Solve for martingale progression in CRAPS

Question

In craps the house has a 1.45% edge (51.45% chance to lose). If you martingale+1 (double up + 1 unit, every time you lose) and do this for 15 outcomes, then double up the initial bet for 30 more outcomes you can almost 4x your money

The odds of 4 losses in a row would be .5145^4 = about 7%

But what are the odds I will make it 40 rolls(outcomes) without losing more then 4 in a row?

Answer 1

We can treat this as a Markov chain. We have $6$ states. For $k=0,1,2,3,4$ state $k$ means that we currently have a string of $k$ losses. State $5$ means that at some time, a string of $5$ losses has occurred, so state $5$ is absorbing. If the probability of a loss is $p$ then the transition matrix is $$\begin{bmatrix} 1-p&p&0&0&0&0\\ 1-p&0&p&0&0&0\\ 1-p&0&0&p&0&0\\ 1-p&0&0&0&p&0\\ 1-p&0&0&0&0&p\\ 0&0&0&0&0&1 \end{bmatrix}$$ and the initial distribution is $$X=\begin{bmatrix}1&0&0&0&0&0\end{bmatrix}$$ We simply have to compute $XP^{45}$ and read off the answers.

I wrote a python script to do this:

import numpy as np
p = (1+.0145)/2
P = np.zeros((6,6))
for i in range(5):
    P[i,0] = 1-p
    P[i,i+1] = p
P[5,5] = 1
X = np.array([1,0,0,0,0,0])
Y = [email protected]_power(P, 45)
print(Y)

This produced the output

[0.2337404  0.12075448 0.06238393 0.03222866 0.0166499  0.53424264]

so the probability of $5$ losses in a row is $$.53424264$$