/*
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: 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. 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 beginning with 3
pennies, instead of beginning with one.
In the last example player A had a 30% chance of winning 7 pennies, and
totaling 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. of repeatable tests the odds
of earning 7 pennies is clearly higher if you begin with 3.
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!
Now for the source code. You can actually prove everything I have
written by running a few simple test cases. In the program when you
set the initial beans to 5, and set 1, 2, 3, 4, or 5 beans as your
goal, the output should look like this:
5:1 83.5%
5:2 71.6%
5:3 62.6%
5:4 56.1%
5:5 50%
But if you only play with 5 beans every time and only go after 1 bean
with those 5 each game, then your output will look like this;
5:1 = 83.5%
5:1 X 5:1 = 69%
5:1 X 5:1 X 5:1 = 58%
5:1 X 5:1 X 5:1 X 5:1 = 48%
5:1 X 5:1 X 5:1 X 5:1 X 5:1 = 40%
So there all all the proof you need. Which experiment would you rather
play?
*/
#include <stdio.h>
#include <stdlib.h>
main ()
{
double r;
long int M;
double x;
int y;
int z;
int count;
int seed = 10000;
srand (seed);
M = 2;
int score = 0;
//Score keeps track of the number of beans won every game
int games = 0;
// games keeps track of the number of games we have played before
//losing all of the beans, which is equal to score.
int beans1 = 0;
// Initial value set to zero and defined within the loop
int wins = 0;
int lost = 0;
int quit = 0;
int init = 0;
printf ("Initial Beans: ");
scanf ("%d", &init);
printf ("Stop after winning X number of beans: ");
scanf ("%d", &quit);
for (int cnt = 0; cnt < 10000; cnt++)
{
// We play 10,000 rounds
int count = 0;
beans1 = init + score;
// Beans gets defined here, as starting with 3 beans
// and having a 0 bonus score (It changes as you
// win more beans per round)
int beans2 = 1;
// The program attempts to win just one
// bean for every game.
while (beans1 != 0 && beans2 != 0)
// The battle begins
{
r = ((double) rand () / ((double) (RAND_MAX) + (double) (1)));
x = (r * M);
y = (int) x;
z = y + 1;
// A coin is flipped and is either 1 or 2 in value
if (z == 1)
{
// Heads wins.
beans1++;
// Beans1 gains one bean from Beans2
beans2--;
}
if (z == 2)
{
// Tails loses
beans1--;
// Beans2 gains one bean from Beans1
beans2++;
}
count++;
// We keep track of the number of rounds in the battle
}
if (beans1 score + init)
{
// If beans1 is greater than the initial value
// of beans plus the total number of beans
// that have been won so far in this game, then
// the score goes up, and we go on to the next
// game. We check this at the end of every game.
score++;
games++;
}
if (beans1 <= 0)
{
//If beans1 has lost the game and doesn't
//have anymore beans then we know the
//game is over, so we reset score, and reset
//games.
printf ("Lost at: %d beans , %d games.\n", score + init, games);
// And we print out the total number of
// games played on this trial and show the
// total score plus the initial value of beans.
lost++;
score = 0;
games = 0;
}
if (score >= quit)
{
wins++;.
printf ("Won at: %d beans , %d games.\n", score + init, games);
beans1 == 0;
score = 0;
games = 0;
}
}
printf ("Total Won: %d/%d\n", wins, wins + lost);
}

You gotta love a title like that. -William Hughes

Richard Heathfield wrote:
Ben Pfaff said:
>
>"William Hughes" <wpihughes@hotmail.comwrites:
>>
You gotta love a title like that. -William Hughes
>The best part is the suggestion that a "final" theory needs a
>version 2.0. Sort of like the naming of the "Final Fantasy
>Umpteen" series.
>
I remember reading about a program which rejoiced in the title "The Last
Program". I don't recall enough details to track down any Web references,
but I do recall the article saying that sales of Version 2 were not
encouraging. :-)

A company I worked for { long long timeAgo ; } bought a copy. My
recollection is that it was unusably tedious for anything non-trivial
and tedious and pointless for anything trivial

Cheers,

William.

Yep, it was a program generator, written in BASIC, generating BASIC.

IIRC it asked you simple questions like: what do you want to see on the
screen, what do you want to input, what calculation do you want to do
on the numbers, what file to write them to.

The real test of a program generator is of course, was it written in
itself? If it's not general enough for that, caveat emptor.