View Full Version : Math Question

Koury

2014-01-03, 05:30 AM

Alright, I have a question. As you may have guessed, from the thread title, it's math related.

OK, say an event has a 50% chance of being a success, and 50% chance of being a failure. Say the event will occur 50 times, unless there are 4 failures in a row. I know how to calculate the odds of any given total amount of successes or failures (say, the chance of 32 successes), but not how to calculate the odds of 4 failures in a row occurring at any point in the sequence.

So how do I do that? How would it change if I needed to know 5 failures in a row, or 6? What if the chance of success/failure varied, or the amount of tries?

Basically, teach me the formula I need here.

Let me know if my question is unclear here. Also, thanks in advance for any help given here.

jonc83

2014-01-03, 07:07 AM

Goodness, my rusty old brain...

First you need the number of permutations of successes and failures. By my reckoning there are 2^50 of these.

Then you need the number which contain your search string. In this case FFFF.

If we have FFFFXXXXXXXXXXXXXXXXX...X where X can be either then there are 2^46 that start FFFF. If we shift the FFFF along and put an X before it we can get 47 positions for FFFF so have 47*2^46 sets with FFFF in.

However, this includes some double accounting which needs to be eliminated.

My battery is at 4%. Look here and I'll try to help again later.

http://stackoverflow.com/questions/6790620/probability-of-3-character-string-appearing-in-a-randomly-generated-password

Jimorian

2014-01-03, 07:27 AM

jonc83 has the first part. Like he says, eliminating the repeats is the tough part. (I'm assuming you're looking for a sequence of 50 with at least one set of 4 fails in a row, and not the total number of times it happens?)

One way to approach it is to take the simplest version, then expand it, and see if a formula drops out of that.

So, for example, in a sequence of 4 chances, your FFFF sequence has exactly a 1 in 16 chance of happening. If you have 5 events, you could have FFFFS, SFFFF, and FFFFF as qualifying sequences, so 3 out of 32.

6 events has:

FFFFSS

SFFFFS

SSFFFF

FFFFFS

FFFFSF

FFFFFF

SFFFFF

FSFFFF

so 7(?) out of 64.

[EDIT: changed X's to S's because I realized I was marking different results (being specifically Successes) than jonc83's either/or X's)

I'd almost say that if you can generalize for 7 you have your formula, but another factor might kick in once you get beyond simple overlapping sequences that satisfy your criteria.

Brother Oni

2014-01-03, 07:33 AM

Koury says that the event will repeat 50 times unless 4 failures in a row, so given that the sequence terminates after FFFF, wouldn't that take care of the repeats or greater than 4 F states problem?

Jimorian

2014-01-03, 07:45 AM

I think that would work, but the formula is probably more complicated.

Another approach is to look at the inverse possibilities. How many sequences have at most 3 Failures in a row?

Again at the simplest, if you were looking for 2 in a row to trigger the result, then there are only 2 sequences out of 2^50 that don't trigger (either FSFS... or SFSF...), so the chance approaches 1:1.

thubby

2014-01-03, 08:25 AM

well the odds of 4 fails in 4 rolls is .0625

this 4 length sequence can occur only 47 ways in the 50 length sequence. you can find this by using the first heads as a sort of marker as the beginning of the sequence. the last possible sequence can only occur starting on coin 47 because the rest must be heads for it to happen.

with that in mind, you can use the binomial distribution with x=0 then take the inverse of that.

the formula would be n!/(x!*(n-x)!) *p^x*(1-p)^(n-x)

where n=trials

x=sucesses (in this case defined as 4 fails in a row never happening)

p=probability of #fails in a row (.0625)

since x is zero, it simplifies to (1-p)^n=.048

that means there's a 4.8% chance that 4 heads will not happen in 50 flips

the inverse, 95.2%, are the odds it will.

do note that this only works so simply because we're looking at a pattern of pure success/failure. checkered patterns (HTH vs HHT) get really ugly really fast.

I think that would work, but the formula is probably more complicated.

my formula covers it, but in principle the question is asking "the odds of AT LEAST 1 success"

in texts this would be P(x>=1)

which is the the equivalent of 1-P(x<1) as my calculation would prove were you to do the non-inverse method 47 times.

thubby's reasoning sounds pretty solid... though just from random guessing the results feels a bit high/low.

Then again, I'm a lousy mathematician and would have just run an algorithm to get a rough estimate of the chance :smalltongue:

Eulalios

2014-01-03, 10:33 AM

Another perspective, subtly different from thubby's formula, probably wrong:

EDIT: ninja'd, the following merely a different explanation of thubby's formula:

Given a path of N 50/50 pass/fail branches,

Q whether X consecutive failures will early terminate the path.

Once you pass the (N-X)th branch, you're in it all the way.

So - what are the odds of failing X times in a row? (0.5)^X.

What are the odds of obtaining a first failure within (N-X) branches from start?

1 - (0.5)^(N-X).

What are the odds of early termination?

( 1 - (0.5)^(N-X) ) * (0.5)^X.

For N=50, X=4, we're looking at ( ~ 1 ) * (1 / 16) or roughly 6.25 % chance that you won't get to the 50th step.

Jay R

2014-01-03, 12:21 PM

Eulalios and thubby's solutions assume that the various possibilities are independent. But four failures ending at 4 and four failures ending at 5 are not independent, since they both depend on failures in positions 2, 3, and 4.

I solved it as a recursive function.

You eliminate the repeats by determining the probability that you get four failures at the n-th trial.

Let F(n) = the probability that the first sequence of exactly 4 failures happens at the n-th trial.

G(n) = the probability that the a sequence of 4 failures happens sometime in the first n trials

Then G(n) = F(1) + F(2) + ... + F(n). And you want G(50).

Obviously, F(1) = F(2) = F(3) = 0. There aren't 4 failures in only 1, 2, or 3 tries.

F(4) = 1 /16, since all 4 trials must be failures.

F(5) = 1/32, since the sequence must be SFFFF.

F(6) = 2/64 = 1/32, since the sequence must be FSFFFF or SSFFFF.

Note that after the fourth trial, to get the first sequence of 4 failures, the last five in the sequence must be ...SFFFF. And in the preceding trials, there cannot be 4 in a row.

So for any larger n,

F(n) = (1-F(n-5)) * Prob(SFFFF) = (1-F(n-5)) / 32

Therefore a sequence of F(1) to F(50) can be quickly produced in Excel. The sum of this sequence is G(50) = 0.827414008

This was solved for 4 failures. To solve for x failures,

F(1) = F(2) = ... =F(x-1) = 0

F(x) = 1/(2^x)

F(x+1) = 1/2^(x+1))

And the recursive formula is:

F(n) = (1-F(n-x-1)) / (2^(x+1))

Sequence of F(n):

F(1)= 0

F(2)= 0

F(3)= 0

F(4)= 0.0625

F(5)= 0.03125

F(6)= 0.03125

F(7)= 0.03125

F(8)= 0.03125

F(9)= 0.029296875

F(10)= 0.028320313

F(11)= 0.02734375

F(12)= 0.026367188

F(13)= 0.025390625

F(14)= 0.024475098

F(15)= 0.023590088

F(16)= 0.022735596

F(17)= 0.021911621

F(18)= 0.021118164

F(19)= 0.020353317

F(20)= 0.019616127

F(21)= 0.01890564

F(22)= 0.018220901

F(23)= 0.017560959

F(24)= 0.016924918

F(25)= 0.016311914

F(26)= 0.015721112

F(27)= 0.015151709

F(28)= 0.014602929

F(29)= 0.014074026

F(30)= 0.013564278

F(31)= 0.013072994

F(32)= 0.012599503

F(33)= 0.012143161

F(34)= 0.011703348

F(35)= 0.011279464

F(36)= 0.010870933

F(37)= 0.010477199

F(38)= 0.010097725

F(39)= 0.009731995

F(40)= 0.009379512

F(41)= 0.009039795

F(42)= 0.008712383

F(43)= 0.008396829

F(44)= 0.008092704

F(45)= 0.007799594

F(46)= 0.007517101

F(47)= 0.007244839

F(48)= 0.006982438

F(49)= 0.006729541

F(50)= 0.006485804

Sequence of G(n):

G(1)= 0

G(2)= 0

G(3)= 0

G(4)= 0.0625

G(5)= 0.09375

G(6)= 0.125

G(7)= 0.15625

G(8)= 0.1875

G(9)= 0.216796875

G(10)= 0.245117188

G(11)= 0.272460938

G(12)= 0.298828125

G(13)= 0.32421875

G(14)= 0.348693848

G(15)= 0.372283936

G(16)= 0.395019531

G(17)= 0.416931152

G(18)= 0.438049316

G(19)= 0.458402634

G(20)= 0.478018761

G(21)= 0.4969244

G(22)= 0.515145302

G(23)= 0.532706261

G(24)= 0.549631178

G(25)= 0.565943092

G(26)= 0.581664205

G(27)= 0.596815914

G(28)= 0.611418843

G(29)= 0.625492869

G(30)= 0.639057147

G(31)= 0.652130141

G(32)= 0.664729644

G(33)= 0.676872805

G(34)= 0.688576153

G(35)= 0.699855617

G(36)= 0.71072655

G(37)= 0.721203748

G(38)= 0.731301473

G(39)= 0.741033469

G(40)= 0.750412981

G(41)= 0.759452776

G(42)= 0.768165159

G(43)= 0.776561988

G(44)= 0.784654692

G(45)= 0.792454286

G(46)= 0.799971387

G(47)= 0.807216226

G(48)= 0.814198664

G(49)= 0.820928204

G(50)= 0.827414008

Razanir

2014-01-03, 12:22 PM

First off, PROBABILITY! There is a difference between probability and odds. Odds is successes/failures, probability is successes/possible outcomes.

Minirant aside, I think this would be a negative multinomial distribution, where the probabilities are geometric.

Explanation of the geometric distribution:

Suppose you have a probability p of a success. The probability that the first success will be on the nth trial is p*(1-p)^(n-1).

Using this, we can find to probability of getting so many failures before the next success.

P(N=1|p=.5)=.5

P(N=2|p=.5)=.25

P(N=3|p=.5)=.125

P(N=4|p=.5)=.0625

P(N=5|p=.5)=.03125

(The capital Ns are a statistics idiosyncrasy. Capital letters refer to variables, lowercase to specific values)

Except as opposed to the geometric distribution where n an increase indefinitely, our distribution can only go up to 5 trials (corresponding to 4 failures in a row and the experiment ending). So we have to adjust our probabilities be dividing by P(n<=5)=.96875.

P(N=1)=.516129

P(N=2)=.258065

P(N=3)=.129032

P(N=4)=.064516

P(N=5)=.032258

So now we have the probabilities of any given success having so many preceding it. The next step is building a negative multinomial distribution for the number of trials.

NOTE: All the math from here to the end of my post ignores the 50 total trials restriction. Messing with the distribution to enforce that is, well, nightmarish.

The negative multinomial is much more complicated than the geometric. As in it involves big-pi notation (the product equivalent of sigma) and the gamma function (the generalization of factorials to all numbers). So I won't even bother giving the formula. Look it up on Wikipedia if you really want to know. But what matters is that if we have m+1 possible outcomes to an event, it gives the probability that:

1) The kth failure is on the nth trial, and

2) That the other n-k trials are divided among the other m outcomes a specified way.

So the outcomes we have are:

*A success is followed immediately by a success. p=.516129

*There is one failure in between. p=.258065

*2 failures. p=.129032

*3 failures. p=.064516

*4 failures in a row, and the experiment ends. p=.032258

The EASIEST things to calculate are the probability of so many successes and the expected number of successes/failures/trials.

Probability of n successes:

This one just follows a geometric distribution from before. The probability of a success (in terms of the geometric) is .032258. So the probability of n successes (in terms of your experiment) is just .032258*.967742^n

Expected number of successes:

Using the geometric again, the expected number of failures (in terms of the geometric) is 1/p=1/.032258=31. So in terms of your experiment, you can expect to see 31 successes.

Expected number of failures:

Back to the negative multinomial. If p_0 is the probability of a failure, k_0 is the number of probabilities before you stop the experiment, and p is the probability of a given event, the expected number of occurrences of that event is k_0*p/p_0.

In this case, k_0 = 1 (because we stop as soon as we get four failures in a row) and p_0 = .032258.

E(0 failures between successes)=.516129/.032258=16

E(1 failure)=.258065/.032258=8

E(2 failures)=.129032/.032258=4

E(3 failures)=.064516/.032258=2

E(4 failures)=1

Yes, I realize those aren't the "exact" values. They actually are. I derounded.

So the expected number of failures (in terms of your experiment) is 16*0+8*1+4*2+2*3+1*4=0+8+8+6+4=26

Expected number of trials:

26+31=57... Umm... Yeah, this is a bit weird. But remember, my math with the negative multinomial was ignoring the bit about trials stopping at 50. It really would have made the math nightmarish.

TL;DR/Summary

So there's a lot of math in this post. But hey, you asked for math help. :smalltongue:

Ignoring the rule about trials stopping at 50, the probability of n successes is .032258*.967742^n. The probability of n failures or trials, though, is tough to calculate.

EDIT: (After being ninja'd by Jay R)

I haven't looked over your math, but I can tell you that's not a valid pdf or cdf. Divide all your values by G(50), so G will actually range from 0 to 1.

Jay R

2014-01-03, 01:30 PM

EDIT: (After being ninja'd by Jay R)

I haven't looked over your math, but I can tell you that's not a valid pdf or cdf. Divide all your values by G(50), so G will actually range from 0 to 1.

Then look over the math, or at least the definitions.

G(n) is the probability that there as at least 4 failures in n trials. That is a cdf of a random variable indicating how many trials it takes to get the first four failures in a row. But it isn't complete. The first set of four failures could be at 51, or 100, or 1,000,000+. (The pdf is F(n), also incomplete.) I stopped at 50, but the sequence doesn't.

Dividing all the values by G(50) would give a Bayesian conditional probability - the probability that there are four failures in n tries, given that there are four failures in the first 50. That's not the problem we were asked to solve.

Koury

2014-01-03, 01:58 PM

However, this includes some double accounting which needs to be eliminated.

My battery is at 4%. Look here and I'll try to help again later.

http://stackoverflow.com/questions/6790620/probability-of-3-character-string-appearing-in-a-randomly-generated-password

That this double accounting problem is a thing is both something that makes complete sense and something I would never have realized.

So, for example, in a sequence of 4 chances, your FFFF sequence has exactly a 1 in 16 chance of happening. If you have 5 events, you could have FFFFS, SFFFF, and FFFFF as qualifying sequences, so 3 out of 32.

6 events has:

FFFFSS

SFFFFS

SSFFFF

FFFFFS

FFFFSF

FFFFFF

SFFFFF

FSFFFF

so 7(?) out of 64.

[EDIT: changed X's to S's because I realized I was marking different results (being specifically Successes) than jonc83's either/or X's)

I'd almost say that if you can generalize for 7 you have your formula, but another factor might kick in once you get beyond simple overlapping sequences that satisfy your criteria. Well technically a few of those aren't possible under this scenario. In the 5 trials, there are less than 32 total outcomes (FFFFS isn't possible since 4 F's end it) and only two ways it can fail (FFFFF would be the same as FFFFS).

well the odds of 4 fails in 4 rolls is .0625

this 4 length sequence can occur only 47 ways in the 50 length sequence. you can find this by using the first heads as a sort of marker as the beginning of the sequence. the last possible sequence can only occur starting on coin 47 because the rest must be heads for it to happen.

with that in mind, you can use the binomial distribution with x=0 then take the inverse of that.

the formula would be n!/(x!*(n-x)!) *p^x*(1-p)^(n-x)

where n=trials

x=sucesses (in this case defined as 4 fails in a row never happening)

p=probability of #fails in a row (.0625)

since x is zero, it simplifies to (1-p)^n=.048

that means there's a 4.8% chance that 4 heads will not happen in 50 flips

the inverse, 95.2%, are the odds it will.

do note that this only works so simply because we're looking at a pattern of pure success/failure. checkered patterns (HTH vs HHT) get really ugly really fast.

my formula covers it, but in principle the question is asking "the odds of AT LEAST 1 success"

in texts this would be P(x>=1)

which is the the equivalent of 1-P(x<1) as my calculation would prove were you to do the non-inverse method 47 times.

Another perspective, subtly different from thubby's formula, probably wrong:

EDIT: ninja'd, the following merely a different explanation of thubby's formula:

Given a path of N 50/50 pass/fail branches,

Q whether X consecutive failures will early terminate the path.

Once you pass the (N-X)th branch, you're in it all the way.

So - what are the odds of failing X times in a row? (0.5)^X.

What are the odds of obtaining a first failure within (N-X) branches from start?

1 - (0.5)^(N-X).

What are the odds of early termination?

( 1 - (0.5)^(N-X) ) * (0.5)^X.

For N=50, X=4, we're looking at ( ~ 1 ) * (1 / 16) or roughly 6.25 % chance that you won't get to the 50th step.

Eulalios and thubby's solutions assume that the various possibilities are independent. But four failures ending at 4 and four failures ending at 5 are not independent, since they both depend on failures in positions 2, 3, and 4.

I solved it as a recursive function.

You eliminate the repeats by determining the probability that you get four failures at the n-th trial.

Let F(n) = the probability that the first sequence of exactly 4 failures happens at the n-th trial.

G(n) = the probability that the a sequence of 4 failures happens sometime in the first n trials

Then G(n) = F(1) + F(2) + ... + F(n). And you want G(50).

Obviously, F(1) = F(2) = F(3) = 0. There aren't 4 failures in only 1, 2, or 3 tries.

F(4) = 1 /16, since all 4 trials must be failures.

F(5) = 1/32, since the sequence must be SFFFF.

F(6) = 2/64 = 1/32, since the sequence must be FSFFFF or SSFFFF.

Note that after the fourth trial, to get the first sequence of 4 failures, the last five in the sequence must be ...SFFFF. And in the preceding trials, there cannot be 4 in a row.

So for any larger n,

F(n) = (1-F(n-5)) * Prob(SFFFF) = (1-F(n-5)) / 32

Therefore a sequence of F(1) to F(50) can be quickly produced in Excel. The sum of this sequence is G(50) = 0.827414008

This was solved for 4 failures. To solve for x failures,

F(1) = F(2) = ... =F(x-1) = 0

F(x) = 1/(2^x)

F(x+1) = 1/2^(x+1))

And the recursive formula is:

F(n) = (1-F(n-x-1)) / (2^(x+1))

Sequence of F(n):

F(1)= 0

F(2)= 0

F(3)= 0

F(4)= 0.0625

F(5)= 0.03125

F(6)= 0.03125

F(7)= 0.03125

F(8)= 0.03125

F(9)= 0.029296875

F(10)= 0.028320313

F(11)= 0.02734375

F(12)= 0.026367188

F(13)= 0.025390625

F(14)= 0.024475098

F(15)= 0.023590088

F(16)= 0.022735596

F(17)= 0.021911621

F(18)= 0.021118164

F(19)= 0.020353317

F(20)= 0.019616127

F(21)= 0.01890564

F(22)= 0.018220901

F(23)= 0.017560959

F(24)= 0.016924918

F(25)= 0.016311914

F(26)= 0.015721112

F(27)= 0.015151709

F(28)= 0.014602929

F(29)= 0.014074026

F(30)= 0.013564278

F(31)= 0.013072994

F(32)= 0.012599503

F(33)= 0.012143161

F(34)= 0.011703348

F(35)= 0.011279464

F(36)= 0.010870933

F(37)= 0.010477199

F(38)= 0.010097725

F(39)= 0.009731995

F(40)= 0.009379512

F(41)= 0.009039795

F(42)= 0.008712383

F(43)= 0.008396829

F(44)= 0.008092704

F(45)= 0.007799594

F(46)= 0.007517101

F(47)= 0.007244839

F(48)= 0.006982438

F(49)= 0.006729541

F(50)= 0.006485804

Sequence of G(n):

G(1)= 0

G(2)= 0

G(3)= 0

G(4)= 0.0625

G(5)= 0.09375

G(6)= 0.125

G(7)= 0.15625

G(8)= 0.1875

G(9)= 0.216796875

G(10)= 0.245117188

G(11)= 0.272460938

G(12)= 0.298828125

G(13)= 0.32421875

G(14)= 0.348693848

G(15)= 0.372283936

G(16)= 0.395019531

G(17)= 0.416931152

G(18)= 0.438049316

G(19)= 0.458402634

G(20)= 0.478018761

G(21)= 0.4969244

G(22)= 0.515145302

G(23)= 0.532706261

G(24)= 0.549631178

G(25)= 0.565943092

G(26)= 0.581664205

G(27)= 0.596815914

G(28)= 0.611418843

G(29)= 0.625492869

G(30)= 0.639057147

G(31)= 0.652130141

G(32)= 0.664729644

G(33)= 0.676872805

G(34)= 0.688576153

G(35)= 0.699855617

G(36)= 0.71072655

G(37)= 0.721203748

G(38)= 0.731301473

G(39)= 0.741033469

G(40)= 0.750412981

G(41)= 0.759452776

G(42)= 0.768165159

G(43)= 0.776561988

G(44)= 0.784654692

G(45)= 0.792454286

G(46)= 0.799971387

G(47)= 0.807216226

G(48)= 0.814198664

G(49)= 0.820928204

G(50)= 0.827414008

First off, PROBABILITY! There is a difference between probability and odds. Odds is successes/failures, probability is successes/possible outcomes.

Minirant aside, I think this would be a negative multinomial distribution, where the probabilities are geometric.

Explanation of the geometric distribution:

Suppose you have a probability p of a success. The probability that the first success will be on the nth trial is p*(1-p)^(n-1).

Using this, we can find to probability of getting so many failures before the next success.

P(N=1|p=.5)=.5

P(N=2|p=.5)=.25

P(N=3|p=.5)=.125

P(N=4|p=.5)=.0625

P(N=5|p=.5)=.03125

(The capital Ns are a statistics idiosyncrasy. Capital letters refer to variables, lowercase to specific values)

Except as opposed to the geometric distribution where n an increase indefinitely, our distribution can only go up to 5 trials (corresponding to 4 failures in a row and the experiment ending). So we have to adjust our probabilities be dividing by P(n<=5)=.96875.

P(N=1)=.516129

P(N=2)=.258065

P(N=3)=.129032

P(N=4)=.064516

P(N=5)=.032258

So now we have the probabilities of any given success having so many preceding it. The next step is building a negative multinomial distribution for the number of trials.

NOTE: All the math from here to the end of my post ignores the 50 total trials restriction. Messing with the distribution to enforce that is, well, nightmarish.

The negative multinomial is much more complicated than the geometric. As in it involves big-pi notation (the product equivalent of sigma) and the gamma function (the generalization of factorials to all numbers). So I won't even bother giving the formula. Look it up on Wikipedia if you really want to know. But what matters is that if we have m+1 possible outcomes to an event, it gives the probability that:

1) The kth failure is on the nth trial, and

2) That the other n-k trials are divided among the other m outcomes a specified way.

So the outcomes we have are:

*A success is followed immediately by a success. p=.516129

*There is one failure in between. p=.258065

*2 failures. p=.129032

*3 failures. p=.064516

*4 failures in a row, and the experiment ends. p=.032258

The EASIEST things to calculate are the probability of so many successes and the expected number of successes/failures/trials.

Probability of n successes:

This one just follows a geometric distribution from before. The probability of a success (in terms of the geometric) is .032258. So the probability of n successes (in terms of your experiment) is just .032258*.967742^n

Expected number of successes:

Using the geometric again, the expected number of failures (in terms of the geometric) is 1/p=1/.032258=31. So in terms of your experiment, you can expect to see 31 successes.

Expected number of failures:

Back to the negative multinomial. If p_0 is the probability of a failure, k_0 is the number of probabilities before you stop the experiment, and p is the probability of a given event, the expected number of occurrences of that event is k_0*p/p_0.

In this case, k_0 = 1 (because we stop as soon as we get four failures in a row) and p_0 = .032258.

E(0 failures between successes)=.516129/.032258=16

E(1 failure)=.258065/.032258=8

E(2 failures)=.129032/.032258=4

E(3 failures)=.064516/.032258=2

E(4 failures)=1

Yes, I realize those aren't the "exact" values. They actually are. I derounded.

So the expected number of failures (in terms of your experiment) is 16*0+8*1+4*2+2*3+1*4=0+8+8+6+4=26

Expected number of trials:

26+31=57... Umm... Yeah, this is a bit weird. But remember, my math with the negative multinomial was ignoring the bit about trials stopping at 50. It really would have made the math nightmarish.

TL;DR/Summary

So there's a lot of math in this post. But hey, you asked for math help. :smalltongue:

Ignoring the rule about trials stopping at 50, the probability of n successes is .032258*.967742^n. The probability of n failures or trials, though, is tough to calculate.

EDIT: (After being ninja'd by Jay R)

I haven't looked over your math, but I can tell you that's not a valid pdf or cdf. Divide all your values by G(50), so G will actually range from 0 to 1.

Aaand four posts with four different answers. This math is certainly above my head, but I feel like it's only just so. (Of course, I'm probably wrong about that. :smallannoyed:) It's like having a word on the tip of your tongue, except it's not a word, it's a mathematical concept. Frustrating.

That being said, thanks for all the help so far and the walkthroughs of your answers. I appreciate the time and effort they can take.

Also, for what it's worth, this is all helping me, and this isn't a theoretical question. Of course, the real world reason I want these numbers, the numbers aren't so nice to work with.

It's actually 52% success rate (so far, still testing it, only 71 examples so far), four failures ends the whole thing and I need 40 successes total.

At that point it shifts to 5 failures in a row needed to end the sequence. I need 94 more successes (134 total from the beginning) here without 5 failures in a row. If I get to 61 more (101 total) I can start over from scratch at least.

After those 94 successes (134 total), if I still haven't had 5 failures in a row, it changes a little again. I now need 165 more trials (not just successes) without 5 failures in a row. If that happens, then the whole thing goes to 6 failures in a row to terminate.

Yep. Overall, I'd say "very low" is the closest I could reliably estimate the probability of getting to the "6 failures in a row" part happening.

ETA: How much difference does adding to the amount of failures needed make? Is 5 failures needed to end significantly less likely than 4? Only slightly? The answer is (% chance of 4 failures occurring)(% chance of failure), right? So with a 50% fail rate, 5 fails is 50% less likely than 4 fails.

Razanir

2014-01-03, 04:52 PM

Then look over the math, or at least the definitions.

G(n) is the probability that there as at least 4 failures in n trials. That is a cdf of a random variable indicating how many trials it takes to get the first four failures in a row. But it isn't complete. The first set of four failures could be at 51, or 100, or 1,000,000+. (The pdf is F(n), also incomplete.) I stopped at 50, but the sequence doesn't.

Dividing all the values by G(50) would give a Bayesian conditional probability - the probability that there are four failures in n tries, given that there are four failures in the first 50. That's not the problem we were asked to solve.

Ah. Yep, my bad.

reaverb

2014-01-03, 08:51 PM

Aaand four posts with four different answers. Jay R is correct. I independently verified his answer (using a different recursive function and Python instead of Excel) and got the same result.

(From spoiler, emphasis mine)

It's actually 52% success rate (so far, still testing it, only 71 examples so far), four failures ends the whole thing and I need 40 successes total. Wait, does this means failures don't count toward the goal (40) ? That's a different problem from the one given, and I don't think that any of the given formulas would generalize well to it.

Also, if you don't mind, what kind of problem is this modeling?

ETA: How much difference does adding to the amount of failures needed make? Is 5 failures needed to end significantly less likely than 4? Only slightly? The answer is (% chance of 4 failures occurring)(% chance of failure), right? So with a 50% fail rate, 5 fails is 50% less likely than 4 fails. No, the number of tolerable failures has a more complex relationship with the total chance of failure.

Assuming the problem in the first post except for the number of failures:

3 failures: 98.3% chance of the entire sequence failing

4 failures: 82.7% chance (as Jay R said)

5 failures: 55.2% chance

6 failures: 31.5% chance

7 failures: 16.5% chance

BaronOfHell

2014-01-04, 11:18 AM

Jay R is correct. I independently verified his answer (using a different recursive function and Python instead of Excel) and got the same result.

I can verify it as well.

Found all possible 'FFFF' combinations at 20 throws (and some lower), found a formula that got the same results. Used the formula on 50 throws. Compared 10000 attempts with the expected values given by the formula for 20 & 50 throws. Values matched to 1/100th of the expected in both cases. (48% for 20 throws & 83% for 50 throws)

Sadly the formula I got wasn't very pretty as it's recursive and I don't know how to simplify it. It was something along the lines of S(X) = 2**(X - 4) + S(X - 4) + S(X - 3) + S(X - 2) + S(X - 1), X >= 4, where S(X) is the amount of 'FFFF' combinations at X throws.

Where S(0) = S(1) = S(2) = S(3) = 0

Edit: That the loop terminates doesn't do anything to the probability. Unless I misunderstood the problem.

Grozomah

2014-01-05, 07:11 PM

I started on my own since i thought Jay R made a mistake but i ended up at the same solution. So i guess four of us now came to that conclusion.

It's actually 52% success rate (so far, still testing it, only 71 examples so far), four failures ends the whole thing and I need 40 successes total.

At that point it shifts to 5 failures in a row needed to end the sequence. I need 94 more successes (134 total from the beginning) here without 5 failures in a row. If I get to 61 more (101 total) I can start over from scratch at least.

After those 94 successes (134 total), if I still haven't had 5 failures in a row, it changes a little again. I now need 165 more trials (not just successes) without 5 failures in a row. If that happens, then the whole thing goes to 6 failures in a row to terminate.

Firstly the numbers crunch for the 0.48 fail probability:

0, 0, 0, 0.0530842, 0.0254804, 0.0254804, 0.0254804, 0.0254804, \

0.0241278, 0.0234785, 0.0228293, 0.02218, 0.0215308, 0.020916, \

0.0203178, 0.0197361, 0.0191709, 0.0186223, 0.0180893, 0.0175716, \

0.0170688, 0.0165803, 0.0161058, 0.0156448, 0.0151971, 0.0147622, \

0.0143397, 0.0139293, 0.0135307, 0.0131435, 0.0127673, 0.0124019, \

0.012047, 0.0117023, 0.0113674, 0.011042, 0.010726, 0.0104191, \

0.0101209, 0.00983125, 0.00954989, 0.00927659, 0.0090111, 0.00875322, \

0.00850272, 0.00825938, 0.00802301, 0.0077934, 0.00757037, 0.00735372

0, 0, 0, 0.0530842, 0.0785646, 0.104045, 0.129525, 0.155006, \

0.179134, 0.202612, 0.225441, 0.247621, 0.269152, 0.290068, 0.310386, \

0.330122, 0.349293, 0.367915, 0.386005, 0.403576, 0.420645, 0.437225, \

0.453331, 0.468976, 0.484173, 0.498935, 0.513275, 0.527204, 0.540735, \

0.553878, 0.566646, 0.579048, 0.591095, 0.602797, 0.614164, 0.625206, \

0.635932, 0.646351, 0.656472, 0.666304, 0.675853, 0.68513, 0.694141, \

0.702894, 0.711397, 0.719657, 0.72768, 0.735473, 0.743043, 0.750397

-- massive erased section because i realised for about 5 times in a row that the problem is more complicated than it seems --

So you want the probability of actually getting 40 successes without critical failure? That complicates things.

If you're happy with approximate values do a Monte Carlo and iterate it enough times. If you don't know how, I can do it for you (send me an e-mail with all the specific conditions).

Now if you want an analytical solution you're in for a treat.

You do know that the best (shortest) scenario is 40 events, all successes with the probablity of 1/2^40 and the worst case (longest) scenario is 160 events (FFFSFFFS ... FFFS) with the probability of 1/2^160.

For comparison the age of the universe in seconds is about 2^60. Behold the power of exponentiation for it is trouly the power of powers.

The only other path I see is calculating the number of all possible permutations for each F={0, ..., 120} and then somehow figuring out the number where fails are in groups of 4 or more.

I'll think about this some more. Do the Monte Carlo in the meanwhile.

Grozomah

2014-01-05, 07:44 PM

Can't go to sleep before i solve this at least partially. :annoyed:

So I ran the simulation, looking for 40 successes with 4 fails breaking the chains.

For a hundred thousand iterations of the experiment the following number of non-critical fails appear:

7764

7775

7751

7772

7758

I do have to test the random generator though, i get some overly consistent results when doing a million iterations, but it would appear that the success rate is around 7% in this case.

thubby

2014-01-05, 07:48 PM

Can't go to sleep before i solve this at least partially. :annoyed:

So I ran the simulation, looking for 40 successes with 4 fails breaking the chains.

For a hundred thousand iterations of the experiment the following number of non-critical fails appear:

7764

7775

7751

7772

7758

I do have to test the random generator though, i get some overly consistent results when doing a million iterations, but it would appear that the success rate is around 7% in this case.

law of large numbers. you would expect a convergence as you add more results.

Grozomah

2014-01-06, 05:02 AM

Thubby not really. You can excpect a relative convergence, since the number of events increases linearly with the number of repetitions - N, and the standard deviation increases as sqrt(N). So yes, you get a better precision, but still excpect larger deviations at large numbers.

My random generator was not-random somehow so i changed it to rand3 from numerical recipes and got the following results.

7671, 7401, 7474, 7679, 7745, 7552, 7639, 7531, 7556, 7542

The law of large numbers also implies an approximately gaussian distribution with std dev = sqrt(M) if M is the mean. In this case the mean is 7579 and std dev = 103, for the theoretical appoximation of 87. At least the order of magnitude is about correct, unlike in the previous case.

And for a million repetitions i get the following:

75840, 75602, 75457, 75702, 75793, 75466, 75312, 75384, 75410, 75190

with mean 75515, sqrt(mean) 274 and stdev 212. Which seems acceptable.

thubby

2014-01-06, 06:00 AM

standard deviation decreases with more samples because n is in the denominator. as n goes to infinity, s approaches 0.

i.e. there is no deviation if you have unlimited data.

a million repetitions is likely pushing the limits of your rng. computers literally can't do random, after all.

Grozomah

2014-01-06, 07:18 AM

standard deviation decreases with more samples because n is in the denominator. as n goes to infinity, s approaches 0.

i.e. there is no deviation if you have unlimited data.

Seven frozen hells, this churmed my brain way longer than it should. But i think you are mistaken. Because one is the number of experiments ran in the bach (e.g. 100 000 or a million) and the other is the number of batches (10 in both cases). The first increases the standard deviation, since there is a larger range, the second is the one you mention and it does, indeed decrease the std.

Jay R

2014-01-06, 09:49 AM

The variance of the sum of n identically distributed random variables increases, and is n * sigma^2.

Therefore the standard deviation of the sum also increases, and is sqrt(n) * sigma.

To find the average of the n random variables, you will divide sum by n. So its variance and standard deviation shrink with large n.

Therefore the variance of the average is sigma ^2 / n,

and the standard deviation of the average is sigma / sqrt(n).

Example: The variance of a single (six-sided) die roll is 2.9167, and the standard deviation is 1.7078 (to 4 places).

So the variance of the sum of 10,000 dice is 29,167, and the standard deviation is 170.78

The variance of the average of 10,000 die rolls is .0002167, and the standard deviation is 0.017078.

Dr. Gamera

2014-01-06, 10:31 AM

mumble mumble Markov chain mumble mumble...

Koury

2014-01-06, 05:34 PM

Wait, does this means failures don't count toward the goal (40) ? That's a different problem from the one given, and I don't think that any of the given formulas would generalize well to it.

Also, if you don't mind, what kind of problem is this modeling?

Yes, failures don't count toward the goal for a while. Well, I could make it so they do, but that almost doubles the money I'd have to put up.

Which leads into the "What kind of problem is this?" question.

Short answer: I came across an investment platform I was previously unaware of and am trying to understand the risk here.

I came across an article about trading binary options and pair options and have been looking into it.

Simply put, you decide whether you think a given stock or currency will go up or down in value over a given time.

Let's choose a currency. Say EUR/USD. It's current value is 1.36329. This means that €1 = $1.36329 US.

Now we choose a duration for our trade to last. This comes in a variety of options, from 5 minutes to 150 days. Let's go with 30 minutes.

Now we choose whether we think the value will rise or fall over that period of time.

Finally, we choose the amount of money we will put up. Based on our choices of duration, we are given the exact amount we will receive if we have chosen correctly.

For a 30 minute trade, the amount is usually 75%. I say usually because I have seen it at 76% occasionally. For 1 hour trades, it is usually 80%. Most other durations fall in this range. The payout of any given trade is known at the time you enter into it and can not change.

Continuing the example above, we put up $20 on a half hour trade. We believe the EUR/USD value will fall, and choose the option indicating that. A half hour later, we check the trade and see it has concluded and the value has indeed fallen. We receive a payout of $35.

In the event you have chosen incorrectly, you receive nothing at the end of the trade.

Now, as for where the 4 failures thing comes in, that's a part that I've added. If you read the example, you would see that a successful $12 trade would return $21, a $9 profit. Say this trade failed, however. Well, for my next trade I would increase the amount to $28. The return, if successful, would be $49 and a profit of $21. After accounting for the $12 lost in the first trade, I would still be in the black $9. My third trade would go back to the $12.

If the second trade is also a failure, my third trade would be for $66. This give a return of $115.50 and, once accounting for the two previous losses, a profit of $9.50.

With an unlimited bank account, this process can (practically) never fail. However, I do not currently possess an unlimited bank account. So let's look at the numbers here.

{table=head]

Trade Amount |

Total Traded |

Return Amount |

Overall Profit

12 | 12 | 21.00 | 9.00

28 | 40 | 49.00 | 9.00

66 | 106 | 115.50 | 9.50

154 | 260 | 280.00 | 9.50

359 | 619 | 628.25 | 9.25

838 | 1457 |1575.00 | 9.50 [/table]

Well, to get to the 4th trade amount there, you'd need to put up at least $260. To get to the 5th, you'd need $619. I'm not comfortable risking $619 on this, but can stomach $260 since that amount would largely be cash I was given at Christmas. This is there the "4 failures in a row ends the process" comes from. To get to $619 from $260, you need $395. Each success giving ~$9. I need 40 successes until I can have my 5th failure. This is the point with the most overall risk in this whole thing. Then another 32 successes until I have enough cash to pull out my initial $260 while maintaining my 5 failure amount.

From this point on, I have no "real" risk anymore. If I go to $0, it's just profits lost. Of course, getting to this point isn't some simple task.

Also, the whole "increase the amount to cover for old losses" thing is a thing I dreamed up back when I really wanted to play a lot of blackjack at casinos. It's fine when it takes like 2 minutes between trials, but here, where trials last much longer, it is a little impractical. $9 over 2 hours isn't terribly great.

The solution to this whole "slow income" issue is, of course, more money. At some point I would like to transfer to the following table of trades:

{table=head]

Trade Amount |

Total Traded |

Return Amount |

Overall Profit

14 | 14 | 24.50 | 10.50

46 | 60 | 80.50 | 20.50

120 | 180 | 210.00 | 30.00

294 | 474 | 514.50 | 40.50

699 | 1173 |1223.25 | 50.25

1644 | 2817 |2877.00 | 60.00 [/table]

This table allows for a much more steady income rate. Winning the third trade, is the same profit:time (basically) as winning the first trade 3 times. But I'm not really willing to sacrifice my 5th failure to jump to this table. After pulling my initial cash out, I would need 62 more successes to get to the 5th line of this table from the 5th line of the last one.

However, once I am on this table, it is no longer # of successes without X failures I need, but # of trials without X failures. Getting from the 5th line on the second table to the 6th line will take 165-169 trials (if it ends SFFFFS).

I would, from this point, probably continue to minimize risk by moving along the table toward 10 failures, taking cash out when I needed it (rent, etc). Not too often, and only in chunks, since it cost $25 to take cash out. Once there, I would probably save up another $1100 and start a second "line" going, starting from 5 failures on the second table.

Anyway, I'm rambling.

tl;dr: Fool throws money in fire, is certain money tree will grow from ashes.

BaronOfHell

2014-01-06, 05:41 PM

I only skimmed your post, but the main problem with "betting more so I cover my loss + can win the same as before" is that the amount of money to bet raises exponentially.

Koury

2014-01-06, 05:48 PM

I only skimmed your post, but the main problem with "betting more so I cover my loss + can win the same as before" is that the amount of money to bet raises exponentially.

Yeah, I covered that in my post. Basically, I choose a point that I'm comfortable with, and live with it. For me, it's the amount that can afford me 4 failures. If I go to $0, I go to $0. I wouldn't keep pouring money in if things go against me.

The higher points only come in to play if I somehow make enough for my profits to cover it.

BaronOfHell

2014-01-06, 05:52 PM

Sorry I missed that point.

Koury

2014-01-06, 05:54 PM

Sorry I missed that point.

It's OK that you didn't read that whole wall of text. :smallwink:

reaverb

2014-01-07, 12:29 AM

tl;dr: Fool throws money in fire, is certain money tree will grow from ashes.Well, you don't need any fancy mathematics to show that.

Here's the formula for the expected return of binary options:

[money_won_if_right * success_rate] - [money_lost_if_wrong * (1 - success_rate)]

First, let's assume the success_rate of 52% is correct. Next, we'll assume that binary option is for a single dollar, which means we can multiply the expected value by the actual dollar amount to get the expected value in a different case. (i.e, the expected value will be in the form of a percentage saying how much of an investment one would expect to get back)

30 min trade: (75% return)

[.75 * .52] - [1.00 * 0.48] = -0.09

1 hour trade: (80% return)

[.80 * .52] - [1.00 * .48] = -0.064

So an investor would expect to lose 9% of their investment in every 30 min. trade and 6.4% of their investment in every hour trade.

You've mentioned Martingale betting system (http://en.wikipedia.org/wiki/Martingale_(betting_system)). However, you've also mentioned it only theoretically works if you have infinite payroll. This is correct.

Effectively, Martingale is playing a shell game with the probabilities. If you use it you can still expect to lose 9% or whatever of your investment for each option. It's just that instead of having a lot of peaks and valleys in your bank roll, you'll be steadily going up until you hit a bad streak and get knocked down to worse than when you started. A 99% chance of getting $1.00 isn't worth it if you lose $200.00 the other 1% of the time.

If you want to know how good at predictions somebody would have to be to break even the formula is basic algebra:

[money_won_if_right * success_rate] - [money_lost_if_wrong * (1 - success_rate)] = 0.00

[money_won_if_right * success_rate] - [money_lost_if_wrong - (money_lost_if_wrong * success_rate)] = 0.00

[money_won_if_right * success_rate] + (money_lost_if_wrong * success_rate) - money_lost_if_wrong = 0.00

(money_won_if_right + money_lost_if_wrong) * success_rate = money_lost_if_wrong

success_rate = money_lost_if_wrong/(money_won_if_right + money_lost_if_wrong)

So with a 80% return you would need to be right [1.00/(0.8+1.00)]= 55.6% of the time. If you get less return, you would need to be right more often.

If somebody theoretically achieved this, there is less variance in a traditional betting strategy than in Martingale.

(Also, 52% success rate out of 71 trials means you got 37 right when the mean is between 36 and 35, so at this point it's basically meaningless.)

Douglas

2014-01-07, 02:26 AM

Schemes for manipulating money-making probabilities like this really are fools' traps. If any such scheme that actually works exists, the collective interest of all humanity typically discovers it very quickly, and it promptly gets exploited until the system that made it possible collapses or adapts to defeat it. To make a reliable profit you'd have to be just about the first discoverer in the world of a particular working scheme, and that is extremely unlikely. In any other case, one way or another it will most likely work out to throwing money away.

I only know of one case where something like this actually worked out, and the fact that it did was so unusual that it became famous. That case is perfect play with card counting in casino blackjack, and even if you manage to find a casino that hasn't taken countermeasures it requires a lot of work, skill, and patience on top of a substantial initial bankroll.

thubby

2014-01-07, 03:21 AM

I only know of one case where something like this actually worked out, and the fact that it did was so unusual that it became famous. That case is perfect play with card counting in casino blackjack, and even if you manage to find a casino that hasn't taken countermeasures it requires a lot of work, skill, and patience on top of a substantial initial bankroll.

the skill part is highly over stated. anyone with a highschool education can do it. the bankroll and dealing with the casino itself, however...

Koury

2014-01-07, 03:36 AM

Martingale betting system (http://en.wikipedia.org/wiki/Martingale_(betting_system)). Thank you for putting a name to this. I thought of it many years ago (well, like 6) and it's nice to have the correct term for it. (While I thought of it independently, I never thought I was somehow the first or anything)

So with a 80% return you would need to be right [1.00/(0.8+1.00)]= 55.6% of the time. If you get less return, you would need to be right more often. And 571/7th % for 75% return.

(Also, 52% success rate out of 71 trials means you got 37 right when the mean is between 36 and 35, so at this point it's basically meaningless.) Oh, yeah, I know 71 trials is meaningless. I'm aiming for around 1000 before I get comfortable with an actual number. Currently at 48/87 trials (and one push, which isn't counted in that 87). One streak of 5 failures, and a couple streaks of 3.

{table=head]Trade |Currency | Time | Number | Majority | Percent | End | Number | Pass

1. | EUR/USD | 2:32 | 1.37222 | Down | 79% | 3:02 | 1.37165 | Yes

2. | GBP/USD | 2:34 | 1.65491 | Down | 80% | 3:04 | 1.65483 | Yes

3. | USD/JPY | 2:35 | 105.35400 | Up | 90% | 3:05 | 105.40100 | Yes

4. | USD/CHF | 2:37 | 0.89753 | Down | 92% | 3:07 | 0.89802 | No

5. | AUD/USD | 2:38 | 0.88840 | Up | 65% | 3:08 | 0.88514 | No

6. | USD/CAD | 2:39 | 1.06321 | Up | 51% | 3:09 | 1.06497 | Yes

7. | EUR/GBP | 2:40 | 0.82913 | Down | 82% | 3:10 | 0.82814 | Yes

8. | EUR/JPY | 2:41 | 144.56000 | Up | 73% | 3:11 | 144.47700 | No

9. | GBP/JPY | 2:42 | 174.34300 | Up | 94% | 3:12 | 174.45100 | Yes

10. | EUR/CHF | 2:43 | 1.23205 | Down | 63% | 3:13 | 1.23125 | Yes

11. | EUR/CAD | 2:44 | 1.45922 | Down | 58% | 3:14 | 1.45849 | Yes

12. | GBP/CHF | 2:45 | 1.48602 | Up | 61% | 3:15 | 1.48752 | Yes

13. | EUR/USD | 9:37 | 1.36683 | Down | 70% | 10:07 | 1.36691 | No

14. | GBP/USD | 9:38 | 1.64368 | Down | 59% | 10:08 | 1.64384 | No

15. | USD/JPY | 9:39 | 104.80200 | Up | 94% | 10:08 | 104.79200 | No

16. | USD/CHF | 9:40 | 0.89986 | Down | 90% | 10:10 | 0.89957 | Yes

17. | AUD/USD | 9:41 | 0.89183 | Up | 52% | 10:11 | 0.89133 | No

18. | USD/CAD | 9:42 | 1.06380 | Up | 83% | 10:12 | 1.06379 | No

19. | EUR/GBP | 9:43 | 0.83138 | Down | 87% | 10:13 | 0.83165 | No

20. | EUR/JPY | 9:44 | 143.23900 | Up | 83% | 10:14 | 143.31200 | Yes

21. | GBP/JPY | 9:45 | 172.33200 | Up | 75% | 10:15 | 172.33900 | Yes

22. | EUR/CHF | 9:47 | 1.22994 | Up | 62% | 10:17 | 1.23010 | Yes

23. | EUR/CAD | 9:48 | 1.45386 | Up | 56% | 10:18 | 1.45449 | Yes

24. | GBP/CHF | 9:49 | 1.47929 | Up | 59% | 10:19 | 1.47911 | No

25. | EUR/USD | 10:21 | 1.36727 | Down | 71% | 10:51 | 1.36691 | Yes

26. | GBP/USD | 10:22 | 1.64423 | Down | 81% | 10:52 | 1.64401 | Yes

27. | USD/JPY | 10:23 | 104.78800 | Up | 95% | 10:53 | 104.67600 | No

28. | USD/CHF | 10:24 | 0.89949 | Down | 91% | 10:54 | 0.89954 | No

29. | AUD/USD | 10:25 | 0.89166 | Down | 51% | 10:55 | 0.89207 | No

30. | USD/CAD | 10:26 | 1.06382 | Up | 84% | 10:56 | 1.06553 | Yes

31. | EUR/GBP | 10:27 | 0.83160 | Down | 91% | 10:57 | 0.83166 | No

32. | EUR/JPY | 10:28 | 143.21900 | Up | 85% | 10:58 | 143.06300 | No

33. | GBP/JPY | 10:29 | 172.22300 | Up | 62% | 10:59 | 172.02500 | No

34. | EUR/CHF | 10:30 | 1.22984 | Up | 87% | 11:00 | 1.22975 | No

35. | EUR/CAD | 10:31 | 1.45498 | Down | 81% | 11:01 | 1.45597 | No

36. | GBP/CHF | 10:32 | 1.47866 | Up | 57% | 11:02 | 1.47882 | Yes

37. | EUR/USD | 11:10 | 1.36669 | Down | 70% | 11:40 | 1.36603 | Yes

38. | EUR/USD | 11:15 | 1.36680 | Down | 70% | 11:45 | 1.36575 | Yes

39. | GBP/USD | 11:20 | 1.64319 | Down | 72% | 11:50 | 1.64306 | Yes

40. | USD/CHF | 11:25 | 0.89753 | Down | 99% | 11:55 | 0.90045 | No

41. | EUR/USD | 11:41 | 1.36603 | Down | 72% | 12:11 | 1.36550 | Yes

42. | EUR/USD | 11:46 | 1.36593 | Down | 71% | 12:16 | 1.36559 | Yes

43. | GBP/USD | 11:51 | 1.64303 | Down | 67% | 12:21 | 1.64284 | Yes

44. | USD/CHF | 11:56 | 0.90041 | Down | 99% | 12:26 | 0.90066 | No

45. | EUR/USD | 12:12 | 1.36565 | Down | 67% | 12:42 | 1.36538 | Yes

46. | EUR/USD | 12:17 | 1.36553 | Down | 66% | 12:47 | 1.36532 | Yes

47. | GBP/USD | 12:22 | 1.64282 | Down | 62% | 12:52 | 1.64271 | Yes

48. | EUR/USD | 12:44 | 1.36550 | Down | 68% | 13:14 | 1.36657 | No

49. | EUR/USD | 12:48 | 1.36533 | Down | 69% | 13:18 | 1.36671 | No

50. | GBP/USD | 12:53 | 1.64265 | Down | 56% | 13:23 | 1.64506 | No

51. | USD/CHF | 12:27 | 0.90066 | Down | 97% | 12:57 | 0.90020 | Yes

52. | USD/JPY | 13:01 | 104.76200 | Up | 93% | 13:31 | 104.70700 | No

53. | EUR/USD | 13:15 | 1.36663 | Down | 65% | 13:45 | 1.36676 | No

54. | EUR/USD | 13:19 | 1.36669 | Down | 66% | 13:49 | 1.36665 | Yes

55. | GBP/USD | 13:25 | 1.64500 | Down | 65% | 13:55 | 1.64500 | Push

56. | USD/JPY | 13:32 | 104.70700 | Up | 93% | 14:02 | 104.80200 | Yes

57. | EUR/USD | 13:46 | 1.36670 | Down | 68% | 14:16 | 1.36676 | No

58. | EUR/USD | 13:50 | 1.36656 | Down | 69% | 14:20 | 1.36657 | No

59. | GBP/USD | 13:57 | 1.64504 | Down | 72% | 14:27 | 1.64438 | Yes

60. | USD/JPY | 14:03 | 104.80600 | Up | 91% | 14:33 | 104.79600 | No

61. | EUR/USD | 14:17 | 1.36683 | Down | 74% | 14:47 | 1.36605 | Yes

62. | GBP/USD | 15:26 | 1.64477 | Up | 93% | 15:56 | 1.64468 | No

63. | USD/JPY | 15:58 | 104.84400 | Up | 92% | 16:28 | 104.79700 | No

64. | USD/JPY | 16:29 | 104.79700 | Up | 92% | 16:59 | 104.81400 | Yes

65. | USD/JPY | 17:02 | 104.82500 | Up | 93% | 17:32 | 104.85600 | Yes

66. | USD/JPY | 17:33 | 104.85600 | Up | 94% | 18:03 | 104.77100 | No

67. | USD/JPY | 18:04 | 104.74800 | Up | 92% | 18:34 | 104.30400 | No

68. | USD/JPY | 18:36 | 104.23900 | Up | 93% | 19:06 | 104.26800 | Yes

69. | GBP/JPY | 1:08 | 171.63000 | Up | 96% | 1:38 | 171.67500 | Yes

70. | GBP/JPY | 1:39 | 171.70200 | Up | 100% | 2:09 | 171.69800 | No

71. | GBP/JPY | 2:10 | 171.74800 | Up | 68% | 2:40 | 171.90500 | Yes

72. | EUR/USD | 11:05 | 1.36063 | Up | 50% | 11:35 | 1.36065 | Yes

73. | GBP/USD | 11:08 | 1.64304 | Up | 93% | 11:38 | 1.64262 | No

74. | USD/JPY | 11:09 | 104.45600 | Up | 94% | 11:39 | 105.52400 | Yes

75. | USD/CHF | 11:10 | 0.90415 | Down | 74% | 11:40 | 0.90475 | No

76. | AUD/USD | 11:11 | 0.89792 | Down | 63% | 11:41 | 0.89743 | Yes

77. | USD/CAD | 11:12 | 1.06120 | Down | 57% | 11:42 | 1.06162 | No

78. | EUR/GBP | 11:13 | 0.82788 | Down | 90% | 11:43 | 0.82810 | No

79. | EUR/JPY | 11:14 | 142.07100 | Up | 78% | 11:44 | 142.18900 | Yes

80. | GBP/JPY | 11:15 | 171.59900 | Up | 97% | 11:45 | 171.73500 | Yes

81. | EUR/CHF | 11:16 | 1.22982 | Down | 88% | 11:46 | 1.23057 | No

82. | EUR/CAD | 11:17 | 1.44312 | Down | 66% | 11:47 | 1.44381 | No

83. | GBP/CHF | 11:18 | 1.48528 | Up | 61% | 11:48 | 1.48620 | Yes

84. | USD/JPY | 11:23 | 104.44600 | Up | 93% | 11:53 | 104.61400 | Yes

85. | GBP/JPY | 11:55 | 171.83800 | Up | 97% | 12:25 | 171.95800 | Yes

86. | EUR/CHF | 12:13 | 1.23204 | Down | 89% | 12:43 | 1.23192 | Yes

87. | GBP/USD | 20:59 | 1.63886 | Up | 94% | 21:29 | 1.63942 | Yes

88. | GBP/USD | 21:32 | 1.63940 | Up | 95% | 22:02 | 1.63951 | Yes[/table]

That whole "majority" and "percent" thing is from a tool that tells you what direction the majority is trading and what percent of people are in the majority. So a 75% Up means that 75% of the people trading that option are choosing Up.

I'm including it in my data so that I can look for trends there. The obvious assumption is that high majorities will be right more often, but we will see. My lone 100% is actually a fail and my lone 50% was a pass, so you know.

A Pass is any time the majority was correct.

Schemes for manipulating money-making probabilities like this really are fools' traps.

In any other case, one way or another it will most likely work out to throwing money away. I completely agree with you. I haven't committed any money to this and will not until I've done enough research to be satisfied.

That being said, unless the success rate falls to something, say, below 50%, I'm likely going to do it anyway.

This is for a few reasons, like that the money going toward this is largely unexpected Christmas cash, but mostly because a small part of me has always wanted to say "screw the odds" and try a get rich quick scheme kind of thing. I know it won't work, and I will in no way compromise myself financially by losing the money I put in, but I'll be able to say I tried, which will satisfy me.

And hey, this is, in my mind, at least a step above one of those late night TV "buy my program to get rich" things. :smalltongue:

Douglas

2014-01-07, 03:49 AM

the skill part is highly over stated. anyone with a highschool education can do it. the bankroll and dealing with the casino itself, however...

Can do it? Yes. Can do it absolutely reliably every single time without fail, with the discipline to never deviate from the algorithm on intuition? That's harder, and the margin is small enough that that's what you need to make it work.

That being said, unless the success rate falls to something, say, below 50%, I'm likely going to do it anyway.

This is for a few reasons, like that the money going toward this is largely unexpected Christmas cash, but mostly because a small part of me has always wanted to say "screw the odds" and try a get rich quick scheme kind of thing. I know it won't work, and I will in no way compromise myself financially by losing the money I put in, but I'll be able to say I tried, which will satisfy me.

And hey, this is, in my mind, at least a step above one of those late night TV "buy my program to get rich" things. :smalltongue:

If you're viewing it as akin to paying money (that you can readily afford to lose) for some entertainment, then sure, go right ahead. That's the only really responsible way to do most gambling in my opinion.

Have fun!

Koury

2014-01-07, 03:57 AM

If you're viewing it as akin to paying money (that you can readily afford to lose) for some entertainment, then sure, go right ahead. That's the only really responsible way to do most gambling in my opinion.

Have fun!

Yeah, this. Same reason me and my girlfriend buy scratch tickets three or four times a year. No, we won't win anything, but we got to have fun scratching them together.

I'll only really be disappointed with this if I go to $0 in like the first 10 trades.

reaverb

2014-01-07, 04:42 AM

I only know of one case where something like this actually worked out, and the fact that it did was so unusual that it became famous. That case is perfect play with card counting in casino blackjack, and even if you manage to find a casino that hasn't taken countermeasures it requires a lot of work, skill, and patience on top of a substantial initial bankroll.

Even card counting isn't really comparable. Card counting works because removing cards from a deck without replacing them meddles with the chances of getting a particular card from that deck later. The equivalent of card counting would be if somebody found out how to predict the result of binary options with 90% accuracy once every couple days. (Of course this would fail to the system adaption you mentioned. Once everybody expects an increase the price would rise beforehand until the chances are 50/50 again).

Yes, treating binary options as gambling seems like the healthiest outlook. It certainly lasts longer than, for example, a slot machine.

Grozomah

2014-01-07, 08:22 AM

Well, it's good to see you going into this with the right mentality. Just don't let it drag you in with the promises of vast riches (According to N.N. Taleb around 10 people per year on wall street become millionaires due to nothing but blind dumb luck).

Also, having done some practical financial calculations myself - economy is a way softer science than it wants to be. Meaning that while the math is solid enough all the assumptions are generally wrong always vary considerably more than everyone would like.

The number of samples is way too small. Also jsut because you were 52% succesfull one week doesn't mean you will continue to do so. There are very likely other people with similair systems meaning there may be others exploiting it - working against you. All you need is one bad luck streak to be out. etc.

In short you cannot build a plane out of mud.

But yes, if you compare this to buying and watching a bundle of jersey shore or something similair then i fully support your decision to dabble in finances. Good luck! :)

PS: I'm still working on the monte carlo simulation, i'll keep you posted if i figure anything out.

TuggyNE

2014-01-08, 02:41 AM

Well, it's good to see you going into this with the right mentality. Just don't let it drag you in with the promises of vast riches (According to N.N. Taleb around 10 people per year on wall street become millionaires due to nothing but blind dumb luck).

10 whole people per year? The odds are in my favor! :smalltongue:

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