Computer Simulations.
Quote: IbeatyouracesDon't waste your time. Some people will never learn.
What concerns me is, if JByrd thinks correct math is "horrifically bad math", what does he think is "good math"? Math that suits his needs?
When over a half a dozen members on this forum say "you're wrong" on a math question, shouldn't that be enough? Does he need Wizard to actually come into this thread and agree with us to change his mind? Would it even change his mind?
Quote: ThatDonGuy
Edit: I just ran one, and got two runs where it took 11.92 billion and 15.19 billion tosses, respectively, to get back to 50%. The first one had a heads/tails difference as high as 170,218, and the second as high as 199,861. At one toss per second, the second one took over 480 years.
There's something to think about:
Let "point A" be the starting point, and "point B" the first point where there are 100,000 more heads than tails since A.
Eventually, you "should" get back to zero from A, but that means that there would be 100,000 more tails than heads since B. Let this be "point C".
Getting back to zero from B means you're back to 100,000 more tails than heads since point C. You end up going back and forth between "100,000 more heads than tails" and "100,000 more tails than heads."
Exactly.
Every simulation you will ever run will have a wave form. The losses will grow and grow, reach an apex (in this case 170,218), and then get smaller and smaller until reaching zero.
This will occur every time you run a simulation. Which means Richard A Epstein (and everyone who agrees with that logic) do REALLY, REALLY, REALLY BAD MATH.
Quote: JyBrd0403Exactly.
Every simulation you will ever run will have a wave form. The losses will grow and grow, reach an apex (in this case 170,218), and then get smaller and smaller until reaching zero.
This will occur every time you run a simulation. Which means Richard A Epstein (and everyone who agrees with that logic) do REALLY, REALLY, REALLY BAD MATH.
Not at all. You and Epstein are explaining two different things - both of which, if you ask me, are true.
Epstein et al. say that the differences "tend" to increase - and in my simulations, they do, for no other reason that, the more tosses you have, the more likely it is for there to be a point where, for some number n, the difference = n. No matter what n you choose, such a point "will" exist, just as the difference "will" return to zero. Nothing bad about that math.
You say that, at some undetermined point in the future, the actual fraction of "wins" will match the expected result. Again, with enough tosses, this "should" always happen as well. The only real problem with this is, this is not 100% "fact" unless you have unlimited time - which, unless you (a) become immortal, and (b) develop some form of spacecraft to escape the inevitable point where Earth cannot support life (e.g. because the sun runs out of fuel to heat it), you don't have.
Quote: ThatDonGuyNot at all. You and Epstein are explaining two different things - both of which, if you ask me, are true.
Epstein et al. say that the differences "tend" to increase - and in my simulations, they do, for no other reason that, the more tosses you have, the more likely it is for there to be a point where, for some number n, the difference = n. No matter what n you choose, such a point "will" exist, just as the difference "will" return to zero. Nothing bad about that math.
You say that, at some undetermined point in the future, the actual fraction of "wins" will match the expected result. Again, with enough tosses, this "should" always happen as well. The only real problem with this is, this is not 100% "fact" unless you have unlimited time - which, unless you (a) become immortal, and (b) develop some form of spacecraft to escape the inevitable point where Earth cannot support life (e.g. because the sun runs out of fuel to heat it), you don't have.
The difference may hit 0, but it will never converge on 0. The heads probability will, however, converge on 50%. Jbyrd seems to think they go hand in hand.
Exactly. Two things are true:Quote: CrystalMathThe difference may hit 0, but it will never converge on 0. The heads probability will, however, converge on 50%. Jbyrd seems to think they go hand in hand.
a) The expected difference between heads and tails increases as the number of trials increases; and
b) For an infinite series of flips, there is a 100% probability that the difference between heads and tails returns to zero at some point after the trials begin.
The mistake is in thinking the above is a paradox.
For the first point:
http://mathworld.wolfram.com/RandomWalk1-Dimensional.html
http://mathworld.wolfram.com/Heads-Minus-TailsDistribution.html
For the second:
http://mathworld.wolfram.com/PolyasRandomWalkConstants.html
Quote: ThatDonGuyNot at all. You and Epstein are explaining two different things - both of which, if you ask me, are true.
Epstein et al. say that the differences "tend" to increase - and in my simulations, they do, for no other reason that, the more tosses you have, the more likely it is for there to be a point where, for some number n, the difference = n. No matter what n you choose, such a point "will" exist, just as the difference "will" return to zero. Nothing bad about that math.
You say that, at some undetermined point in the future, the actual fraction of "wins" will match the expected result. Again, with enough tosses, this "should" always happen as well. The only real problem with this is, this is not 100% "fact" unless you have unlimited time - which, unless you (a) become immortal, and (b) develop some form of spacecraft to escape the inevitable point where Earth cannot support life (e.g. because the sun runs out of fuel to heat it), you don't have.
No, that's what's funny, Epstein and I are both explaining the same thing, the Law of Large Numbers. Strike 1.
Epstein says the differences tend to increase "indefinitely", they don't they tend to form a wave, Every simulation you do unless you are immortal and have endless amounts of time, will have this wave. And, that's on your little simulator that makes 480 years in real life take a couple of hours on your computer. Get it, you can simulate billions of years of tosses and it will NEVER do anything else but form that wave.
And now I'm done dealing with you as well. Don't bother responding, I'm blocking you as well. I have no interest in your bad math either.
Wow, if you block half the people on this site the threads actually aren't so stupid. Whoa.
Quote: someoneYes. You keep getting closer and closer to exactly 50%, but you're not guaranteed to actually hit it. Most people will agree that 49.999999999999999999999999999999999999% is very close to 50%. If you need closer you can keep flipping until you get 1000 9's after the decimal place and you can reach these figures with an imbalance between heads and tails.
If you are still struggling to see this, James Grosjean did a blog http://www.gamblingwithanedge.com/the-denominator-where-due-happens a few years ago the explains it very well.
This is another interesting thing. If you actually do the math 49.99999999999999999999% would get you within .00000000000001 of 50%.
100 million trials * 49.99999999999999% = 49,999,999.9999999999
The math doesn't really even match the reality that you can't have a .99999 win. You can't just keep adding 1000 more 9's because in reality you would either be -1 or would go up to 0. You can't have a .999999 win.
So the math you want to do there doesn't really even apply to this situation. 49.999999% would have to be rounded to 50%, just because you can't have .9999999 wins. So, all those 49.9999999% things, where you want to add another 1000 9's after that, would actually have to be rounded off to 50%. In other words you can't really just keep getting closer to 50%, you would actually have to hit 50%, because there is no .9999999 wins.
Quote: JyBrd0403This is another interesting thing. If you actually do the math 49.99999999999999999999% would get you within .00000000000001 of 50%.
100 million trials * 49.99999999999999% = 49,999,999.9999999999
The math doesn't really even match the reality that you can't have a .99999 win. . . . you would actually have to hit 50%, because there is no .9999999 wins.
JB,
First you made me chuckle when you described my response as "horrifically bad math"
Then you made me laugh out loud when you describe ME's response as "horrifically bad math"
Then I fell about laughing when I read the nonsense espoused in your own previous threads where you described the near certainty of making megabucks with d'Alembert
Finally I despaired when you indicated that probabilities are inherently wrong if they are not integers. No. You cannot have 0.9999999 wins. But you can have a 0.99999999 probability of a win.
I cordially invite you to block me. Indeed block your own internet connection, since you know some of the greatest mathematicians to be ignorant fools and clearly you don't want to see the real source of horrifically bad maths and horrifically bad logic to boot.
Save your energy. Go beat up the casinos with your math prowess.
I think we are ALL agreed, though, that a very large n flips of a coin (to simplify) will
result in a close proximity to 50% each.
Now, can we get to this stage-
Probability of < x% heads in n flips? Anyone know the formula for this?
then we might get to-
Probability of <34% reds in n spins. (we will have to consider that for 1 spin, probability is 48.65%, not 50%)
This will tell us the probability we won't win 1/3 of our bets in n spins, which would be a more practical use
of our knowledge. Can anyone help?
http://mathworld.wolfram.com/BinomialDistribution.html
Spreadsheets like excel also have built in formulas for these type of calculations.
Quote: daveylbraHey guys, sorry I didn't mean to cause arguments.
I think we are ALL agreed, though, that a very large n flips of a coin (to simplify) will
result in a close proximity to 50% each.
Now, can we get to this stage-
Probability of < x% heads in n flips? Anyone know the formula for this?
then we might get to-
Probability of <34% reds in n spins. (we will have to consider that for 1 spin, probability is 48.65%, not 50%)
This will tell us the probability we won't win 1/3 of our bets in n spins, which would be a more practical use
of our knowledge. Can anyone help?
I'm not sure why you think winning 1/3 of your bets is going to produce a profit with the labouchere progression. I'm guessing it's because you add 1 number to the line on a loss, and subtract 2 numbers on a win. If you play the system, though, you would need to win 1/2 of your bets, not 1/3.
Just quickly if you started the line with...
1-2
you lose and add 3
1-2-3
you lose again add 4
1-2-3-4
That's 2 losses. So, in order to end the labouchere you would need 2 wins.
You bet 5 and win, subtract the 1 and 4, you're left with...
2-3
You bet 5 again and win, subtract the 2 and 3, and you've cleared the line.
The end result was 2 losses, and 2 wins or 50%.
So, for the labouchere progression you would still need the Law of Large Numbers, a legitimate LLN.
So, if your going to use a computer simulator, make sure that the RNG is not cheating the LLN. You can check at each "large number" intervals to make sure the percentage is getting closer to the mean. If the LLN is being broken then the whole simulation is worthless, but the Simulators won't check this for you, so you have to check it yourself.
OnceDear, please don't respond to my posts,, I've blocked you and tringlomane as well, for your horrific math.
Quote: JyBrd0403I'm not sure why you think winning 1/3 of your bets is going to produce a profit with the labouchere progression. I'm guessing it's because you add 1 number to the line on a loss, and subtract 2 numbers on a win. If you play the system, though, you would need to win 1/2 of your bets, not 1/3.
Just quickly if you started the line with...
1-2
you lose and add 3
1-2-3
you lose again add 4
1-2-3-4
That's 2 losses. So, in order to end the labouchere you would need 2 wins.
You bet 5 and win, subtract the 1 and 4, you're left with...
2-3
You bet 5 again and win, subtract the 2 and 3, and you've cleared the line.
The end result was 2 losses, and 2 wins or 50%.
So, for the labouchere progression you would still need the Law of Large Numbers, a legitimate LLN.
So, if your going to use a computer simulator, make sure that the RNG is not cheating the LLN. You can check at each "large number" intervals to make sure the percentage is getting closer to the mean. If the LLN is being broken then the whole simulation is worthless, but the Simulators won't check this for you, so you have to check it yourself.
OnceDear, please don't respond to my posts,, I've blocked you and tringlomane as well, for your horrific math.
I fully expect you will continue to get responses to your calculations whether you like them or not. This forum does not allow bad advice or information to go unchallenged. It's part of its basic purpose. So block anyone you like, preferably without making a public point of it, but don't think you have the right to stop them from commenting.
Uh oh, better watch out or JyBrd might block you for "horrific commenting."Quote: beachbumbabsI fully expect you will continue to get responses to your calculations whether you like them or not. This forum does not allow bad advice or information to go unchallenged. It's part of its basic purpose. So block anyone you like, preferably without making a public point of it, but don't think you have the right to stop them from commenting.
I wonder if he's blocked the Wizard yet...
Quote: beachbumbabsI fully expect you will continue to get responses to your calculations whether you like them or not. This forum does not allow bad advice or information to go unchallenged. It's part of its basic purpose. So block anyone you like, preferably without making a public point of it, but don't think you have the right to stop them from commenting.
Just letting them know that I will not be able to read their posts, so they are not directing their posts at me, as OnceDear had earlier.
Just like I'm blocking you now BBB, so you don't have to bother sending another posts directed at me, like the one above. I won't be reading it.
I thought it would be nice to let you know that, my bad.
Quote: MathExtremistUh oh, better watch out or JyBrd might block you for "horrific commenting."
Quote: JyBrd0403Just like I'm blocking you now BBB, so you don't have to bother sending another posts directed at me, like the one above. I won't be reading it.
Man, you can't make this up.
Quote: MathExtremistMan, you can't make this up.
Oh please..... Please Mike, Please wade in with your Wizardy Wisdom and show JyBrd403 some maths, any maths, just so we can get the spectacle of Wizard being blocked on his own forum.
From wikipedia, labouchere system - "Theoretically, because the player is cancelling out two numbers on the list for every win and adding only one number for every loss, the player needs to have his proposition come at least 33.34% to eventually complete the list. For example, if the list starts with seven numbers and the player wins five times and loses three (62.5% winning percentage) the list is completed and the player wins the desired amount, if the list starts with seven numbers and the player wins 43,600 times and loses 87,193 times (33.34% winning percentage) the list completes and the player wins."
Similarly, I understood Labouchere to require one third plus one bets, out of the total bets, to win.
BUT, suppose I start with 1,1,1,1,1. I lose the first 9 bets.
I have 1,1,1,1,1,2,3,4,5,6,7,8,9, 10. I have lost 54 units. (2+3+4+5+6+7+8+9+10)
I then win 6 bets. I have 3,4. I have won 52 units in those 6 bets. (11+10+9+8+7+7)
I have won 6 out of 15 bets. This is a third plus one. But I am still clearly down!
Can someone please explain?? Is there a certain number of total bets required before the one third plus one law 'kicks in?'
Very true. But what I've often wondered is this: Does a baseball team have a memory? If an average baseball team lost 8 in a row would it be more likely than usual to win one of the next 4 games? Assuming no super pitchers are involved. It would seem like they would want very badly and try very hard to have a winning game and end their losing streak. More so than their opposing team would want to keep their losing streak alive. I really don't know the answer to this. It would take an awful lot of data mining to answer the question. But if I had to express an opinion I would say that they are more likely than usual to win one of their next four games.
Quote: OnceDearOh please..... Please Mike, Please wade in with your Wizardy Wisdom and show JyBrd403 some maths, any maths, just so we can get the spectacle of Wizard being blocked on his own forum.
This would be hilarious.
Quote: daveylbraTo everyone:
From wikipedia, labouchere system - "Theoretically, because the player is cancelling out two numbers on the list for every win and adding only one number for every loss, the player needs to have his proposition come at least 33.34% to eventually complete the list. For example, if the list starts with seven numbers and the player wins five times and loses three (62.5% winning percentage) the list is completed and the player wins the desired amount, if the list starts with seven numbers and the player wins 43,600 times and loses 87,193 times (33.34% winning percentage) the list completes and the player wins."
Similarly, I understood Labouchere to require one third plus one bets, out of the total bets, to win.
BUT, suppose I start with 1,1,1,1,1. I lose the first 9 bets.
I have 1,1,1,1,1,2,3,4,5,6,7,8,9, 10. I have lost 54 units. (2+3+4+5+6+7+8+9+10)
I then win 6 bets. I have 3,4. I have won 52 units in those 6 bets. (11+10+9+8+7+7)
I have won 6 out of 15 bets. This is a third plus one. But I am still clearly down!
Can someone please explain?? Is there a certain number of total bets required before the one third plus one law 'kicks in?'
I don't think the "one-third plus one" rule is entirely accurate. If you add one number when you lose and remove two when you win, then you will have the same number as when you started if you have one-third wins; one additional win removes an additional two numbers, but if the remaining numbers (caused by earlier losses) total more than the sum of your initial numbers (as happened in your case; you started with five 1s, and are now left with 3,4), then you are still behind.
Quote: daveylbraTo everyone:
From wikipedia, labouchere system - "Theoretically, because the player is cancelling out two numbers on the list for every win and adding only one number for every loss, the player needs to have his proposition come at least 33.34% to eventually complete the list. For example, if the list starts with seven numbers and the player wins five times and loses three (62.5% winning percentage) the list is completed and the player wins the desired amount, if the list starts with seven numbers and the player wins 43,600 times and loses 87,193 times (33.34% winning percentage) the list completes and the player wins."
Similarly, I understood Labouchere to require one third plus one bets, out of the total bets, to win.
BUT, suppose I start with 1,1,1,1,1. I lose the first 9 bets.
I have 1,1,1,1,1,2,3,4,5,6,7,8,9, 10. I have lost 54 units. (2+3+4+5+6+7+8+9+10)
I then win 6 bets. I have 3,4. I have won 52 units in those 6 bets. (11+10+9+8+7+7)
I have won 6 out of 15 bets. This is a third plus one. But I am still clearly down!
Can someone please explain?? Is there a certain number of total bets required before the one third plus one law 'kicks in?'
I'm not very familiar with the labouchere system, I've heard of it but never really crunched the numbers. So, I played around with it a little bit, and it is truly a bogus system.
Let's say after losing 9 in a row, you don't immediately win 9 in a row back, but instead you win 1 lose 1 for a while. You would end up clearing the line, but you wouldn't make a profit.
1-1-1-1-1-2-3-4-5-6-7-8-9-10-11 -54 units.
Win 1 +12 units
1-1-1-1-2-3-4-5-6-7-8-9-10 -42 units
Lose 1 -11 units
1-1-1-1-2-3-4-5-6-7-8-9-10-11 -53 units.
Win 1 +12 units
1-1-1-2-3-4-5-6-7-8-9-10 -41 units.
Lose 1 -11 units
1-1-1-2-3-4-5-6-7-8-9-10-11 -52 units.
You've cleared 2 of the 1's from your line, but only made +2 units (from -54 units to -52). If you continue to win 1 lose 1 you would clear the line, but you would not make a profit.
In other words, simply clearing the line DOES NOT guarantee you will make a profit, as they say it will. That is completely bogus as well.
Just really a bad system all together.
"While he made great strides in mathematics and physics, d'Alembert is also famously known for incorrectly arguing in Croix ou Pile that the probability of a coin landing heads increased for every time that it came up tails. In gambling, the strategy of decreasing one's bet the more one wins and increasing one's bet the more one loses is therefore called the D'Alembert system, a type of martingale."
https://en.wikipedia.org/wiki/Jean_le_Rond_d'Alembert
( Number of losses + Numbers originally on list ) / 2 <= Number of wins
results in the list being completed.
My idea is that we could have a long list of 1111...s, to go for a 'worst case scenario' number of wins, but we stop whenever
we are up. But I cannot work out the formula for how many wins to losses for x numbers on list would be required, or even if
it is calculatable.
Can anyone help? Wizard, perhaps??
