Simply Risk of Ruin Question

[Nottom]

Suppose I like to play -EV games and go to a casino to play some Roulette (double zero).

If I take $50 and decide to stay until I double up or go broke what are the odds that I walk away with $100 assuming I play the following ways:
a) always bet $10 on black
b) always bet $1 on black
c) always bet $10 on the first row of 12
d) always bet $1 on 00

Even if you don't want to do the math, how would you go about finding these results?

[pzhon]

[ QUOTE ]

If I take $50 and decide to stay until I double up or go broke what are the odds that I walk away with $100 assuming I play the following ways:
a) always bet $10 on black


[/ QUOTE ]
(20/18)^5 - 1
--------------- ~ 0.371263
(20/18)^10 -1

See this derivation (http://www.mathpages.com/home/kmath084.htm).

[ QUOTE ]

b) always bet $1 on black


[/ QUOTE ]
(20/18)^50 - 1
----------------- ~ 0.00512735
(20/18)^100 - 1

[ QUOTE ]

c) always bet $10 on the first row of 12


[/ QUOTE ]
That's more complicated since you will sometimes end up with $110 and sometimes end up with $100.

P(end with 100) = 0.269917
P(end with 110) = 0.144882
P(end with 0) = 0.585201

[ QUOTE ]

d) always bet $1 on 00


[/ QUOTE ]
You reach at least 100 with probability 0.408746.

I set up a system of 99 linear equations and solved them.

Mathematica code:
<ul type="square">
Table[p[ii] ==
37/38 If[ii == 1, 0, p[ii - 1]] +
1/38 If[ii &gt;= 65, x[ii + 35], p[ii + 35]], {ii, 1, 99}
Solve[%, Table[p[ii], {ii, 99}]]
[/list]
The solution was a linear combination of the dummy variables x[100] through x[134]. The coefficient of x[k] was the probability of ending with $k.