weaselman
Joined: Jul 11, 2010
• Posts: 2349
May 8th, 2012 at 2:50:48 PM permalink
Quote: blaxius

but I'll just say one last thing: I never mentioned unlimited bankroll and unlimited maximum bet.

Oh? In that case, I misunderstood. So, what are the limits you are proposing for the challenge then?

Quote: blaxius

By the way, I've doing some research, and with the systems that I was using that had a 16% chance of winning and attempt, I would have 0,00615232472362199% chance of winning 11 times out of 20. Is that correct?? The formula I used was the binomial distribution: P(X=20)= (20(n) choose 11(k))(4/25)^11(1-4/25)^(20-11).
I don't understand much of it, anyone here can correct me if I am wrong?

If 4/25 is the probability of a single win, then this formula gives you the probability of winning exactly 11 times out of 20. I doubt that is the number you are looking for though. What you want is the probability of winning 11 or more times, which would be P(11)+P(12)+P(13)...+P(20).
"When two people always agree one of them is unnecessary"
blaxius
Joined: May 2, 2012
• Posts: 38
May 8th, 2012 at 3:09:33 PM permalink
Quote: weaselman

Oh? In that case, I misunderstood. So, what are the limits you are proposing for the challenge then?

As the original challenge, infinite play money and \$5 min - \$5000max bet. What I meant is that I never mentioned unlimited maximum bet. This plus unlimited bankroll would be ridiculous. With that martingale would win 100% of the time.

Quote: weaselman

If 4/25 is the probability of a single win, then this formula gives you the probability of winning exactly 11 times out of 20. I doubt that is the number you are looking for though. What you want is the probability of winning 11 or more times, which would be P(11)+P(12)+P(13)...+P(20).

Then I would have to sum that 0.006% with 9 more results that would be smaller, even if all were 0.006% that would give me 0.055% the most, which still is pretty low... I though the odds weren't that bad =/

Thanks for the help.
thecesspit
Joined: Apr 19, 2010
• Posts: 5936
May 8th, 2012 at 3:14:54 PM permalink
Quote: blaxius

I am not trying to code a solution, I am trying to beat a challenge here. Frankly, I've been taking a look at the link you sent me: http://www.mathpages.com/home/kmath084/kmath084.htm, but I don't understand most of it.

Code a solution to beat a challenge. I'm saying that the challenge you are trying to beat is -probably- impossible based on the starting requirements set, because the risk of ruin (and amount lost) is for each round/step outweighs the profits on success. If you have a round that wins 50.5% of the time and loses 49.5%, but the amount lost is more than the amount won, all you've done is refine up the problem from being a single spin to a group of spins, but the same laws apply.

I understand you are trying to get in such that you can win 11 from 20 'runs' over 200,000 spins, and this is why I said 'probably'. I think the bet size restrictions will always get you.

As for the article, try the wiki page on Gambler's ruin as a first pass. If that's not clear, I'd suggest going backwards into probability theory until you can work forwards again.
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
blaxius
Joined: May 2, 2012
• Posts: 38
May 8th, 2012 at 3:24:11 PM permalink
Quote: thecesspit

Code a solution to beat a challenge. I'm saying that the challenge you are trying to beat is -probably- impossible based on the starting requirements set, because the risk of ruin (and amount lost) is for each round/step outweighs the profits on success. If you have a round that wins 50.5% of the time and loses 49.5%, but the amount lost is more than the amount won, all you've done is refine up the problem from being a single spin to a group of spins, but the same laws apply.

I understand you are trying to get in such that you can win 11 from 20 'runs' over 200,000 spins, and this is why I said 'probably'. I think the bet size restrictions will always get you.

It's not the bet size restrictions that are getting me, it's the starting bankroll. But I am assuming that if I were to try the challenge with 1 billion spins the starting bankroll should increase as well, at least to cover a few rounds of losses in a row so that my system could start increasing my net with no risk of hitting \$0.

Quote: thecesspit

As for the article, try the wiki page on Gambler's ruin as a first pass. If that's not clear, I'd suggest going backwards into probability theory until you can work forwards again.

I think I will try that another day, trying to understand the binomial distribution has already exhausted me.
thecesspit
Joined: Apr 19, 2010
• Posts: 5936
May 8th, 2012 at 3:34:16 PM permalink
Bet size restrictions or bankroll. It doesn't matter which. Bankroll can be counted in the number of minimal unit bets (\$5000 is just 1000 \$5 bets).
"Then you can admire the real gambler, who has neither eaten, slept, thought nor lived, he has so smarted under the scourge of his martingale, so suffered on the rack of his desire for a coup at trente-et-quarante" - Honore de Balzac, 1829
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
May 8th, 2012 at 4:05:12 PM permalink
Quote: blaxius

I think I will try that another day, trying to understand the binomial distribution has already exhausted me.

It is not that hard to understand.
Binomial Probability Distribution

It helped me a year ago. Even has a free calculator. Now I just use Excel.

Good Luck to you!
Sally
I Heart Vi Hart
blaxius
Joined: May 2, 2012
• Posts: 38
May 8th, 2012 at 4:09:30 PM permalink
Quote: mustangsally

It is not that hard to understand.
Binomial Probability Distribution

It helped me a year ago. Even has a free calculator. Now I just use Excel.

Good Luck to you!
Sally

Thanks, that's one of the pages that I came up with and I used that calculator to know the chances I said before, but as it happened to me, it only gives you the exact number, not the sum of all that are need. Could you tell me the formula you use on Excel?

Thanks again!
weaselman
Joined: Jul 11, 2010
• Posts: 2349
May 8th, 2012 at 4:25:16 PM permalink
Quote: thecesspit

Bet size restrictions or bankroll. It doesn't matter which. Bankroll can be counted in the number of minimal unit bets (\$5000 is just 1000 \$5 bets).

He is talking about limited bet size (1000 units), but unlimited bankroll, so, if you lose one sequence, you can continue betting.
It is not that outrageous actually (and not entirely unrealistic - you lose your entire "session bankroll" sometimes, but still come back to casino and play after that).
1000 unit (let's say 1023) is enough to place 10 martingale bets. In a single zero roulette, you'd lose about every 605th sequence. The average length of a sequence would be 2.11 spins. So, you'd need to play about 95K of them to complete 200K spins. To end up in the positive territory, you have to lose less than 94 of them. The probability of that happening is about 1 in 826,753 if I am not mistaken in my huge number math ...
"When two people always agree one of them is unnecessary"
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
May 8th, 2012 at 4:29:35 PM permalink
The calculator that page links to shows the cumulative probabilities.
Both greater than or less than.

Excel is real easy to use with the BINOMDIST function (up to 2007)
as one can just make a large table in a few minutes or less.

Here is Microsoft page for the function
Excel statistical functions: BINOMDIST

I can show you how I do it much later tonight, that is US California time.
I Heart Vi Hart
blaxius
Joined: May 2, 2012
• Posts: 38
May 8th, 2012 at 4:35:17 PM permalink
Quote: weaselman

He is talking about limited bet size (1000 units), but unlimited bankroll, so, if you lose one sequence, you can continue betting.
It is not that outrageous actually (and not entirely unrealistic - you lose your entire "session bankroll" sometimes, but still come back to casino and play after that).
1000 unit (let's say 1023) is enough to place 10 martingale bets. In a single zero roulette, you'd lose about every 605th sequence. The average length of a sequence would be 2.11 spins. So, you'd need to play about 95K of them to complete 200K spins. To end up in the positive territory, you have to lose less than 94 of them. The probability of that happening is about 1 in 826,753 if I am not mistaken in my huge number math ...

You are right, that's what I meant. But we have to stick to the terms of the challenge.

Quote: mustangsally

The calculator that page links to shows the cumulative probabilities.
Both greater than or less than.

Excel is real easy to use with the BINOMDIST function (up to 2007)
as one can just make a large table in a few minutes or less.

Here is Microsoft page for the function
Excel statistical functions: BINOMDIST

I can show you how I do it much later tonight, that is US California time.

I really appreciate your help, but I can't today. I will read that later, thanks.