May 13th, 2020 at 11:09:06 AM
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Using European roulette (single zero roulette), only flat betting, about how many roulette spins does it take to where a player is pretty much guaranteed to be in negative territory? Can a player be in the positive after 10,000 spins flat betting on a single number (the same number each spin)? How about 20,000 spins? 50,000 spins? 250,000 spins? What is the cut-off point?

Does the variance or anything change if the player is instead betting on even money bets or anything else? I assume it takes the same amount of spins whether the player is betting $1 on red every spin vs. $1 on number 1 every spin.

Would the number of results to practically guarantee a loss in baccarat only need to be half of that of european roulette since the house edge is about half the amount (banker bet= 1.06%, player bet=1.24%, european roulette bets=2.70%)

Do the same rules apply to craps pass and don't pass bets? (pass bet=1.41%, don't pass=1.36%)

What about playing a low house edge version of blackjack using basic strategy?

Does it take the same number of results to practically guarantee a loss if using progression? For example, if you only double the amount you bet sometimes. Can it take more results to wipe out systems that use steeper progressions?

Does the variance or anything change if the player is instead betting on even money bets or anything else? I assume it takes the same amount of spins whether the player is betting $1 on red every spin vs. $1 on number 1 every spin.

Would the number of results to practically guarantee a loss in baccarat only need to be half of that of european roulette since the house edge is about half the amount (banker bet= 1.06%, player bet=1.24%, european roulette bets=2.70%)

Do the same rules apply to craps pass and don't pass bets? (pass bet=1.41%, don't pass=1.36%)

What about playing a low house edge version of blackjack using basic strategy?

Does it take the same number of results to practically guarantee a loss if using progression? For example, if you only double the amount you bet sometimes. Can it take more results to wipe out systems that use steeper progressions?

Last edited by: quickpick on May 13, 2020

May 13th, 2020 at 12:59:02 PM
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I've been cleaning up on negative counts in BJ lately, and positive count losses offset my wins.

May 13th, 2020 at 1:42:40 PM
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Define "pretty much guaranteed."

Here are the probabilities I get for being ahead on even-money bets after a certain number of spins (assume no "portage" or "prison" - i.e. all bets lose on 0):

10,000: 1 / 300

20,000: 1 / 15,609

30,000: 1 / 724,530

40,000: 1 / 31,987,468

50,000: 1 / 1,372,521,331

Note that for a double-zero wheel, the probability of being ahead after 10,000 spins jumps to 1 / 15,225,644.

Here are the probabilities I get for being ahead on even-money bets after a certain number of spins (assume no "portage" or "prison" - i.e. all bets lose on 0):

10,000: 1 / 300

20,000: 1 / 15,609

30,000: 1 / 724,530

40,000: 1 / 31,987,468

50,000: 1 / 1,372,521,331

Note that for a double-zero wheel, the probability of being ahead after 10,000 spins jumps to 1 / 15,225,644.

May 13th, 2020 at 1:54:11 PM
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Thank you. What are the probabilities of being ahead using the "la partage" rule (half back rule), where you get half the amount of money you bet back (on even-money bets) when the ball lands on zero? Also, what are the probabilities of being ahead if there was no house edge, pretending there is no zero, like flipping a coin and getting paid even-money?

May 13th, 2020 at 2:09:32 PM
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Note: This is Just for 10,000 spins.

"half back rule": ~ 2/23 chance of being ahead

"no house edge" (hypothetical game): 50% chance of being ahead.

"half back rule": ~ 2/23 chance of being ahead

"no house edge" (hypothetical game): 50% chance of being ahead.

May 13th, 2020 at 2:17:09 PM
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Quote:quickpickThank you. What are the probabilities of being ahead using the "la partage" rule (half back rule), where you get half the amount of money you bet back (on even-money bets) when the ball lands on zero? Also, what are the probabilities of being ahead if there was no house edge, pretending there is no zero, like flipping a coin and getting paid even-money?

For zero house edge, if there are an odd number of spins, the probability is exactly 1/2. If there are an even number, it is 1/2 minus the probability of the number of red spins equalling the number of black spins, which is about 1/40 for 1000 spins.

I don't know if there is an easy way to calculate the probabilities if partage is used. I can simulate it:

10,000: 1 / 12

20,000: 1 / 38

30,000: 1 / 110

40,000: 1 / 330

May 13th, 2020 at 2:21:42 PM
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Thank you. Can I get the probabilities of being ahead, betting red every spin, using the half-back rule for other large amounts of results, such as 50,000 results, 60,000 results, 70,000 results, 80,000 results, 90,000 results, 100,000 results, etc.? I appreciate it.

May 13th, 2020 at 3:33:19 PM
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This sounds like homework.

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http://www.DaveMillerGaming.com/ —————————————————————————————————————
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May 13th, 2020 at 4:12:29 PM
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Quote:quickpickThank you. Can I get the probabilities of being ahead, betting red every spin, using the half-back rule for other large amounts of results, such as 50,000 results, 60,000 results, 70,000 results, 80,000 results, 90,000 results, 100,000 results, etc.? I appreciate it.

Anything higher than 40,000, I can't generate enough results fast enough to do an accurate simulation.

May 13th, 2020 at 5:27:16 PM
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This is basic statistics. The expectation for the game with partage is -1/74 and the standard deviation is 0.98972. So, for instance, the expectation after 100,000 bets is to be down 1,351 units +/- 313. To be at zero is 4.32 SDs above expectations, the probability of which is 1 in 126,870.Quote:ThatDonGuy

I don't know if there is an easy way to calculate the probabilities if partage is used. I can simulate it:

Last edited by: Ace2 on May 13, 2020

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