slackyhacky
Joined: Jan 18, 2012
• Posts: 348
March 13th, 2017 at 11:20:08 PM permalink
Let's assume I came up with a money management program that over the long run, beat the house.

(Please don't comment about how we can't make that assumption, etc, etc. I am asking you to make the assumption - just pretend for a second please.)

How many trials (dice rolls) would it take to show that the system is in fact a winning system?

Said another way - if I thought I had a winning system, and I simulated it 1000 times and it wins (more bank roll than starting bank roll) - would that be enough rolls to prove it? How about 2000 rolls? 5000?

I guess another way to say it, is given that the house edge with optimal strategy is 1.36 - 1.41% - how many rolls would it take with a 99% probability that bank roll is less than starting? If we get that number, could one use that number to then say - if I can show a positive bankroll after that number of rolls, there is only a 1% chance it isn't a winning system?
odiousgambit

Joined: Nov 9, 2009
• Posts: 7241
March 14th, 2017 at 3:01:51 AM permalink
3 standard deviations would give you that

but it's not easy to calculate unless you flat-bet

"Baccarat is a game whereby the croupier gathers in money with a flexible sculling oar, then rakes it home. If I could have borrowed his oar I would have stayed." .......... Mark Twain
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 2933
March 14th, 2017 at 7:04:06 AM permalink
You also need to define "win" in specific terms. If your first bet won, your bankroll would be larger than when you started; would you stop betting at that point?

If the "system" is "always bet 1", then, if the bet is red on a single-zero roulette wheel and your starting bankroll is 100, the probability of being ahead at any point before losing the entire 100 is about 90%.

In fact, if the probability p of winning a single bet < 1/2, then, as your initial bankroll increases, the probability of being ahead B bets before losing the entire bankroll approaches ( (p / (1 - p) )B. In the single-zero roulette case, p = 9/19, so 1 - p = 10/19, p / (1 - p) = 9/10, and the probability of being ahead by 1 bet before busting approaches 9/10 as your initial bankroll increases, being ahead by 2 bets before busting approaches 81/100, being ahead by 3 bets before busting approaches 729/1000, and so on.

If I am crunching my numbers correctly, for the single-zero roulette even-money flat bet system, if you try it 100 times, you are expected to win 90 times, and the standard deviation is 3 (I am not sure if there is an "easy" way to calculate this; in 100 trials, assuming the probability of being ahead at some point before being behind by 100 is 0.9, then the probability of n out of the 100 trials being wins is (100)C(n) x 0.9n x 0.1100-n ; some Excel number crunching shows the mean is 90 and the standard deviation is 3), so, in this case, a 99% success rate would be +3 standard deviations.
slackyhacky
Joined: Jan 18, 2012
• Posts: 348
March 14th, 2017 at 7:23:21 AM permalink
Thanks.

Could you do the same for craps.

Assuming an overall house edge of 1.41,

What is the probability of being ahead (positive bank roll) after 1000 rolls? 5000? 10000?
RS
Joined: Feb 11, 2014
• Posts: 5225
March 14th, 2017 at 8:38:49 AM permalink
Quote: slackyhacky

Thanks.

Could you do the same for craps.

Assuming an overall house edge of 1.41,

What is the probability of being ahead (positive bank roll) after 1000 rolls? 5000? 10000?

What bets are you making? Every bet has a different EV and SD, as well as if you're doing a combination of bets.

I'd determine what the N0 is for a game. N0 (in a -EV game) is the number of rolls/hands/slot spins/etc. played such that 1 standard deviation = expected value. After playing 1 N0 rounds, you'll be within 1 SD 68.2% of the time.

Just figure out how many rounds you gotta play such that EV = 3 SD's. Or just skip straight to it and figure it out for 4 or 5 SD's.
"should of played 'Go Fish' today ya peasant" -typoontrav
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 2933
Thanks for this post from:
March 14th, 2017 at 9:05:59 AM permalink
Quote: slackyhacky

Assuming an overall house edge of 1.41,

What is the probability of being ahead (positive bank roll) after 1000 rolls? 5000? 10000?

Do you mean ahead at any point, or just ahead at the end of the set of rolls?

If you mean being ahead at the end of the set of rolls, I get:
100: 40.47%
200 39.34%
500: 35.94%
1000: 31.64%
Yes, the chance of being ahead decreases with more rolls; this is because you are exposing more bets to the house edge.

The probability of being ahead at any point before reaching N rolls is going to get closer and closer to p / (1 - p) as N increases.
Wait a minute...that's actually the probability of being ahead before being N behind. What you want to know is a different problem.
However, after some number crunching, I get a probability of 90.6835% of being ahead at some point in 100 come-outs.

BetsProbability
10090.6835%
20092.943%
30093.9362%
40094.5228%
50094.9194%
60095.2093%
70095.4324%
80095.6105%
90095.7566%
100095.8789%
110095.9832%
120096.0732%
130096.1518%
140096.2212%
150096.2829%
160096.3382%
170096.388%
180096.4331%
190096.4743%
200096.5119%

Note that p = 1/2 - HE/200; in this case, the house edge is 1.41%, so p = 1/2 - 1.41/200 = 0.49295, and p / (1 - p) = about 97.22%.

Note: do not PM me with requests
Last edited by: unnamed administrator on Mar 14, 2017
slackyhacky
Joined: Jan 18, 2012
• Posts: 348
March 14th, 2017 at 12:50:44 PM permalink
Thatdonguy-

I don't mean at any point - I mean at the end of the x number of rolls.

Thanks for the numbers.

I'm having a hard time following the last post.

The spreadsheet was for the probability of being ahead at any point right?
bbvk05
Joined: Jan 12, 2011
• Posts: 382
March 14th, 2017 at 1:01:42 PM permalink
If you invented a money management system that defied the fundamental math of a game I'm not sure any number of trials would prove it, but I'm sure you'd be very famous with all the attention you'd attract in the mathematics world with your discovery. But if you had billions of quality simulations that showed something notable then you'd have my attention.

Your whole premise is flawed though. Money management just rearranges variance. You might be able to create a system that wins 99% of the time, but that 1% is going to ream you. The hard work has already been done: you can easily walk into a casino and do a roulette martingale tomorrow with less than a 1% chance of losing--- but God help you if it does.
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 2933
March 14th, 2017 at 1:19:26 PM permalink
Quote: slackyhacky

I don't mean at any point - I mean at the end of the x number of rolls.

Thanks for the numbers.

Note that I calculated based on 100 (for example) bets - not 100 rolls. The 100 rolls problem is harder as you have to account for different numbers of rolls per bet.

Quote: slackyhacky

I'm having a hard time following the last post.

The spreadsheet was for the probability of being ahead at any point right?

Correct - at first, I listed the formula for being ahead at any point before being a particular number behind, before noticing that you wanted to know the probability of being ahead at any point before a certain number of bets (which is what is on the spreadsheet).
slackyhacky
Joined: Jan 18, 2012
• Posts: 348
March 14th, 2017 at 1:29:59 PM permalink
Quote: ThatDonGuy

If you mean being ahead at the end of the set of rolls, I get:
100: 40.47%
200 39.34%
500: 35.94%
1000: 31.64%
Yes, the chance of being ahead decreases with more rolls; this is because you are exposing more bets to the house edge.

What is the equation you used to figure this out?

Quote: ThatDonGuy

Note that p = 1/2 - HE/200; in this case, the house edge is 1.41%, so p = 1/2 - 1.41/200 = 0.49295, and p / (1 - p) = about 97.22%.

What does this math represent? What does 49% mean? What does 97% mean?