Expected value of a roulette strategy

Question

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.

Answer 1

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 $\left(1-\left(\frac{20}{38}\right)^4\right)^{15}\approx 0.302$

Answer 2

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):
    Q=P@Q
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])

produced

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.

Answer 3

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).