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December 7th, 2011, 06:55 AM  #1 
Newbie Joined: Apr 2008 Posts: 8 Thanks: 0  Statistic variation problem
Hi, I would have a statistically perfect coin, which would have a 50% chance of landing on heads. If the coin lands on heads, I'd win. If the coin lands on tails, I would lose, so my friend who I'm playing against wins. To make it more interesting, we would be placing bets. Every toss, we would bet an equal amount, that amount is our variable, and I get to chose how much we bet every toss. I've been told when I lose, chances that I win next toss will become slightly greater. The idea is to bet slightly more money when such a chance is greater than 50%, and bet less when the chance is lower than 50%. I've crudely tested this, and I over time, I was making a profit. The only measurable variable is the amount of times that I won/lost in the past. In function of that, we'd need to calculate the chance of winning next toss. And in function of that chance, the next bet should be calculated. I've figured that the latter function would look like one of these two (I'm not sure which); With on the xaxis the chance of winning, and on the yaxis the amount of times the 'base' bet should be multiplied with. I'm leaning towards the exponential graph, I can't explain why. (Just realized I drew my axis wrong, I'm sorry.) But the main problem remains; calculating the chance of winning next round. If we take our x being the amount of times won minus the amount of times lost; Code: x = (# of times won)  (# of times lost) That would be of great help. Thank you in advance, CX 
December 7th, 2011, 01:22 PM  #2  
Global Moderator Joined: May 2007 Posts: 6,821 Thanks: 722  Re: Statistic variation problem Quote:
 
December 7th, 2011, 02:48 PM  #3  
Newbie Joined: Apr 2008 Posts: 8 Thanks: 0  Re: Statistic variation problem Quote:
Supposedly the Monty Hall problem has something to do with it, but I understand none of that. So here's a brief simulation I did with a simple formula; Code: next bet = base bet + x x = (# of times won)  (# of times lost) Code: int baseBet = 10; int totalMoney = 100; int var = 0; int bet; Random r = new Random(); System.out.println("#\tout\tvar\tbet\ttotal"); for (int i = 0; i < 1000; i++) { bet = baseBet + var; if (r.nextBoolean()) { // Win totalMoney += bet; var; System.out.println(i + "\tWin\t" + var + "\t" + bet + "\t" + totalMoney); } else { // Lose totalMoney = bet; var++; System.out.println(i + "\tLose\t" + var + "\t" + bet + "\t" + totalMoney); } } Code: # out var bet total 0 Win 1 10 110 1 Lose 0 9 101 2 Lose 1 10 91 3 Win 0 11 102 4 Lose 1 10 92 5 Win 0 11 103 ... 995 Win 2 13 576 996 Win 1 12 588 997 Win 0 11 599 998 Win 1 10 609 999 Win 2 9 618 Code: # Money 0 600 1 600 2 488 3 488 4 552 5 312 ... 995 552 996 200 997 138 998 1528 999 258 Average profit: 31 Is 1000 statistically significant in order to make my point? I'll run some larger numbers now. (Will take a long time, since I didn't implement multithreading in this quick experiment yet.) 1 000 000 experiments throwing 10 000 coins: Average profit: 2 Average profit: 4 Average profit: 7 Average profit: 2 Average profit: 8 I bet you'd like to see the look on my face when I saw that first number pop up. They seem statistically irrelevant after throwing 1000 coins (which may or may not be due to a simplified formula, rather than the one in the OP). I just realized I am indeed very wrong and profit seems to be determined by utter chance. There goes my plan to win over casino's xD. This has been a great opportunity for me to learn to not try to outsmart the system. Criticism saved the day. Consider this thread closed. Thank you for your help, CX  
December 8th, 2011, 02:45 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,821 Thanks: 722  Re: Statistic variation problem
Two comments. I can't understand the data you are showing  I am not familiar with the code language. Monte Hall problem has nothing to do with your problem. 

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