Martingale Paradox Claim

[CoreyWhite]

So it has been a year since I proposed my own million dollar challenge, and was banned because I offered to buy up to $5,000 lottery tickets for potential psychics.

And since that time I have been growing as a psychic, but although I haven't struck it rich at the lottery, I have been learning and working as a free tarot card reader and psychic entertainer. In my spare time I've even been solving a few crimes. And doing volunteer work at churches and nursing homes.

And now I've finally come up with a new claim that is scientifically verifiable, because I realize that ordinary psychic abilities aren't enough to impress James Randi (or even winning the lottery).

I've uploaded this experiment to Wikipedia now, and we have already covered it on this forum in the Fun in the Casino thread. I noticed that everyone on the forum was skeptical of my betting strategy, but I've created computer models to prove it is empirically verifiable. So if James Randi wants to challenge this betting strategy I'm willing to play this game with him.

He can bring one million dollars in pennies, and I can bring $5,000 in pennies. We play according to the rules of this game and winner takes all! 8)

Well, I doubt he will want to play, but getting a post like this on Wikipedia is like winning the million dollar challenge to me, and just as good. Because almost no one in the world has understood the math behind this idea, and people who believe in this theory are under sharp criticism. So it may be one small step for me, but it is one giant leap for the education of mankind.

The Martingale Paradox

It is possible to use martingale probability theory to beat some games of chance. In a fair game of coin toss, where the odds reach an equilibrium of 50/50 chain reactions do occur. This can be explained using martingale probability theory, but in simpler terms it only shows an example of how order emerges out of chaos.

Example: One player has 3 pennies, and another player has only 1 penny. A fair coin is tossed every round to determine if a penny is won or lost for either player. The odds are 3/4 that player A (Who begins with 3 pennies) will win the game. This is entirely different than the martingale betting strategy, because only 1 penny is bet for each round of the game.

Because there are 3 ways player A may win, and only one way player B can win, player A has a concrete advantage. Player B, only wins in the event that the coin is tossed in his favor 3 times in a row, while player A can win on the first throw. Or he can win after losing the first coin toss, or he can win after losing the second coin toss. So the odds are 75% that he (or she) will win in this game.

Upon further analysis it is possible to calculate the average number of coin flips before player A is likely to win. The equation k(n-k) works for perfectly fair games according to martingale probability theory to solve this problem. In this case 3(4-3) solves the problem, so on average it takes 3 coin flips for player A to win.

To show that chain reactions occur you only have to move from the probability of winning the first game, and multiply it by the probabilities of winning the following games. For example, if 3 pennies are used to play this game in an attempt to win one penny, the odds are 3/4. And once that penny is collected there is now a 4/5th chance of winning another penny.

So statistics tells us that there is a (3/4) * (4/5) * (5/6) * (7/8) * (8/9) * (9/10) = 30% chance of the 3 pennies growing into a pile of 10.

But in repeatable tests you will find that on average there is not a net win or loss in this game. If there is a 75% chance of winning 1 penny, and a 25% chance of losing 3. The two odds cancel each other out, to create an equilibrium in 50/50 games.

And at the same time we can see that despite the fact that the initial value of coins reaches an equilibrium when the pattern is extended to any length, we can show a concrete advantage to begining with 3 pennies, instead of begining with one.

In the last example player A had a 30% chance of winning 7 pennies, and totalling 10 in all. If we started with only one penny then player A would just have to total 8 pennies in order to earn 7. So lets look at the math:

(1/2) * (2/3) * (3/4) * (4/5) * (5/6) * (6/7) * (7/8) = 12.5%

So we can cleary see that even though winning 7 pennies has the same expected value as losing 1 penny. Outside of repeatable tests the odds of earning 7 pennies is clearly higher if you begin with 3.

This experiment is empirically verifiable, and was first published on usenet by a member of the sci.physics and sci.math groups. The theory was titled "Corey's Punctuated Theory Of Chain Reactions Within An Equilibrium Of Chaos". Because it is mathematics it shouldn't be considered independent research, and can be proven by referencing the books in the martingale probability theory wikipedia page.

Retrieved from Wikipedia

[Cuddles]

This has already been shown to be wrong in the thread you mentioned. I should also point out that anyone can post anything they like on wikipedia so I'm not sure why you would think of that as some lind of achievment. When you get it in the Britannica I'll be impressed.

[Soapy Sam]

I do not understand your purpose here.
If you do have a mathematically verifiable gambling strategy it is, by definition, not paranormal and of no interest at all to JREF.

If you do not, then it is of no interest to JREF.

So good luck at the casino.

[Overman]

Heads I win, Tails you lose.

[petre]

I got as far as the first sentence. The reason for his banning was easily determined in under 60 seconds of searching. Further, I found none of the information described on the Wikipedia site.

It seems clear this poster believes he can alter reality by simply stating what he believes to be true. I find this is not the case.

[IXP]

Mr White,

Do you really not understand why your betting strategy cannot win any money in the long run?

What you state about 2 people, one starting with 1 coin and one starting with 3 coins are not equally likely to get to 4 coins. This is trivially obvious. However, this does not change the odds of winning or losing money over many trials.

Please try simulating the following:

For each trial start with 3 pennies. Play a 50/50 game until either one of the following occurs: you have 4 pennies, or you have lost all 3. In the first case, count a win of 1, in the second case a loss of 3.

Add up the wins and losses over a large number of trials. You will find that your average win will do a random walk around 0 instead of continually growing like you seem to think.

IXP

[TobiasTheViking]

Originally Posted by CoreyWhite

So it has been a year since I proposed my own million dollar challenge, and was banned because I offered to buy up to $5,000 lottery tickets for potential psychics.

Ok, the rest of your message is irrelevant now that you have been shown to being a liar.

For those who don't have the information here it is.

Corey started his career here by posting this message.

Originally Posted by CoreyWhite

I Corey White am offering to buy up to $5,000 in Ohio Lottery Tickets. Any psychic on the JREF Forum who wishes to accept CoreyWhite's Million Dollar Challenge can play up to 100 lottery numbers, from any of the Ohio Lottery Games. If your numbers are played after I have informed you that I bought the ticket, I will write you a check for the prize money!!!

SWEET DOOD!

It was, however, quickly uncovered that Corey couldn't be trusted, based, amongst others, on the following message

Quote:

Acording to my source, James Randi has stolen the million dollar
prize. I don't think James Randi qualifies to take his own challenge, but
after he takes the prize money to Vegas he may prove he has psychic powers
after all.

In other news, I am starting a new $1,000,000 prize.
It will be a website
where psychics everywhere can register to play, and I will put up $5,000 to
buy lottery tickets with. The top lotto numbers the psychics picks will be
played, and we will see if we can win $1,000,000 for the site.
Happy Thanks-Giving

http://groups.google.com/group/sci.s...ca01986270f03a

This resulted in him being banned.

Shortly thereafter he was allowed to use his account for 7 days so he could either appologize, or come up with evidence.. that resulted in the following post.

Originally Posted by CoreyWhite

>I have discussed the changes you have made against Mr. Randi with the
>JREF staff, and we have decided to allow you to use your forum account
>for 7 days effective immediately. During that time, we ask that you
>provide any evidence you might have for the claims you have made
>regarding Mr. Randi's illegal activity.
>
>
>
>If you fail to do so, your account will be permanently banned and the
>James Randi Educational Foundation will pursue legal action against
>you.
>
>
>
>Jeff Wagg
>
>JREF Webmaster


I am formally appologizing to James Randi, and hope he accepts the fact that words are not against the law even if I don't have proof of this claim. There is no law against a joke, even if there is hurt feelings.

I will remove myself from this forum, but would rather not meet with your lawyer in a court of law, evidence aside. Thanks for this time to appologize, it is a nice offer.

As apologies goes, that one is not only bad, it is gut wrenchingly awfull.

Any further corespondence with this ... individual ... seems irrelevant with these facts at hand. Especially because he have been told this earlier today, and still continue with the same lies.

His account was NOT closed after the 7 day period. But he thought it was, which resulted in him coming to the channel on skepticsrock and complaining about him not being alowed to post on the forum. Because Randi(and the forum at large) was afraid of him. In the channel it was found out who he was, and what he had done. He was told his account was not closed, and he left.

He slandered back then, he slandered in the channel earlier today, and he slandered in the opening message here.

He is a compulsive liar and it is no good to spend any time on him.

[Bradk3]

Meow.

[T'ai Chi]

My quick impressions:

Expected winnings for each player:

expected value for player 1=
P(win)*$win - P(lose)*$lose =
3/4*1 penny - 1/4*3 pennies = 0.

expected value for player 2=
P(win)*$win - P(lose)*$lose =
1/4*3 pennies - 3/4*1 penny = 0.

[T'ai Chi]

Here's a quick TI calculator program I wrote for it:

penny()
#initialize winning counts
0->winp1
0->winp2
#initialize banks
0->bankp1
0->bankp2
#initialize first time played
1->t

Lbl start
#initialize number of pennies each player has at the start
3->player1
1->player2

Lbl a
ClrIO
Disp player1, player2
Disp approx(winp1/t), approx(winp2/t)
Disp bankp1, bankp2
Disp t
#if someone wins
If player1=0: Goto b
If player2=0: Goto b
#the flip, 1 is heads
rand(2)-1->p1
rand(2)-1->p2
If p1=1: player2-1->player2
If p1=1: player1+1->player1
If p2=1: player1-1->player1
If p2=1: player2+1->player2
Goto a

Lbl b
If player2=0: winp1+1->winp1
If player2=0: bankp1+1->bankp1
If player2=0: bankp2-1->bankp2
If player1=0: winp2+1->winp2
If player1=0: bankp2+3->bankp2
If player1=0: bankp1-3->bankp1
t+1->t
Goto start

EndPrgm

[CoreyWhite]

What are your conclusions after running your program? I wrote a program in C and concluded that even though the expected value of the two games were the same, one probability was higher. So you had an advantage playing the first game. Expected value isn't the only kind of value.

[Jekyll]

Originally Posted by Soapy Sam

I do not understand your purpose here.
If you do have a mathematically verifiable gambling strategy it is, by definition, not paranormal and of no interest at all to JREF.

If you do not, then it is of no interest to JREF.

So good luck at the casino.

Mind you if he can win with that stratergy it would be ing paranormal.

[CoreyWhite]

I can also explain the laws of nature with these prinicples. If we look at the equation for gravity on earth, which accelerates at 9.8 m/s we can derive an acceptable answer from the earlier equations. The gravity equation I am using is sqrt(2*n/9.8).

In this example we are dropping a ball from 4.9 meters, and you can see it takes one second to land.

t = sqrt( ( 2(4.9 m) ) / ( 9.8 m/s^2 ) ) = 1 s

So here is my gravity theory. We are using the quadratic formula to
solve: 2*n/9.8 = k(n-k) , for k. (The formula k(n-k) finds the average number of coin flips).

k=(1/14) (7n +- sqrt(49 n^2 - 40 n)).


So now an example...


We are dropping a ball from 10 meters above the ground. So we plug 10
meters into n to solve for k.


k=(1/14) (7n +- sqrt(49 n^2 - 40 n))
k=9.791574237


My question to calculate the average number of coin flips in my game is
k(n-k), so we plug in k & n:


k*(10-k) = 2.040816327 = average number of coin flips


Now we take the square root of the average number of flips to get the
actual time it takes to land:


sqrt(avg flips) = 1.428571429 = number of seconds to land.


Now finally to factor in a problem with my equation we say that if k is
9.791574327, that means our large gravity pile is that many pennies.
And our small gravity pile is exactly 0.208425673 pennies!

[GzuzKryzt]

How does this relate to the JREF Challenge, Mr. White?

[T'ai Chi]

Originally Posted by CoreyWhite

What are your conclusions after running your program? I wrote a program in C and concluded that even though the expected value of the two games were the same, one probability was higher.

I believe one can conclude that without even running the program.

Quote:

So you had an advantage playing the first game. Expected value isn't the only kind of value.

I believe if you play the game once, then stop forever, you'll have an advantage being player 1. But mostly games are talked about in terms of repeatedly playing them, and keeping track of results.

[T'ai Chi]

Originally Posted by GzuzKryzt

How does this relate to the JREF Challenge, Mr. White?

It is in the JREF challenge forum- that's how it relates. Duh!

[T'ai Chi]

Corey, are you saying

player 1's earnings are greater than player 2's earnings

for a large number of trials of the game?

My program confirms what IXP said

Quote:

Add up the wins and losses over a large number of trials. You will find that your average win will do a random walk around 0 instead of continually growing like you seem to think.

and I will post the full program and a pic of player 1's earnings for 1000 trials. It fluctuates above and below the earnings = 0 line.

[CoreyWhite]

The theory should hold for a small number of trials, and maybe a finite number of trials. Why not run a test and see?

[nathan]

Originally Posted by CoreyWhite

I can also explain the laws of nature with these prinicples. If we look at the equation for gravity on earth, which accelerates at 9.8 m/s we can derive an acceptable answer from the earlier equations. The gravity equation I am using is sqrt(2*n/9.8). ...

WARNING: Internet may contain nuts

[geni]

Originally Posted by Cuddles

This has already been shown to be wrong in the thread you mentioned. I should also point out that anyone can post anything they like on wikipedia

Not so as this individual found out (still not sure if I regret telling him to try publishing in Social Text).

[IXP]

Originally Posted by CoreyWhite

What are your conclusions after running your program? I wrote a program in C and concluded that even though the expected value of the two games were the same, one probability was higher. So you had an advantage playing the first game. Expected value isn't the only kind of value.

Corey,

First player starts 3 coins and tries to win 1
Second player starts with 1 coin and tries to win 1

Yes, the first player has higher probablity of succeeding, but when he fails, he loses 3. The second player has a lower probability of succeeding, but when he fails, he lose only 1. That is the part you are ignoring.

IXP

[Calcas]

Originally Posted by nathan

WARNING: Internet may contain nuts

Best line of the day...LOL

[GzuzKryzt]

CoreyWhite, do you intend to apply for the JREF Challenge with what you claim in this thread?

If not, how does it relate to the JREF Challenge, i.e. why did you post here?

[T'ai Chi]

I see what I believe Corey is seeing, but I'm not sure if I'm just not running enough trials, or if my program is doing exactly what I want it to do for that matter.

I ran 2397 trials (that's just when I stopped my program), and graphed player 1's winnings (see attached pic), and while it dipped below the winnings = 0 line a few times, it never crosses it again after 673 trials.

Now maybe for some larger number of trials it comes back and maybe dips down to average out?

Here's the program:

penny()
#initialize player 1 earning vector
{0}->ep1
#and trial vector
{1}->x
#initialize winning counts
0->winp1
0->winp2
#initialize banks
0->bankp1
0->bankp2
#initialize first time played
1->t
#initialize number of coins each player has at start
3->coinsp1
1->coinsp2
Lbl start
#initialize number of pennies each player has at the start
coinsp1->player1
coinsp2->player2
Lbl a
ClrIO
Disp player1, player2
Disp approx(winp1/t), approx(winp2/t)
Disp bankp1, bankp2
Disp t
#if someone wins
If player1=0: Goto b
If player2=0: Goto b
#the flip, 1 is heads
rand(2)-1->p1
rand(2)-1->p2
If p1=1: player2-1->player2
If p1=1: player1+1->player1
If p2=1: player1-1->player1
If p2=1: player2+1->player2
Goto a
Lbl b
If player2=0: winp1+1->winp1
If player2=0: bankp1+coinsp2->bankp1
If player2=0: bankp2-coinsp2->bankp2
If player1=0: winp2+1->winp2
If player1=0: bankp2+coinsp1->bankp2
If player1=0: bankp1-coinsp1->bankp1
t+1->t
bankp1->ep1[t]
t->x[t]
Goto start
EndPrgm

and the pic was obtained from graphing the points (x[t],ep1[t]), for t=1 to t=number of trials

[digithead]

Originally Posted by CoreyWhite

Example: One player has 3 pennies, and another player has only 1 penny. A fair coin is tossed every round to determine if a penny is won or lost for either player. The odds are 3/4 that player A (Who begins with 3 pennies) will win the game. This is entirely different than the martingale betting strategy, because only 1 penny is bet for each round of the game.

Because there are 3 ways player A may win, and only one way player B can win, player A has a concrete advantage. Player B, only wins in the event that the coin is tossed in his favor 3 times in a row, while player A can win on the first throw. Or he can win after losing the first coin toss, or he can win after losing the second coin toss. So the odds are 75% that he (or she) will win in this game.

I don't think you're conceptualizing the problem correctly and your initial probabilities are wrong given the information you've provided...

You have two players, one has three chances to win (i.e., 1 flip in his favor out of three) while the other needs three favorable flips in a row. I'm assuming independent trials hence player A's probability is p(x1=1 or x2=1 or x3=1)=p(x1=1)+p(x1=0 and x1=1)+p(x1=0 and x2=0 and x3=1)=.5+.5^2+.5^3=.875 while player B's probability is p(x1=1 and x2=1 and x3=1)=.5^3=.125. Note that player B winning intersects player A winning as player A wins on player B's first toss.

I'm assuming that both player A and player B can win at the same time. So from the gambler's ruin theorem we can calculate the probability that a player will go broke. The theorem gives us:

p(player A will go broke)=(1-r^(b))/(1-r^(a+b)) where r=p/(1-p), p=probability of a favorable outcome, a=player A's beginning fortune, and b=available winnings.

From this, we can easily see that any available winnings significantly greater than player A's assets make it nearly a certainty that player A will go broke...

Plugging in r=.875/.125=7, a=3 cents, b=100 dollars then p(player A will go broke)=94%

So if a pot is greater than 1000 dollars, it's nearly a certainty in this case that player A will go broke. One can also see that any pot over 11 cents and player b will go broke...

Some gamblers realized that it was inevitable for a player to go broke so this allows one to figure out the probability of winning a certain amount of money to figure out optimal stopping times. Hence, gambler's ruin is part of martingale theory. Markov chains are a related concept...

Sorry, but Vega$ is built on this principle, so unless you can figure out how to break the bank, it's inevitable that you will lose in the long run. Unless you're psychic, then you wouldn't need probability theory...

[CoreyWhite]

If the computer model is accurate then I feel we should apply for the JREF challenge, and take the money before anyone begins to accept this as something in the common knowledge. Because right now it is paranormal.

8)

[nathan]

Originally Posted by CoreyWhite

If the computer model is accurate then I feel we should apply for the JREF challenge, and take the money before anyone begins to accept this as something in the common knowledge. Because right now it is paranormal

Ok then, please apply

[HappyCat]

Originally Posted by T'ai Chi

I see what I believe Corey is seeing, but I'm not sure if I'm just not running enough trials, or if my program is doing exactly what I want it to do for that matter.

I ran 2397 trials (that's just when I stopped my program), and graphed player 1's winnings (see attached pic), and while it dipped below the winnings = 0 line a few times, it never crosses it again after 673 trials.

Now maybe for some larger number of trials it comes back and maybe dips down to average out?

I decided to duplicate this experiment. Here is the results that I got.
http://img15.imgspot.com/?u=/u/06/25...1158520618.jpg (sorry for the labeling on the X axis, I couldn't figure out how to change that in Open Office. There are 10000 trials there, though)
After 10000 plays of the game, person A is ahead 20 pennies. Also, I don't understand part of your program.

rand(2)-1->p1
rand(2)-1->p2
If p1=1: player2-1->player2
If p1=1: player1+1->player1
If p2=1: player1-1->player1
If p2=1: player2+1->player2

In this, you are selecting 2 random numbers, and modifying player1 and player2 based on the outcome of those numbers. It seems selecting 2 numbers is unnecesary, because if p1 = p2, then the values remain unchanged. While this does mathematically come out to be correct (since it only adds a bunch of trials where money doesn't change hands), I think it would make more sense to say

rand(2)-1->coinflip
If coinflip = 1: player2-1->player2 else player2+1->player2
If coinflip = 1: player1+1->player1 else player1-1->player1

[Ririon]

Originally Posted by HappyCat

I decided to duplicate this experiment. Here is the results that I got.

...

I think it would make more sense to say

...

...and you should also have stopped about half way through...

[CoreyWhite]

I also noticed something different about your program than my origonal experiment. It seems you are only testing 3 vs 1, and 1 vs 1. And you program isn't designed to run a chain reaction. You could do a test to see how far you can take these chain reactions on average. After you win 1 penny you just add it to the pile of 3, and repeat the experiment with 4 pennies. And do the same when you start with 1.

[TobiasTheViking]

Originally Posted by CoreyWhite

I also noticed something different about your program than my origonal experiment. It seems you are only testing 3 vs 1, and 1 vs 1. And you program isn't designed to run a chain reaction. You could do a test to see how far you can take these chain reactions on average. After you win 1 penny you just add it to the pile of 3, and repeat the experiment with 4 pennies. And do the same when you start with 1.

Just wondering, are you gonna take back your statement that Randi have stolen the 1 million dollars?

And are you going to properly apologize for making that blatant lie?

[digithead]

TC & Happycat,

Correct me if I'm wrong here, but once a player's bankroll goes to 0, isn't the game is over? Any simulation should stop at that point...

Also, it's not clear from the original post if player A and player B are competing against each other or against some imaginary house. If they aren't competing, then by gambler's ruin, the house just needs a big enough bankroll to bankrupt each of them. If they are competing against each other, then by the same theorem, player A will beat player B everytime because he has both a higher bankroll and a higher probability than player B...

And trust me, there's nothing paranormal about this. It's just a mathematical exercise...

-digithead

[HappyCat]

Originally Posted by CoreyWhite

I also noticed something different about your program than my origonal experiment. It seems you are only testing 3 vs 1, and 1 vs 1. And you program isn't designed to run a chain reaction. You could do a test to see how far you can take these chain reactions on average. After you win 1 penny you just add it to the pile of 3, and repeat the experiment with 4 pennies. And do the same when you start with 1.

I would appreciate you showing how you arrived at person A having a 4/5ths chance of winning a round if person A starts with 4 pennies and person B starts with 1. I think that your reasoning is flawed when you say 3 pennies in person A's bank, 4 pennies over all, therefore 3/4 chance of winning. I agree with a 3/4ths probability of a win in this instance, but I think you used a flawed method to arrive at the correct answer. The way I see it, if A has 3 and B has 1, then on the first flip, A has a 50% chance of eliminating B. if A loses the first flip, then both players have the same number of pennies, and therefore the same chances of winning at that point, which is 50%. So 50% of the time player A wins right away, and of the cases where he does not win right away, he wins half of those too. Which would yield 75% chance of winning over all. I don't think the same argument can be made for A having 4 and B having 1, however, so if you don't mind, please show your work, so I can see how you arrived at your conclusion.

Originally Posted by digithead

TC & Happycat,
Correct me if I'm wrong here, but once a player's bankroll goes to 0, isn't the game is over? Any simulation should stop at that point...

I was assuming that a running debt would be calculated over time. Otherwise, neither player could get larger than 4 pennies,and that isn't a very good experiment (since you are right, A's larger bank would mean that more often than not, A would end with 4 pennies and B would end with 0). The point of this experiment is to show that there is nothing paranormal going on here. The expected gain of either player is 0, which differs from what CoreyWhite is claiming, which is essentially that the probability of each successive coin flip is influenced by the previous flip. I would say that gaining only 20 cents in 10000 flips would be pretty unusual if A had some kind of significant advantage.

[GzuzKryzt]

Originally Posted by CoreyWhite

If the computer model is accurate then I feel we should apply for the JREF challenge, and take the money before anyone begins to accept this as something in the common knowledge. Because right now it is paranormal.

8)

CoreyWhite, please explain to the forum members what "is paranormal" about this.

Also, I'd like to hear your reply to TobiasTheCommie's questions in his last post.

[CoreyWhite]

Originally Posted by GzuzKryzt

CoreyWhite, please explain to the forum members what "is paranormal" about this.

Also, I'd like to hear your reply to TobiasTheCommie's questions in his last post.

If you can't see what's paranormal about this challenge then there is something WRONG WITH YOU! 8)

And as for Tobias, James Randi sent me a letter saying he ran off with the million and some hot mamma. That's where the joke came from!

[CoreyWhite]

Here is the e-mail:

I've just stolen the million dollars and run off to Latvia with Sophia
Loren, so I can't answer your e-mail...

James Randi.

[digithead]

Originally Posted by HappyCat

I would appreciate you showing how you arrived at person A having a 4/5ths chance of winning a round if person A starts with 4 pennies and person B starts with 1. I think that your reasoning is flawed when you say 3 pennies in person A's bank, 4 pennies over all, therefore 3/4 chance of winning. I agree with a 3/4ths probability of a win in this instance, but I think you used a flawed method to arrive at the correct answer. The way I see it, if A has 3 and B has 1, then on the first flip, A has a 50% chance of eliminating B. if A loses the first flip, then both players have the same number of pennies, and therefore the same chances of winning at that point, which is 50%. So 50% of the time player A wins right away, and of the cases where he does not win right away, he wins half of those too. Which would yield 75% chance of winning over all. I don't think the same argument can be made for A having 4 and B having 1, however, so if you don't mind, please show your work, so I can see how you arrived at your conclusion.


I was assuming that a running debt would be calculated over time. Otherwise, neither player could get larger than 4 pennies,and that isn't a very good experiment (since you are right, A's larger bank would mean that more often than not, A would end with 4 pennies and B would end with 0). The point of this experiment is to show that there is nothing paranormal going on here. The expected gain of either player is 0, which differs from what CoreyWhite is claiming, which is essentially that the probability of each successive coin flip is influenced by the previous flip. I would say that gaining only 20 cents in 10000 flips would be pretty unusual if A had some kind of significant advantage.

I think we're all tripped up by the incredibly poor description that CW gave in his original post describing his game so we're all making assumptions and imputing information at this point about what we think he meant...

Also, I agree with you that player A has a 50% chance of winning on his first toss. However to calculate his true probability of winning in the first round you need to add in the disjoint events of (toss1=0 and toss2=1) and (toss1=0 and toss2=0 and toss3=1) because there are three different ways player A can win which brings player A's probability of winning to 87.5%. The only way player a can win is if (toss1=1 and toss2=1 and toss3=1)...

And you're quite correct that this not paranormal. It's someone (CW) with an apparently poor grasp of mathematics and logic fooling themselves into thinking that they're clever...

And given his last post, I predict he'll soon be suspended then banned. Anybody want to lay odds on that?

[T'ai Chi]

Originally Posted by HappyCat

Also, I don't understand part of your program.
...
In this, you are selecting 2 random numbers, and modifying player1 and player2 based on the outcome of those numbers. It seems selecting 2 numbers is unnecesary, because if p1 = p2, then the values remain unchanged. While this does mathematically come out to be correct (since it only adds a bunch of trials where money doesn't change hands), I think it would make more sense to say

Yup, your way certainly makes more sense to do, thanks.

[TobiasTheViking]

Originally Posted by CoreyWhite

Here is the e-mail:

I've just stolen the million dollars and run off to Latvia with Sophia
Loren, so I can't answer your e-mail...

James Randi.

I'm going to inquiere as to whether or not that is a lie. Even if it isn't it is clearly a joke.

You never said if your withdraw your statement that he stole the million dollars.

You never made a proper appology, do you intend to?

[TobiasTheViking]

Hm, considering that i doubt you have the investigation skills to come up with that message i think i will asume it is genuine. Certain aspects of it look to be genuine Randi.

It is still clearly a joke, and not a reason for the attack you made.

Also, when you were asked for evidence why didn't you supply the email?