How many results are needed to guarantee loss

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?

I've been cleaning up on negative counts in BJ lately, and positive count losses offset my wins.

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.

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?

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.

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

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.

This sounds like homework.

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.

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.

Quote: ThatDonGuy

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

However, I can figure out a way to compute it:
5000: 1 / 6
10,000: 1 / 12
15,000: 1 / 21
20,000: 1 / 38
25,000: 1 / 65
30,000: 1 / 111
40,000: 1 / 318
50,000: 1 / 886
60,000: 1 / 2433
70,000: 1 / 6613
80,000: 1 / 17,822
90,000: 1 / 47,729
100,000: 1 / 127,162

And no, I'm not going to figure out any of the other numbers - it took long enough as it is to get these

Quote: ThatDonGuy

However, I can figure out a way to compute it:
5000: 1 / 6
10,000: 1 / 12
15,000: 1 / 21
20,000: 1 / 38
25,000: 1 / 65
30,000: 1 / 111
40,000: 1 / 318
50,000: 1 / 886
60,000: 1 / 2433
70,000: 1 / 6613
80,000: 1 / 17,822
90,000: 1 / 47,729
100,000: 1 / 127,162

And no, I'm not going to figure out any of the other numbers - it took long enough as it is to get these

Do these exact same probabilities apply for betting one unit on one number (straight bet with a 35:1 payout) every spin if you got half back when the ball lands on zero? I am assuming that the probabilities would apply to any bet (if the half back rule was in effect).

Quote: quickpick

Do these exact same probabilities apply for betting one unit on one number (straight bet with a 35:1 payout) every spin if you got half back when the ball lands on zero? I am assuming that the probabilities would apply to any bet (if the half back rule was in effect).

I doubt it, as the variance is different, and, as Ace2 has shown, these numbers come very close to the numbers you get when you apply statistical methods to it.

Also, what casino offers roulette with partage on anything other than even-money bets?

Quote: ThatDonGuy

I doubt it, as the variance is different, and, as Ace2 has shown, these numbers come very close to the numbers you get when you apply statistical methods to it.

Also, what casino offers roulette with partage on anything other than even-money bets?

It's pretty weird if it isn't, right?. I can see the probabilities being different in the short run, e.g., after one spin, even-money bets would have a much higher chance of being ahead than betting on one number.

Imagine two players betting the same bet for 10,000 spins. One player bets $1 on red every spin, while the other player bets $1 on the number one every spin. If they are both expected to lose about 1.35% of their bets each spin, why wouldn't they be expected to have almost the exact, if not the exact, same amount of $ after 10,000spins (or have the same chance of being ahead after 10,000 spins?)?

I am not concerned whether casinos offer half-back on inside bets or not; I just want to know.

Quote: quickpick

It's pretty weird if it isn't, right?. I can see the probabilities being different in the short run, e.g., after one spin, even-money bets would have a much higher chance of being ahead than betting on one number.

Imagine two players betting the same bet for 10,000 spins. One player bets $1 on red every spin, while the other player bets $1 on the number one every spin. If they are both expected to lose about 1.35% of their bets each spin, why wouldn't they be expected to have almost the exact, if not the exact, same amount of $ after 10,000spins (or have the same chance of being ahead after 10,000 spins?)?

I am not concerned whether casinos offer half-back on inside bets or not; I just want to know.

Expected to have the same amount after 10,000 spins? Yes.

Have the same chance of being ahead? No. In fact, that's the "secret" of systems like Martingale - the chance of being ahead is much greater than with flat betting, but the amount you lose when you do lose is much greater.

This is why "variance" and "standard deviations" play an important part in gambling.

Look at it this way. If you wanted to make a single $1,000,000 bet on red, you’d be lucky to find a casino anywhere that would take that bet. However, if you want to make a thousand $1,000 bets on red (over 1,000 spins), every decent sized casino will take that action.

The expectation is for the casino to hold $1,000,000 / 74 = $13,500 in both scenarios. However, there are only three possibilities in scenario one: casino wins $1,000,000, wins $500,000 (partage), or loses $1,000,000

In scenario 2, the standard deviation is about $31,500, meaning the casino can be about 95% confident the result will be within $63,000 of expectations. So they could still lose some money, but it’s an acceptable risk level. An 18/37 chance of losing $1,000,000, however, would not be acceptable at most casinos

In some cases, variance is more important than the edge

Quote: ThatDonGuy

Expected to have the same amount after 10,000 spins? Yes.

Have the same chance of being ahead? No. In fact, that's the "secret" of systems like Martingale - the chance of being ahead is much greater than with flat betting, but the amount you lose when you do lose is much greater.

This is why "variance" and "standard deviations" play an important part in gambling.

According to my current thinking, if you bet $1 on one number with a 35:1 payout for 10,000 spins, you should have the same chance of being ahead as someone who bets $1 on an even-money bet for 10,000 spins. Assume you get half back every time the ball lands on zero, even if there aren't casinos that offer half back on inside bets. Correct me if I am wrong, but doesn't the math just balance out. I can only imagine there being a very slight difference in the probabilities of being ahead after 10,000 spins, 15,000 spins, 20,000 spins, 25,000 spins, 30,000 spins, 40,000 spins, 50,000 spins, 60,000 spins, 70,000 spins, 80,000 spins, 90,000 spins, and 100,000 spins.

Forget martingale and progressions; I am just speaking of flat betting, $1 each spin.

Can you please simulate or calculate this: the chance of being ahead after X amount of spins if you are just betting one unit on one number with a 35:1 payout with the half-back rule in effect (you get half-back when the ball lands on zero; nevermind whether casinos offer this or not). Please calculate or simulate this for 10,000 spins, 15,000 spins, 20,000 spins, 25,000 spins, 30,000 spins, 40,000 spins, 50,000 spins, 60,000 spins, 70,000 spins, 80,000 spins, 90,000 spins, and 100,000 spins. Thank you.

I know that casinos don't offer "la partage" on anything other than even-money bets; just pretend that they do offer "la partage" on single number bets for this example.

If you have a 25 hand buy-in and you bet Player while the Banker wins 50 to 25 in one shoe of Baccarat.

Quote: quickpick

Can you please simulate or calculate this: the chance of being ahead after X amount of spins if you are just betting one unit on one number with a 35:1 payout with the half-back rule in effect (you get half-back when the ball lands on zero; nevermind whether casinos offer this or not). Please calculate or simulate this for 10,000 spins, 15,000 spins, 20,000 spins, 25,000 spins, 30,000 spins, 40,000 spins, 50,000 spins, 60,000 spins, 70,000 spins, 80,000 spins, 90,000 spins, and 100,000 spins. Thank you.

Here's what I get:

Spins	Success
5000	43.49%
10000	40.83%
15000	38.82%
20000	37.15%
25000	35.72%
30000	34.46%
35000	33.25%
40000	32.17%
45000	31.17%
50000	30.22%
60000	28.54%
70000	26.99%
80000	25.62%
90000	24.35%
100000	23.18%

Thanks for the results ThatDonGuy.