Given $1 bet on boxcars at craps. True odds 35:1 against, pays 30 to 1
Iím misapplying or misinterpreting the binomial distribution.
Good math: EV = +30(1/36) - 1(35/36) = -5/36, terrible bet
My goofed up binomial distribution: P(boxcars)= p = 1/36; Trials = n = 25; Number of successes = x = 1
===> P(X=1) ~ 0.353; P(X>1) ~ 0.152; P(X>=1) ~ 0.505
So now I think Iíve contradicted the known good math in that it looks like walking up to a craps table with a single green chip gives me a slightly less than 50% chance of losing it and a slightly better than 50% chance of gaining 30 or more. Obviously, this is wrong. How did I misapply or misinterpret the binomial distribution? Please help me unstick myself. Thank you.
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I don't understand what you mean by, "A single green chip." The 50%+ does not represent the probability of doing it in one roll; it represents the probability of doing it, at least once, in 25 rolls. If you walk up with a single green chip, then you will lose it 35/36 times by betting on Midnight.
This Binomial Distribution calculator:
Agrees with your math.
Thank you for your time and attention.
There's a (35/36)^25 = 49.4% chance you lose all 25 bets.
That means that your average net win, assuming you win at least one bet, is (0.494 * 25 - 3.47) / (1 - 0.494) = $17.59. Using the same assumption, your expectation is to win (17.59 + 25) / 31 = 1.37 bets. 25 is a very small sample, but the average net win can also be roughly approximated as: (1/36 * 5/36 * 25 * 2 / π)^.5 * 31 = $20.32