Yahoo Answers: Answers and Comments for Another coin problem: a superfair coin making patterns? [Mathematics]
Copyright © Yahoo! Inc. All rights reserved.
https://answers.yahoo.com/question/index?qid=20080829051554AAtao1f
From Ben
enUS
Fri, 29 Aug 2008 05:15:54 +0000
3
Yahoo Answers: Answers and Comments for Another coin problem: a superfair coin making patterns? [Mathematics]
292
38
https://answers.yahoo.com/question/index?qid=20080829051554AAtao1f
https://s.yimg.com/zz/combo?images/emaillogous.png

From Dr D: While everyone thinks of a theoretical approac...
https://answers.yahoo.com/question/index?qid=20080829051554AAtao1f
https://answers.yahoo.com/question/index?qid=20080829051554AAtao1f
Fri, 29 Aug 2008 11:16:15 +0000
While everyone thinks of a theoretical approach, let me just give the results of a monte carlo simulation of this.
Over 100,000 trials, I'm getting on average that THT ends the sequence 52500, while TTH ends it 47500. The ratio is approximately 1.11.
*EDIT*
I'm surprised no one else is answering. I'm really interested to know if an analytical solution exists for this problem.
*EDIT*
No I rounded those values. Of course with monte carlo simulations, you'll never get exactly the same answer twice. At least with mine, the best you can do is average over numerous trials. But to 2 decimal places, the ratio is roughly 1.11. If we take that as 10/9, then that gives P(THT) = 10/19 and P(TTH) = 9/19. But it's hard to say for sure those are exact values. I doubt it.
Here are some results I got. Based on 1,000,000 trials, I'm getting that THT ends the sequence 525,245 times.
So let p = prob that THT ends the sequence.
We may say that
p^ = 0.525245
p ~ N(p^, p^q^/n)
A 95% confidence interval for p would be
0.525245 ± 9.788x10^(4)
or (0.524266, 0.526224).
0.5 is over 50 standard deviations away from 0.525245.
So I think it is safe to conclude that THT definitely has a higher probability of ending the sequence than TTH.

From Anonymous: After your initial pertabations, it will quick...
https://answers.yahoo.com/question/index?qid=20080829051554AAtao1f
https://answers.yahoo.com/question/index?qid=20080829051554AAtao1f
Thu, 26 May 2016 06:41:42 +0000
After your initial pertabations, it will quickly converge into a normal guasian curve. Remember that the 'normal' standard deviation varies with 1/sqrt(n). Once you get big n's it won't matter. For example, at 1000 flips, we would expect an s.d of 16 or 500 +/ 16. At that point, using your 'fair' coin, if you were one s.d. out, your odds would only be (1484/1000) or .516. In other words, you are only marginally more likely to flip a head v. a tail. It would barely change your percentage. At 100, the sd would be 5, and if you were one sd out, your precentage using the fair coin would be .55 At 10, the sd would be 1.6 and your percentage using the fair coin at one sd out would be .66 In other words, your coin would be very self correcting initially, but as you go out it would resemble a normal gausian curve. _______________________ Edit: Having pondered this, I think I am wrong. It never converges to the standard gaussian curve. It is narrower. In the abovecase of 1000 tries, if the coin flips were off by one sd ~16, your function would change the odds to .516, which means that in another 1000 tries, if the odds stayed the same, the mean would be 516, i.e. your expected answer would center on 1000 instead of 986. You can show this for all number of tries and all deviations. My guess is that it would cause the gausian curve to become uniformily narrower by some factor such 2 or sqr 2, or maybe 3 as suggested by John C, above. I will give it some more thought. _________________________ In the interest of science, I did 250 runs of 1000 flips on an excel spreadsheet. Avg: 500.16 sd: 9.33 variance: 87.05 3 looks like a good number (i.e. 1/3rd variance of a regular coin flip, 1/sqr(3) s.d.) ........................... I find it interesting that a random walk appears to be balanced exactly by a 1/r field such that it (appears to) always be narrower by 1/sqr (3) than an regular distribution. Great question.