March 14th, 2017 at 2:10:11 PM
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Quote:slackyhackyWhat is the equation you used to figure this out?

If p is the probability of winning a particular even-money bet, then the probability of being ahead after N bets is:

(N)C(N) x p

+ (N)C(N-1) x p

+ (N)C(N-2) x p

+ (N)C(N-3) x p

+ ...

+ (N)C(K) x p

where (A)C(B) is the number of combinations of A things taken B at a time (this is also written as C(A,B), or Combin(A,B))

p is the probability of winning a single bet

K is the smallest integer > N/2 (e.g. if N = 100, K = 51; if N = 101, K is also 51)

There's no way to "simplify" this that I know of; you need to use a spreadsheet to calculate each value and then add them up.

Quote:slackyhackyWhat does this math represent? What does 49% mean? What does 97% mean?Quote:ThatDonGuyNote 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%.

49.295% is the probability of winning the bet; if you know the house edge on a "win-or-lose" bet, the probability of winning = 1/2 - HE/2.

I shouldn't have used percent with 0.9722; the value is p / (1 - p), and is used in the formula I posted earlier in the thread about calculating the probability of being ahead at any point before losing your entire bankroll.

March 14th, 2017 at 8:56:54 PM
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Quote:ThatDonGuyIf p is the probability of winning a particular even-money bet, then the probability of being ahead after N bets is:

(N)C(N) x p^{N}

+ (N)C(N-1) x p^{N-1}x (1 - p)

+ (N)C(N-2) x p^{N-2}x (1 - p)^{2}

+ (N)C(N-3) x p^{N-3}x (1 - p)^{3}

+ ...

+ (N)C(K) x p^{K}x (1 - p)^{N-K}

where (A)C(B) is the number of combinations of A things taken B at a time (this is also written as C(A,B), or Combin(A,B))

p is the probability of winning a single bet

K is the smallest integer > N/2 (e.g. if N = 100, K = 51; if N = 101, K is also 51)

There's no way to "simplify" this that I know of; you need to use a spreadsheet to calculate each value and then add them up.

49.295% is the probability of winning the bet; if you know the house edge on a "win-or-lose" bet, the probability of winning = 1/2 - HE/2.

I shouldn't have used percent with 0.9722; the value is p / (1 - p), and is used in the formula I posted earlier in the thread about calculating the probability of being ahead at any point before losing your entire bankroll.

Holy crap. I guess this is why you didn't want me to PM you.

March 14th, 2017 at 9:03:39 PM
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Quote:bbvk05If 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.

Maybe - but not if the system when it looses doesn't loose that big.

Speaking of roulette, I'm pretty sure I am way ahead on that electronic 5 cent single 0 roulette machine at 4 Queens. I just play black while I"m counting. When a 0 doesn't roll for 50 rolls, I start betting 1 bet and addd one bet to the 0 until it hits. I know this is faulty since it seems to rely on the gamblers fallacy - but I've been pretty lucky so far. It is rather boring, but much less boring then watching my wife play the pinball machine. One time it went 150 times before hitting. I'm sure a much larger streak is highly probable. When I get to 99, I have to convert the machine to 25 cent and do it that way.

March 14th, 2017 at 9:52:42 PM
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Quote:slackyhackyMaybe - but not if the system when it looses doesn't loose that big.

Speaking of roulette, I'm pretty sure I am way ahead on that electronic 5 cent single 0 roulette machine at 4 Queens. I just play black while I"m counting. When a 0 doesn't roll for 50 rolls, I start betting 1 bet and addd one bet to the 0 until it hits. I know this is faulty since it seems to rely on the gamblers fallacy - but I've been pretty lucky so far. It is rather boring, but much less boring then watching my wife play the pinball machine. One time it went 150 times before hitting. I'm sure a much larger streak is highly probable. When I get to 99, I have to convert the machine to 25 cent and do it that way.

Then the system loses more often. You can't escape the house edge by makeing more bets that have a house edge.

Your roulette thing is rank superstition. That can be fun but it has no value. You're just following the "it's due" theory.

March 14th, 2017 at 9:52:43 PM
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Quote:slackyhackyMaybe - but not if the system when it looses doesn't loose that big.

Speaking of roulette, I'm pretty sure I am way ahead on that electronic 5 cent single 0 roulette machine at 4 Queens. I just play black while I"m counting. When a 0 doesn't roll for 50 rolls, I start betting 1 bet and addd one bet to the 0 until it hits. I know this is faulty since it seems to rely on the gamblers fallacy - but I've been pretty lucky so far. It is rather boring, but much less boring then watching my wife play the pinball machine. One time it went 150 times before hitting. I'm sure a much larger streak is highly probable. When I get to 99, I have to convert the machine to 25 cent and do it that way.

Then the system loses more often. You can't escape the house edge by makeing more bets that have a house edge.

Your roulette thing is rank superstition. That can be fun but it has no value. You're just following the "it's due" theory.

March 14th, 2017 at 10:25:47 PM
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Okay,

So I couldn't figure out your formula -

Instead, I took your 4 points, plotted them in excel, did an equation fit - then used that formula to get some more data. It isn't exact, but I think it is close.

at 16000 roles, there is a 0.3% chance you will be winning.

at 32000 roles, there is a .003% chance you will be winning.

So - if I came up with system that was ahead after 32000 roles, the likelyhood that it is a loosing system is 0.003% chance.

That's pretty cool. Now I just need to come up with a system that will do that.

So I couldn't figure out your formula -

Instead, I took your 4 points, plotted them in excel, did an equation fit - then used that formula to get some more data. It isn't exact, but I think it is close.

at 16000 roles, there is a 0.3% chance you will be winning.

at 32000 roles, there is a .003% chance you will be winning.

So - if I came up with system that was ahead after 32000 roles, the likelyhood that it is a loosing system is 0.003% chance.

That's pretty cool. Now I just need to come up with a system that will do that.

March 15th, 2017 at 8:20:47 AM
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I couldn't find a definition of "rank superstition" anywhere on the internet so I have no idea what that means.

I still think it is better described by gamblers fallacy - or the fallacy of the maturity of chance.

What do you mean it has no value? I'm not sure if your point of saying that. Did you think that I believe I could sell the idea - this meaning the idea has value?

Of course it has no value. Let me give you some hints to my statement that should have given you the clue that I know it doesn't have value - I used the word "faulty". I also acknowledged that my thinking is prey to falllacy.

If you meant that - yes, you agree with me that my method of play has no value - ok then.

But you seemed to be trying to point out to me as if I didn't recognize it already - which in that case, your statement about value has no value.

I still think it is better described by gamblers fallacy - or the fallacy of the maturity of chance.

What do you mean it has no value? I'm not sure if your point of saying that. Did you think that I believe I could sell the idea - this meaning the idea has value?

Of course it has no value. Let me give you some hints to my statement that should have given you the clue that I know it doesn't have value - I used the word "faulty". I also acknowledged that my thinking is prey to falllacy.

If you meant that - yes, you agree with me that my method of play has no value - ok then.

But you seemed to be trying to point out to me as if I didn't recognize it already - which in that case, your statement about value has no value.

March 15th, 2017 at 8:38:16 AM
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Quote:slackyhackyI couldn't find a definition of "rank superstition" anywhere on the internet so I have no idea what that means.

I still think it is better described by gamblers fallacy - or the fallacy of the maturity of chance.

What do you mean it has no value? I'm not sure if your point of saying that. Did you think that I believe I could sell the idea - this meaning the idea has value?

Of course it has no value. Let me give you some hints to my statement that should have given you the clue that I know it doesn't have value - I used the word "faulty". I also acknowledged that my thinking is prey to falllacy.

If you meant that - yes, you agree with me that my method of play has no value - ok then.

But you seemed to be trying to point out to me as if I didn't recognize it already - which in that case, your statement about value has no value.

Acknowledging something is faulty then doing it anyway doesn't get you much credit in my book. It's one of the several things people call the gamblers fallacy.

Here's the relevant definition of "rank" that google provided:

Rank (especially of something bad or deficient): complete and utter (used for emphasis).

"rank stupidity" synonyms: downright, utter, outright, out-and-out, absolute, complete, sheer, arrant, thoroughgoing, unqualified, unmitigated, positive, perfect, patent, pure, total; archaicarrant

"rank stupidity"

March 15th, 2017 at 9:19:40 AM
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Quote:slackyhackyOkay,

So I couldn't figure out your formula -

Instead, I took your 4 points, plotted them in excel, did an equation fit - then used that formula to get some more data. It isn't exact, but I think it is close.

at 16000 roles, there is a 0.3% chance you will be winning.

at 32000 roles, there is a .003% chance you will be winning.

So - if I came up with system that was ahead after 32000 roles, the likelyhood that it is a loosing system is 0.003% chance.

That's pretty cool. Now I just need to come up with a system that will do that.

I couldn't calculate the numbers directly after 1050 - it seems that 0.49295

Note that this does not take into account the possibility that your bankroll will run out before you reach 16,000 (or 32,000) bets. With infinite time and bankroll, every system that has the possibility of making money "works"...and I do mean infinite; I have run simulations where it took millions of years in order for a D'Alembert system on a 50-50 bet (i.e. house edge = zero) to make a profit.

March 15th, 2017 at 9:37:05 AM
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Bbvk05,

If you always live your life without doing things that you know are faulty - then you are an amazing person. You should write a book about your travels and choices - that would be an inspiring and wonderful read.

If you always live your life without doing things that you know are faulty - then you are an amazing person. You should write a book about your travels and choices - that would be an inspiring and wonderful read.