September 22nd, 2022 at 10:58:04 AM
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Given $1 bet on boxcars at craps. Never mind, I see my mistake. My apologies and thank you.
Last edited by: MathIsGood on Sep 22, 2022
September 22nd, 2022 at 11:04:36 AM
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Quote: MathIsGoodGiven $1 bet on boxcars at craps. True odds 35:1 against, pays 30 to 1
I’m 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 I’ve 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:
https://stattrek.com/online-calculator/binomial
Agrees with your math.
https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219
September 22nd, 2022 at 11:16:19 AM
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I neglected to back out the cost of all the lost trials the trips one wins. So a little better than 50% of the time one walks out with 30+ minus 24-. Embarrassing mistake, but at least the world makes sense again. Amazing how writing it down clarifies the thinking.
Thank you for your time and attention.
Thank you for your time and attention.
Last edited by: MathIsGood on Sep 22, 2022
September 22nd, 2022 at 11:43:27 AM
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Assuming 25 one dollar bets, your expectation is to lose $25 * 5/36 = $3.47
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
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
It’s all about making that GTA