quickpick
quickpick
Joined: Apr 14, 2020
  • Threads: 5
  • Posts: 12
May 13th, 2020 at 11:09:06 AM permalink
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?
Last edited by: quickpick on May 13, 2020
ChumpChange
ChumpChange
Joined: Jun 15, 2018
  • Threads: 56
  • Posts: 2479
May 13th, 2020 at 12:59:02 PM permalink
I've been cleaning up on negative counts in BJ lately, and positive count losses offset my wins.
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 106
  • Posts: 5065
Thanks for this post from:
quickpick
May 13th, 2020 at 1:42:40 PM permalink
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.
quickpick
quickpick
Joined: Apr 14, 2020
  • Threads: 5
  • Posts: 12
May 13th, 2020 at 1:54:11 PM permalink
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?
ksdjdj
ksdjdj
Joined: Oct 20, 2013
  • Threads: 80
  • Posts: 931
Thanks for this post from:
quickpick
May 13th, 2020 at 2:09:32 PM permalink
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.
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 106
  • Posts: 5065
Thanks for this post from:
quickpick
May 13th, 2020 at 2:17:09 PM permalink
Quote: quickpick

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?


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
quickpick
quickpick
Joined: Apr 14, 2020
  • Threads: 5
  • Posts: 12
May 13th, 2020 at 2:21:42 PM permalink
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.
DJTeddyBear
DJTeddyBear
Joined: Nov 2, 2009
  • Threads: 191
  • Posts: 10548
May 13th, 2020 at 3:33:19 PM permalink
This sounds like homework.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
  • Threads: 106
  • Posts: 5065
May 13th, 2020 at 4:12:29 PM permalink
Quote: quickpick

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.


Anything higher than 40,000, I can't generate enough results fast enough to do an accurate simulation.
Ace2
Ace2
Joined: Oct 2, 2017
  • Threads: 24
  • Posts: 1115
Thanks for this post from:
quickpick
May 13th, 2020 at 5:27:16 PM permalink
Quote: ThatDonGuy



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

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.
Last edited by: Ace2 on May 13, 2020
It痴 all about making that GTA

  • Jump to: