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December 7th, 2011, 06:55 AM   #1
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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 x-axis the chance of winning, and on the y-axis 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)
Then what function would return the probability to win next round?

That would be of great help. Thank you in advance,

CX
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December 7th, 2011, 01:22 PM   #2
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Re: Statistic variation problem

Quote:
Originally Posted by cx gamer

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.





CX
What you've been told is wrong. Each toss is independent. Your chances of winning remain 50% no matter what the history was.
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December 7th, 2011, 02:48 PM   #3
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Re: Statistic variation problem

Quote:
Originally Posted by mathman
What you've been told is wrong. Each toss is independent. Your chances of winning remain 50% no matter what the history was.
Criticism, I like you.

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)
I simulated this with some simple java code;
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);
			}			
		}
This outputted the following;
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
To show this is consistent, I run this same experiment multiple times:
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
While the average profit may not be huge, after throwing 1000 coins, but it seems to usually be positive.

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
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December 8th, 2011, 02:45 PM   #4
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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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