/* 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 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. Outside 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?
If you decide to change the source code here is another experiment you can try:
Modify the program to run 10,000,000 games instead of 10,000. Starting with 3 beans each time with a target of 5000. You will win 5000 beans
273 times, for winnings of 1,365,000 beans in total. And you would lose 3 beans 443323 times for a loss of 1,329,969 beans. So you ended up 35,031 beans ahead.Try changing the seed and you will still be ahead in the long run.
Even if you play 100 million games you will still be ahead.
*/. #include . #include . . . 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