MATH 529 A Martingale Betting Strategy The traditional martingale betting strategy calls for the bettor to double the wager after each loss until finally winning. This strategy ensures that, even with a 1:1 payoff, the bettor will come out ahead in the end. The problem with the strategy is that with the size of the bet growing exponentially in this manner, table limits may be exceeded and/or the bettor may run out of money before finally winning. Here we will show how to avoid these pitfalls while guaranteeing a high probability of making a profit. The key is that the bettor does not have to double the bets! We assume that we are playing a game of chance for which we have probability p > 0 of winning and probability q =1− p of losing. We make an initial bet of $b , for which there is a payoff ratio of R :1 . The scenario is best illustrated and best executed with Roulette. There are various bets that can be made, such as the following four types:

Bet p q Payoff

Single Number 138

3738

35 :1

Block 638

3238

5 :1

Column 1238

2638

2 :1

Red 1838

2038

1:1 Step 1: We first determine our desired probability of winning w , where w is near 1. Here we will take w = 0.99 . Step 2: Determine the number of bets n needed to make the probability of winning within n bets to be at least w . Solution. In a sequence of independent bets, the probability of losing n times in a row is qn . Hence, the probability of at least one win in n bets is 1− qn . We simply solve for n

in the inequality 1− qn ≥ w to obtain ln(1−w)lnq

≤ n . Thus, the required number of bets is

n = ln(1−w)lnq

⎡

⎢⎢

⎤

⎥⎥ .

Example 1. With Column bets, if we want at least a 99% chance of winning within n

bets, then we need to be able to make n = ln(0.01)ln(26 / 38⎡

⎢⎢

⎤

⎥⎥=13 bets. And the actual

probability of winning is 1− (26 / 38)13 ≈ 0.9928 .

The initial bet is $ b . After each loss, we multiply the bet by a fixed constant C . So the bets amounts are $b , $Cb , $C2 b , . . . Step 3. Given an initial bet of $b , determine how much money we need in order to make n bets.

Solution. The amount of money required is M = Ci bi=0

n−1∑ = b×1−C

n

1−C.

Example 2. Starting with a $10 bet. How much money do you need to make 13 bets when (i) doubling the bet after each loss, and (ii) multiplying the bet by 1.6 after each loss.

Solution. (i) With b = $10 and C = 2, we need M =10×1− 213

1− 2=10× (213 −1) = $81,910 .

(ii) With b = $20 and C = 1.6, we only need M = 20×1−1.613

1−1.6= 20×1.6

13 −10.6

≈ $15,000 .

Step 4: Given a fixed amount of cash M , determine the maximum size of the initial bet so that you can make n bets.

Solution. We need b×1−Cn

1−C≤M , which gives b ≤M ×

C −1Cn −1

.

Example 3. Suppose you are ready to risk $25,000. What size initial bet can you make in order to be able to cover 13 bets while increasing bets by a factor of 1.6?

Solution. We must have b ≤ 25000× 1.6−11.613 −1

≈ $33.38 . So we can start with a $30 bet.

Note: The amount of the last possible bet will be Cn−1b , which cannot exceed the allowable maximum bet of the house. For roulette the standard table max is $10,000. In this case, the largest initial bet from Example 3 becomes

b ≤ 100001.612

≈ $35 .

Step 5: Determine a lower bound for the betting factor C that ensures that the payoff from the eventual win is greater than the sum of all previous losses. Solution. The sum of the losses on the first k bets is

b+Cb+ . . . +Ck−1b = b Ci

i=0

k−1∑ = b×1−C

k

1−C= b× C

k −1C −1

.

The next amount bet is Ckb with a payoff of R :1 . So winning this bet earns you

$RCkb . Therefore we need b× Ck −1C −1

< RCkb , or Ck −1C −1

< RCk . This inequality is also

maintained with the less strict inequality Ck −1C −1

<Ck

C −1≤ RCk , which is easier to solve

for C : 1C −1

≤ R , which gives 1R≤C −1 , or finally C ≥1+ 1

R=R+1R

.

Example 4. Using an R = 2 :1 payoff, we need C ≥1.5 to guarantee that we come out ahead when stopping on the first win. Suppose we start with a $20 bet. The following tables illustrate the profits upon the first win:

(i) Using C =1.5 Bet # Cumulative Loss Amount Bet Payoff Net Gain

1 0 20 40 40 2 20 30 60 40 3 50 45 90 40 4 95 67.50 135 40 5 162.5 101.25 202.5 40

(ii) Using C ≈1.6

Bet # Cumulative Loss Amount Bet Payoff Net Gain 1 0 20 40 40 2 20 32 64 44 3 52 51 102 50 4 103 82 164 61 5 185 131 262 77

(ii) Using C = 2

Bet # Cumulative Loss Amount Bet Payoff Net Gain 1 0 20 40 40 2 20 40 80 60 3 60 80 160 100 4 140 160 320 180 5 300 320 640 340

Step 6: What safeguard does the casino use in order to curtail such profits by bettors employing this strategy? (i) They limit the size of bets. (ii) They forbid using martingale strategies. (iii) They choose a payoff ratio that guarantees an average profit on any single bet. With a payoff of $ a , the players’ average gain for a bet of $b is ap− bq . So the

casino chooses a so that ap− bq < 0 . They simply choose a < qp×b . That is, they use a

payoff ratio of R :1 , where R < q / p .

A fair payoff ratio would be R :1 where R = q / p . Then using a = qp×b , we have

ap− bq = 0. That is, on average bettors break even.

Note: Suppose Cq ≤1 . Then C ≤1+ 1q−1=1+1− q

q=1+ p

q<1+ 1

R (because R < q

p, which

makes pq<1R

). But we want C ≥1+ 1R

. Hence, we must have Cq >1 .

Example 5. The roulette wheel has the numbers 1 – 36, 0, and 00; thus; the probabilities are out of 38. The actual payoffs are chosen by R = !q / !p , where !q and !p are out of 36. What are the fair payoffs and actual payoffs for a Block bet? Solution. For a Block, p = 6 / 38 and q = 32 / 38 . The fair payoff ratio is q : p = 32 : 6 = 5. 33 : 1. But then !p = 6/36 and !q = 30/36, hence the actual payoff ratio is !q / !p = 30 : 6 = 5 : 1. (Note: Because the numerator of !q is always less than q , while the numerators of !p and p are always the same, we see that in Roulette the actual payoff ratio !q / !p is less than the fair payoff ratio q : p .) Step 7: We shall make a maximum of n bets while employing the strategy of increasing the bet by a factor of C after each loss, and stopping after the first win. Determine the average net gain. Solution. We let Gk denote the gain after the k th bet, and let T count the number of bets need to obtain the first win. Then T ~ geo(p) but we may not actually get a win for we are stopping after a maximum of n bets. Thus we want E[GT∧n ] , where T ∧n is the minimum of T and n . To compute this average, we look at the range of possible winnings and losses and the probabilities of these events:

Bet # Cumul. Loss Bet Winnings Prob

1 0 b R b p 2 b Cb R Cb q p 3 b+Cb C2 b R C2 b q2 p 4 b+Cb+C2 b C3b R C3b q3 p

n−1 b(1+C + . . . +Cn−3) Cn−2 b R Cn−2 b qn−2 p n b(1+C + . . . +Cn−2) Cn−1b R Cn−1b qn−1 p n b(1+C + . . . +Cn−2) Cn−1b –Cn−1b qn

We now let W denote the winnings and L denote the losses. Then we needE[GT∧n ]= E[WT∧n ]−E[LT∧n ] . First, because Cq ≠1 , we have

E[WT∧n ]= Rb p (Cq)i

i=0

n−1∑ = Rb p×1− (Cq)

n

1−Cq= Rb p× (Cq)

n −1Cq−1

.

Next, as in Step 5, the cumulative loss for losing k bets is

b Ci

i=0

k−1∑ = b× C

k −1C −1

.

The probability of losing k bets is qk p for 1≤ k ≤ n−1 , and qn for k = n . Thus, the average loss is

E[LT∧n ]=bpC −1

(Ck −1)qk

k=1

n−1∑ +

b(Cn −1)qn

C −1=

bpC −1

Cq− (Cq)n

1−Cq−q− qn

1− q

⎛

⎝⎜⎜

⎞

⎠⎟⎟+

b(Cn −1)qn

C −1

=bpC −1

Cq− (Cq)n

1−Cq

⎛

⎝⎜⎜

⎞

⎠⎟⎟+

b((Cq)n − qn )C −1

−b(q− qn )C −1

=b(1− q)C −1

Cq− (Cq)n

1−Cq

⎛

⎝⎜⎜

⎞

⎠⎟⎟+

b((Cq)n − q)C −1

=b

C −1Cq−Cq2 − (Cq)n + q(Cq)n + (Cq)n − q− (Cq)n+1+Cq2

1−Cq

⎛

⎝⎜⎜

⎞

⎠⎟⎟

=b

C −1Cq+ q(Cq)n − q− (Cq)n+1

1−Cq

⎛

⎝⎜⎜

⎞

⎠⎟⎟=

bC −1

(Cq− q)(1− (Cq)n )1−Cq

⎛

⎝⎜⎜

⎞

⎠⎟⎟

= bq (Cq)n −1Cq−1

⎛

⎝⎜⎜

⎞

⎠⎟⎟.

Therefore, the average gain is

E[GT∧n ]= E[WT∧n ]−E[LT∧n ]= (Rb p− bq)(Cq)n −1Cq−1

⎛

⎝⎜⎜

⎞

⎠⎟⎟ .

Step 8: Verify that E[GT∧n ]< 0 . Solution. Because R < q / p , then R p− q < 0 ; thus, is Rb p− bq = b(Rp− q)< 0. And we have seen above that we must have Cq >1 . So (Cq)n >1 also. Hence, E[GT∧n ]< 0 . Example 6. With Column bets at Roulette, where p =12 / 38 , q = 26 / 38 , and R = 2 , if we use C =1.6 , then we have 99.28% chance of winning within n =13 bets. But on the very few occasions when we lose 13 in a row, we lose a lot of money. Our average gain using an initial bet of b = $20 is

E[GT∧n ]= (Rb p− bq)(Cq)n −1Cq−1

⎛

⎝⎜⎜

⎞

⎠⎟⎟= 2×20×12

38− 20× 26

38⎛

⎝⎜

⎞

⎠⎟(1.6×26 / 38)13 −11.6×26 / 38−1

⎛

⎝⎜⎜

⎞

⎠⎟⎟ ≈ −24.93 .

Exercises

1. Derive the average number of plays made while using this martingale strategy. That is, derive E[T ∧n] where T ~ geo(p) . 2. Under the general conditions of this martingale strategy, with parameters p , q , C , and n , derive the average amount of the final bet. 3. In the traditional martingale strategy, we continue doubling the bet until the first win with no limit on how many bets we make. Derive the average amount of the final bet in this scenario.