Simply Risk of Ruin Question
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? |
Re: Simply Risk of Ruin Question
[ 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. [ 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 >= 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. |
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