$ occurs, this is a losing strategy (like every other strategy you could ever possibly devise).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.