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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
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(20/18)^5 - 1
--------------- ~ 0.371263
(20/18)^10 -1
See this
derivation.
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b) always bet $1 on black
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(20/18)^50 - 1
----------------- ~ 0.00512735
(20/18)^100 - 1
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c) always bet $10 on the first row of 12
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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
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d) always bet $1 on 00
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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 >= 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.