Computer Simulations.
Just found your C++ code for simulating a labouchere progression and find it very interesting.
I understand all that you say in regards to the house edge in any game being constant, and all
casino games have a negative expectancy.
However, I also undertand that, in roulette, as the number of spins increase, so does the 'equilibrium',
ie the expected ratio of red/black, high/low numbers, etc. approaches 50/50. Is't this 'predictabiity' something we can make
use of if we can extend a progression over enough spins? ie, if,and I know its a big IF, we could extend a cancellation
progression to 300 spins, we could expect to get over 1/3 of the bets a wins, which is all we would need to be in profit.
(What I do not understand about the laws of probability, is that is this equilibrium a mathematical law, or can
we say that, given billions of spins, would we find a sequence of, say, 300 reds?)
Anyway, what I want to do, is adapt the C++ code to simulate a cancellation starting with 33 11111...s.
Our target would be 1, ie quit whenever we are ahead. My idea is that this should take a maximum of 99 spins,
33 being 1/3 of the total, 99.
Repeated trials of the program would then show us the maaximum bet needed, wouldn't it?
Would it be quite simple to change the code to do this? Would be so grateful if you could help me with this!
Best regards,
Dave.
PS I canot see from your code whether you start from 10 111..s or how many. Also I think you assume a double 0 wheel,
whereas a single 0 simulation would be much better. Sorry I cannot do it myself, but I have no experience with Excel.
I suppose that my idea will be proved to fail, But it would be fascinating to see the results.
Yes, they do, as a percentage or ratio. But does not "make up" or "fix" the difference.
Avoiding green, let's say there have been 100 spins, 45 red and 55 black. After 100 more spins, you'd expect 50 more red and 50 more black, for an end result of 95 red and 105 black.
First you had 45 red and 55 black. That's 45% to 55%.
Then you could expect to get 95 red and 105 black which is 47.5% red and 52.5% black.
Say you do another 200 spins. You'd expect 100 reds and 100 blacks. Resulting in 195 red and 205 black. That's 48.75% to 51.25%.
The percentage gets closer to 50%, but is not going to be helpful for what you're doing....because you can only expect 50% of future spins to result in red or black (excluding green).
Quote: daveylbra
However, I also undertand that, in roulette, as the number of spins increase, so does the 'equilibrium',
ie the expected ratio of red/black, high/low numbers, etc. approaches 50/50. Is't this 'predictabiity' something we can make use of if we can extend a progression over enough spins?
Absolutely correct. As the number of spins increases, our losses will get closer and closer to 2.7% of all money wagered.
No.Quote: daveylbraHowever, I also undertand that, in roulette, as the number of spins increase, so does the 'equilibrium',
ie the expected ratio of red/black, high/low numbers, etc. approaches 50/50. Is't this 'predictabiity' something we can make
use of if we can extend a progression over enough spins?
Quote: Richard A. Epstein, "The Theory of Gambling and Statistical Logic", p. 28The law of large numbers has frequently been cited as the guarantor of an eventual head-tail balance. Actually, in colloquial form, the law proclaims that the difference between the number of heads and the number of tails thrown may be expected to increase indefinitely as the number of trials increases, although by decreasing proportions.
Yes, I looked up The Law Of Large Numbers on wiki, and boy is it complicated. Weak Law, Strong Law, Uniform Law etc.
But, simply put, can we not expect at least 33 reds in 100 spins? I don't know the formula to find the probability of less than 33,
but I'm guessing it will be small. If not, then how about 66 in 200 etc etc.
Math Extremist, I'm surprised you say no. You might be the guy that can help me.
" the difference between the number of heads and the number of tails thrown may be expected to increase indefinitely as the number of trials increases, although by decreasing proportions."
Yes, decreasing proportions, surely more equal than 1/3 to 2/3? and a labouchere progression only requires 1 more than 1/3 wins.
So given enough trials, the labouchere should complete. Yes, the bets may become large, but I'm looking for a way to calculate how large.
I guess I'm missing something, and I'm sorry to labour the point.
Anyway, if anyone can adapt the C++ code (maybe the Wizard can interject) that would be great.
Dave.
It doesn't matter whether a progression "should" complete, or whether the probability will be small. Small is not zero. The only thing that matters is whether, in playing a certain way, you're changing the edge of the game in your favor. Ask yourself what a pattern of variable bet sizes could possibly help toward that end.
That said, I am surprised websites such as wizard of odds and others exist. Anyone who gambles for any length of time must
surely realise eventually we 'cannot even dent the house edge.'
I just want to prove it to myself mathematically. Work out that the amount we lose > how often.
Ie we lose 1 in 100 times, but lose more than 100 units, (as an example) for the simulation on the first post here.
You can dent the house edge, just not with betting patterns. Denting the house edge requires a game-specific strategy, not paying attention to patterns of wins and losses regardless of how they're generated. If you're fading me with an even-money bet on heads or tails of a fair coin, there's nothing I can do to beat you. If you're fading me with an even-money bet on whether each next card in a deck will be red or black, I'll beat you. It's not the pattern of wins/losses that determines whether you can beat a game, it's the game itself.Quote: daveylbraThe more sensible side of me knows that each bet stands alone and is just an amount on one or more numbers.
That said, I am surprised websites such as wizard of odds and others exist. Anyone who gambles for any length of time must
surely realise eventually we 'cannot even dent the house edge.'
I just want to prove it to myself mathematically. Work out that the amount we lose > how often.
Ie we lose 1 in 100 times, but lose more than 100 units, (as an example) for the simulation on the first post here.
It's a good exercise to prove to yourself mathematically that betting patterns don't alter the house edge.
RE: Independent TrialsQuote: RSRE: the more spins -> high/low and red/black etc. approach 50/50
Yes, they do, as a percentage or ratio. But does not "make up" or "fix" the difference.
Avoiding green, let's say there have been 100 spins, 45 red and 55 black. After 100 more spins, you'd expect 50 more red and 50 more black, for an end result of 95 red and 105 black.
First you had 45 red and 55 black. That's 45% to 55%.
Then you could expect to get 95 red and 105 black which is 47.5% red and 52.5% black.
Say you do another 200 spins. You'd expect 100 reds and 100 blacks. Resulting in 195 red and 205 black. That's 48.75% to 51.25%.
The percentage gets closer to 50%, but is not going to be helpful for what you're doing....because you can only expect 50% of future spins to result in red or black (excluding green).
The difference in *raw numbers* of reds and blacks does not ever "forcibly converge", it actually gets bigger as time goes on. If you get 60 reds and 40 blacks after 100 spins, and you spin another ten billion spins, the chances of the difference now being less than 20 are infinitesimal.
The PERCENTAGE, however, will tend to approach exact theoretical values as time increases. That's because the percentages -- and so, the variance in the percentages -- are calculated with a fraction, and the variance in the numerator increases at a speed that is the square root of the speed in which the denominator increases.
So yeah, it never actually forces a regression to the long run. It's more like, if we're on a perfectly flat plain (say, we're in Wyoming :P ) and you walk due north and I walk north-by-northwest, and we each take the same number of steps, then I look due north, the more steps we take, the closer you'll get directly to my center of vision even though, east-west wise, we're actually growing farther apart.
Quote: CharlesMousseau...The PERCENTAGE, however, will tend to approach exact theoretical values as time increases. That's because the percentages -- and so, the variance in the percentages -- are calculated with a fraction, and the variance in the numerator increases at a speed that is the square root of the speed in which the denominator increases...
Exactly. Here's an example. Wins-losses.
45-55
450-550
4500-5500
45000-55000
Our percentages are the same but our losses get larger. And if I go to a higher percentage, I still lose more than the original figure of 45-55
490000-510000.
Quote: IbeatyouracesExactly. Here's an example. Wins-losses.
45-55
450-550
4500-5500
45000-55000
Our percentages are the same but our losses get larger. And if I go to a higher percentage, I still lose more than the original figure of 45-55
490000-510000.
That's not really accurate. I'd say 45k to 55k (in a 50-50 proposition) is less likely than 18 yo's in a row, and it doesn't tend towards 50%.
A better example -- if you flip a coin ten times, you're likely to have a difference between heads and tails of either 0 (5-5), 2 (4-6), or 4 (3-7). You can get 2-8, 1-9, and 0-10, but not super likely.
If you flip a coin 10,000 times, you're less likely to get a small difference of heads to tails (ie: very unlikely to get exactly 5000-5000 or 4999-5001 or 4998-5002) because you have a much wider possible range of results.
But percentage or ratio wise, you can get results like 30% to 70% or 40% to 60% when only doing 10 flips. But in 10,000 flips, no way in hell you'd be getting anywhere near 40%-60%.
Quote: CharlesMousseauThe key difference -- and I *THINK* this is right but if I said something laughable, I'm open to correction.
The difference in *raw numbers* of reds and blacks does not ever "forcibly converge", it actually gets bigger as time goes on. If you get 60 reds and 40 blacks after 100 spins, and you spin another ten billion spins, the chances of the difference now being less than 20 are infinitesimal.
This is the laughable part. If the percentage regresses to the mean, then that also means at some point on a 50/50 game, the wins/ losses would regress to 0, the wins and losses being equal, let's say 1 million to 1 million. The percentage regressed to 50%, and the wins and losses regressed to even amounts of wins & losses. This HAS to happen. If you, let's say, start off with more wins then losses, then eventually you will regress back to even wins & losses. Then if you start having more losses then wins, you will eventually regress back to even amount of wins & losses again. So, it goes high and comes back, then goes low and comes back. Just like a wave. So, yes the law of large numbers "forcibly converges" not only the percentage, but the "raw numbers" as well. That's a FACT.
That's a FACT, for a 50/50 game as well as a 48.5% game. For a 48.5% game the mean is not 50%, it is now 48.5%. So, you may be higher then 48.5%, let's say you are at 49%, or in "raw numbers" you have +50 more wins then you should have at this point. It will regress back to 48.5%, eventually, and when it does you will be exactly even on "raw numbers", 0 higher then you should be, and 0 lower then you should be. That's a FACT.
Quote: IbeatyouracesWrong. Just because there is a 50/50 proposition. It does NOT mean it will even out. Again, that's gamblers fallacy.
Wrong, gamblers fallacy applies to 1 bet, and 1 bet only. Not thousands of bets.
Quote: IbeatyouracesWhat I'm getting at, is there is no tendency to catch up. Use my example earlier. If you guess and win 4 and lose 6 out of 10, you have a 40% win rate and am down 2. Now if you continue and win 4900 and lose 5100 out of 10000, your now at 49% which is much better but down a lot more (200). Again, there is no law that says things will even out and in fact it's more likely they'll get farther apart in a 50/50 game. That's what he's saying. This is talked about on page 199 and 200 of The World's Greatest Blackjack Book. It's probably in many others as well.
Yes, there's the Law of Large Numbers, which says things will eventually even out, the percentage will get closer and closer until finally reaching the mean.
It's much less likely, actually it's nearly impossible (exceptions being made for infinite losses in a row), that they will get farther apart in a 50/50 game. Again, law of large numbers.
Blackjack is not a 50/50 game. A game that's less than 50% would, however, continue to get farther and farther away from 50%. This, funny enough, is due to the Law of Large Numbers. The LLN is what ensures the casino of that their House Edge will produce profits for them in the long run. It also ensures that a computer simulator will produce 48.5% on a single 0 roulette game after thousands of trials.
Quote: JyBrd0403Just to finish the thought here. The LLN states that the average of the results (or win percentage) will become closer to the expected value the more trials performed. So, I did a little math experiment, starting at 10,000 trials with a win percentage of 49.1%, or -90 units from 0. I added a .001 percent increase to the win percentage every 10,000 trials. So, a modest move closer to 50% every 10,000 trials. Here's what I got.
10,000 trial intervals.
10k x .491 = 4,910 wins -90 from 0
20k x .492 = 9,840 wins -160 from 0
30k x .493 = 14,790 wins -210 from 0
40k x .494 = 19,760 wins -240 from 0
50k x .495 = 24,750 wins -250 from 0
60k x .496 = 29,760 wins -240 from 0
70k x .497 = 34,790 wins -210 from 0
80k x .498 = 39,840 wins -160 from 0
90k x .499 = 44,910 wins -90 from 0
As you can see, the longer you play the closer the win percentage gets to 50%, and the closer to 50% you get, the closer you get to 0.
If the percentage continues to get closer to the mean, the "raw numbers" will eventually get closer as well, until finally getting to even.
Quote: IbeatyouracesSo you're saying you're "guaranteed" to get to even on a 50/50 game when down by 1000 because you think things "even out"?
No, that's the Law of Large numbers that guarantees it, not me thinking things will "even out".
Quote: IbeatyouracesThe house makes there profit by paying less than true odds. And in most cases when they win, you pay them better than true odds.
Yes, the LLN ensures that paying less than true odds will make them profits. Otherwise, the 48.5% might shoot a 52%, and you wouldn't have any mathematical law to say that's BS.
"For example, a fair coin toss is a Bernoulli trial. When a fair coin is flipped once, the theoretical probability that the outcome will be heads is equal to 1/2. Therefore, according to the law of large numbers, the proportion of heads in a "large" number of coin flips "should be" roughly 1/2. In particular, the proportion of heads after n flips will almost surely converge to 1/2 as n approaches infinity.
Though the proportion of heads (and tails) approaches 1/2, almost surely the absolute difference in the number of heads and tails will become large as the number of flips becomes large. That is, the probability that the absolute difference is a small number, approaches zero as the number of flips becomes large. Also, almost surely the ratio of the absolute difference to the number of flips will approach zero. Intuitively, expected absolute difference grows, but at a slower rate than the number of flips, as the number of flips grows."
As I highlighted, it's MORE LIKELY the difference will get larger. Not get closer to each other. Only the percentage will. Big difference.
Quote: JyBrd0403This is the laughable part. If the percentage regresses to the mean, then that also means at some point on a 50/50 game, the wins/ losses would regress to 0, the wins and losses being equal, let's say 1 million to 1 million. . . . . That's a FACT.
Wrong. WRONG WRONG!!!!
It's a FACT that you are wrong.
Quote: Ibeatyouraceshttps://en.m.wikipedia.org/wiki/Law_of_large_numbers
"For example, a fair coin toss is a Bernoulli trial. When a fair coin is flipped once, the theoretical probability that the outcome will be heads is equal to 1/2. Therefore, according to the law of large numbers, the proportion of heads in a "large" number of coin flips "should be" roughly 1/2. In particular, the proportion of heads after n flips will almost surely converge to 1/2 as n approaches infinity.
Though the proportion of heads (and tails) approaches 1/2, almost surely the absolute difference in the number of heads and tails will become large as the number of flips becomes large. That is, the probability that the absolute difference is a small number, approaches zero as the number of flips becomes large. Also, almost surely the ratio of the absolute difference to the number of flips will approach zero. Intuitively, expected absolute difference grows, but at a slower rate than the number of flips, as the number of flips grows."
As I highlighted, it's MORE LIKELY the difference will get larger. Not get closer to each other. Only the percentage will. Big difference.
This is why systems players CANNOT beat a game where A) they don't have an edge and B) the odds of winning are less than the odds of losing. Hell, they can't even beat a coin flip!
Quote: Ibeatyouraceshttps://en.m.wikipedia.org/wiki/Law_of_large_numbers
"For example, a fair coin toss is a Bernoulli trial. When a fair coin is flipped once, the theoretical probability that the outcome will be heads is equal to 1/2. Therefore, according to the law of large numbers, the proportion of heads in a "large" number of coin flips "should be" roughly 1/2. In particular, the proportion of heads after n flips will almost surely converge to 1/2 as n approaches infinity.
.
I hope you have better then wiki for this. He says the proportion of heads will almost surely converge to 1/2 as n approaches infinity. If the percentage will almost surely converge, so will the "raw numbers". Your boy has some explaining to do.
Quote: OnceDearWrong. WRONG WRONG!!!!
It's a FACT that you are wrong.
It almost surely converges.
Quote: IbeatyouracesQuote: Ibeatyouraceshttps://en.m.wikipedia.org/wiki/Law_of_large_numbers
"For example, a fair coin toss is a Bernoulli trial. When a fair coin is flipped once, the theoretical probability that the outcome will be heads is equal to 1/2. Therefore, according to the law of large numbers, the proportion of heads in a "large" number of coin flips "should be" roughly 1/2. In particular, the proportion of heads after n flips will almost surely converge to 1/2 as n approaches infinity.
Though the proportion of heads (and tails) approaches 1/2, almost surely the absolute difference in the number of heads and tails will become large as the number of flips becomes large. That is, the probability that the absolute difference is a small number, approaches zero as the number of flips becomes large. Also, almost surely the ratio of the absolute difference to the number of flips will approach zero. Intuitively, expected absolute difference grows, but at a slower rate than the number of flips, as the number of flips grows."
As I highlighted, it's MORE LIKELY the difference will get larger. Not get closer to each other. Only the percentage will. Big difference.
This is why systems players CANNOT beat a game where A) they don't have an edge and B) the odds of winning are less than the odds of losing. Hell, they can't even beat a coin flip!
Bad math.
Quote: JyBrd0403I hope you have better then wiki for this. He says the proportion of heads will almost surely converge to 1/2 as n approaches infinity. If the percentage will almost surely converge, so will the "raw numbers". Your boy has some explaining to do.
As in 50% but never gets to it exactly. And "almost surely" isn't the same as "guaranteed".
Quote: JyBrd0403Then if you start having more losses then wins, you will eventually regress back to even amount of wins & losses again. So, it goes high and comes back, then goes low and comes back. Just like a wave. So, yes the law of large numbers "forcibly converges" not only the percentage, but the "raw numbers" as well. That's a FACT.
No this is wrong.
If you flip a theoretical fair coin it has exactly 50% chance of coming up heads and 50% chance of coming up tails. It doesn't matter if you have won the last 3 flips or lost 15 of the last 17. It has no memory it is always 50/50 proposition.
The law of large numbers doesn't "forcibly" do anything. It is just the observation that after a large enough number of coin flips you will end up arbitrarily close to 50% head and 50% tails. This happens not because the absolute number of heads flipped evens out, but because the total number of flips eventually grows to make the current number of differences irrelevant.
As an example imagine we flipped our coin 10 times and ended with 7 head and 3 tails, giving us a rate of 70% heads. If we then proceed to flip the coin 10000 more times we would expect to average 5000 heads and 5000 tails. After this we would end up 5007 heads and 5003 tails giving us a heads rate of 50.01998%. If we then flipped the coin 1,000,000 more times and ended with another 500,090 heads and 499,910 tails we would end up with a rate of 50.0091% even though the absolute difference of heads and tails went up.
Quote: JyBrd0403The numbers of "raw numbers" cannot continue to grow if the percentage continuously gets closer to the mean. It's bad math.
As I pointed out earlier, 4 wins, 6 losses is 40%. 4900 wins, 5100 losses is 49% but is FARTHER from even. See where you are flawed?
Quote: JyBrd0403The numbers of "raw numbers" cannot continue to grow if the percentage continuously gets closer to the mean. It's bad math.
So, in your 'good maths', this cannot happen. . .
| heads | tails | proportion heads | count difference |
|---|---|---|---|
| 1000 | 1001 | 0.499750125 | 1 |
| 1001 | 1003 | 0.499500998 | 2 |
| 1000005 | 1000010 | 0.49999875 | 5 |
Quote: IbeatyouracesAs in 50% but never gets to it exactly. And "almost surely" isn't the same as "guaranteed".
It's only "almost surely" because you can't dismiss Infinite losses in a row. However, 99.999...% chance of converging I don't think anyone will argue with. It's a mathematical certainty.
It's actually like saying if the percentage converges to 50%, the wins and losses will be equal to one another. That's a 100% correct statement. which mathematicians consider 100% to be 99.999...%
Quote: OnceDearSo, in your 'good maths', this cannot happen. . .
heads tails proportion heads count difference 1000 1001 0.499750125 1 1001 1003 0.499500998 2 1000005 1000010 0.49999875 5
Did it converge to 0 during those 2 million trials? You have to know that for this arguement. If it converged to 50% at any time you would be doing, horrificly bad math.
But, it still doesn't matter, because it will continue to get closer after 2 million trials, until eventually reaching 50%. That's the LLN again.
Quote: someoneNo this is wrong.
It is just the observation that after a large enough number of coin flips you will end up arbitrarily close to 50% head and 50% tails.
What I think you're missing is that after that large enough number of flips where you get close to 50% ( which I believe most mathematicians would agree that 10,000 trials will get you close to 50%). You will continue to get closer and closer (according to the LLN) until reaching 50%.
I think you leave that part of the LLN out of your example, it doesn't end until reaching the mean, in this case 50%. That's the only thing that ends it. Otherwise, you will continue to get closer and closer.
Quote: JyBrd0403. . . Otherwise, you will continue to get closer and closer.
try this. extend this table...
| tosses | Maximum possible count difference between heads and tails | Number of possible outcomes | Number of possible outcomes | Average count difference between heads and tails |
|---|---|---|---|---|
| 0 | 0 | 1* | =(0*0)/1 | 0 |
| 1 | 1 | 2 | =(1*2)/2 | 1 |
| 2 | 2 | 4 | =(2*2+0*2)/4 | 1 |
| 3 | 3 | 8 | =(3*2+1*6)/8 | 1.5 |
| 4 | 4 | 16 | =(4*2+2*8+0*6)/16 | 1.5 |
| 5 | 5 | 32 | =(5*2+3*10+1*20)/32 | 1.875 |
| 6 | 6 | 64 | =(2*6+12*4+30*2+20*0)/64 | 1.875 |
So...
The Maximum possible difference in count between heads and tails increases with count of tosses.
and
The average difference in count between heads and tails increases with count of tosses.
Reconcile that with the difference in count approaching zero, which is your assertion.
* if you toss 0 times you get 1 possible outcome where it lands neither headfs nor tales. Live with it!
Quote: OnceDeartry this. extend this table...
tosses Maximum possible count difference between heads and tails Number of possible outcomes Number of possible outcomes Average count difference between heads and tails 0 0 1* =(0*0)/1 0 1 1 2 =(1*2)/2 1 2 2 4 =(2*2+0*2)/4 1 3 3 8 =(3*2+1*6)/8 1.5 4 4 16 =(4*2+2*8+0*6)/16 1.5 5 5 32 =(5*2+3*10+1*20)/32 1.875 6 6 64 =(2*6+12*4+30*2+20*0)/64 1.875
So...
The Maximum possible difference in count between heads and tails increases with count of tosses.
and
The average difference in count between heads and tails increases with count of tosses.
Reconcile that with the difference in count approaching zero, which is your assertion.
* if you toss 0 times you get 1 possible outcome where it lands neither headfs nor tales. Live with it!
horrifically bad math. Has nothing to do with the question at hand. How many times does it converge to the mean? I think this is the infinite losses in a row argument that I already gave you the 0.000...% chance of. And mathematicians don't even think that 0.000...% exists. Funny huh.
Quote: JyBrd0403I think you leave that part of the LLN out of your example, it doesn't end until reaching the mean, in this case 50%. That's the only thing that ends it. Otherwise, you will continue to get closer and closer.
Yes. 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.
"Just to finish the thought here. The LLN states that the average of the results (or win percentage) will become closer to the expected value the more trials performed. So, I did a little math experiment, starting at 10,000 trials with a win percentage of 49.1%, or -90 units from 0. I added a .001 percent increase to the win percentage every 10,000 trials. So, a modest move closer to 50% every 10,000 trials. Here's what I got.
10,000 trial intervals.
10k x .491 = 4,910 wins -90 from 0
20k x .492 = 9,840 wins -160 from 0
30k x .493 = 14,790 wins -210 from 0
40k x .494 = 19,760 wins -240 from 0
50k x .495 = 24,750 wins -250 from 0
60k x .496 = 29,760 wins -240 from 0
70k x .497 = 34,790 wins -210 from 0
80k x .498 = 39,840 wins -160 from 0
90k x .499 = 44,910 wins -90 from 0
As you can see, the longer you play the closer the win percentage gets to 50%, and the closer to 50% you get, the closer you get to 0."
I think that Excel thing these guys talk about can do this for you.
Quote: JyBrd0403But, you didn't actually do the math did you. If you actually do the math, you get this.
"Just to finish the thought here. The LLN states that the average of the results (or win percentage) will become closer to the expected value the more trials performed. So, I did a little math experiment, starting at 10,000 trials with a win percentage of 49.1%, or -90 units from 0. I added a .001 percent increase to the win percentage every 10,000 trials. So, a modest move closer to 50% every 10,000 trials. Here's what I got.
10,000 trial intervals.
10k x .491 = 4,910 wins -90 from 0
20k x .492 = 9,840 wins -160 from 0
30k x .493 = 14,790 wins -210 from 0
40k x .494 = 19,760 wins -240 from 0
50k x .495 = 24,750 wins -250 from 0
60k x .496 = 29,760 wins -240 from 0
70k x .497 = 34,790 wins -210 from 0
80k x .498 = 39,840 wins -160 from 0
90k x .499 = 44,910 wins -90 from 0
As you can see, the longer you play the closer the win percentage gets to 50%, and the closer to 50% you get, the closer you get to 0."
I think that Excel thing these guys talk about can do this for you.
Haven't we had this conversation before?
Anyway, it seems to me that we are now talking about two different, although related, things.
You can have the win percentage get closer to 50% while the difference between wins and losses differs. For example, if you win 4900 of the first 10,000 tosses, and then win 4950 out of every 10,000 after that, the percentage starts at 49 and increases toward 49.5, but the difference increases by 100 after every 10,000 tosses. Nothing "bad" about that math.
On the other hand, we seem to agree that "eventually" a 50/50 bet will get to the point where wins = losses. However, this makes some assumptions, like the person placing the bets is immortal. "Eventually" is a long, long, long, did I mention "long", long time. Speaking of computer simulations, let me run some to see how long of a run I would need to get back to the break-even point.
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."
Just to repeat it, with emphasis this time:
Quote: Richard A. Epstein, "The Theory of Gambling and Statistical Logic", p. 28The law of large numbers has frequently been cited as the guarantor of an eventual head-tail balance. Actually, in colloquial form, the law proclaims that the difference between the number of heads and the number of tails thrown may be expected to increase indefinitely as the number of trials increases, although by decreasing proportions.
The proportion of heads/tales converges toward 50%, but the absolute difference increases indefinitely. That's because the denominator (the number of trials) grows faster than the numerator (the difference). It's not an inconsistency if you understand the distinction between the absolute difference in outcomes and that difference's proportion of the total number of trials.
Quote: MathExtremistHeh, I wake up to see a four-page argument over whether the quote I published on page 1 is correct.
Just to repeat it, with emphasis this time:Quote: Richard A. Epstein, "The Theory of Gambling and Statistical Logic", p. 28The law of large numbers has frequently been cited as the guarantor of an eventual head-tail balance. Actually, in colloquial form, the law proclaims that the difference between the number of heads and the number of tails thrown may be expected to increase indefinitely as the number of trials increases, although by decreasing proportions.
The proportion of heads/tales converges toward 50%, but the absolute difference increases indefinitely. That's because the denominator (the number of trials) grows faster than the numerator (the difference). It's not an inconsistency if you understand the distinction between the absolute difference in outcomes and that difference's proportion of the total number of trials.
Horrifically bad math. The absolute difference will not continue to grow indefinitely (as I showed in the post above). What will happen is that a wave will form. The "raw numbers" will grow, reach an apex, and come back down to the mean. That will always happen. If the win percentage continues to get closer and closer to the mean, then the "raw numbers" will form a wave. Richard A Epstein is dead wrong, and anyone can use an Excel spreadsheet to show that.
I'm blocking you ME, so don't bother responding to me. I have no desire to read any more of the bad math that you have been spewing out for years now.
Um, no. Epstein is perfectly correct, and so is the statistics department at Cal Berkeley:Quote: JyBrd0403Horrifically bad math. ... Richard A Epstein is dead wrong
Quote: UC Berkeley Glossary of Statistical TermsThe Law of Large Numbers says that in repeated, independent trials with the same probability p of success in each trial, the percentage of successes is increasingly likely to be close to the chance of success as the number of trials increases. More precisely, the chance that the percentage of successes differs from the probability p by more than a fixed positive amount, e > 0, converges to zero as the number of trials n goes to infinity, for every number e > 0. Note that in contrast to the difference between the percentage of successes and the probability of success, the difference between the number of successes and the expected number of successes, n*p, tends to grow as n grows.
http://www.stat.berkeley.edu/~stark/SticiGui/Text/gloss.htm
