20100429, 16:23  #1 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
167A_{16} Posts 
Coin tossing 2
Here is another look at coin tossing.
Which of these three strings is most likely to have been produced by 32 tosses of a coin and not constructed manually? Can anyone come up with some statistics to verify that? a) THHTHTHTHTHTHTHTHTTTHTHTHTHTHTHH b) THHTHTHTHHTHTHHHTTHHTHTTHHHTHTTT c) TTHTTHHTHHTTTTTHTTHHTTHTTTTTTHTH 
20100429, 16:53  #2  
"Mark"
Apr 2003
Between here and the
2^{2}·1,511 Posts 
Quote:
Last fiddled with by rogue on 20100429 at 16:53 

20100429, 16:55  #3 
Nov 2008
2×3^{3}×43 Posts 
Each one has a 1/2^{32} chance of being produced by 32 tosses of a coin. Thus none of them is more likely than the others to appear. C is the one that most *resembles* the majority of 32toss sequences.

20100429, 18:55  #4 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2·3·7·137 Posts 
I was hoping someone wouldn't come up with that answer.
Try this: How probable is it that in a sequence of 32 tosses you would get runs of heads or tails of length r? Then which do you think was not created manually. Last fiddled with by henryzz on 20100429 at 18:55 
20100429, 22:06  #5  
Sep 2008
Kansas
3,181 Posts 
Quote:
The exactly r would be much more difficult since that means the one before the first toss of the run must be the "other" side and the one immediately after the run must also be the "other" side. But there is a slight bonus if the run starts at toss 1 or ends at toss 32. Any takers? 

20100430, 16:59  #6  
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2·3·7·137 Posts 
Quote:


20100513, 01:46  #7 
Dec 2008
Boycotting the Soapbox
2^{4}×3^{2}×5 Posts 
I think what henryzz wants to hear, is that since we expect N/2 runs of length 1, N/4 runs of length 2,...,N/2^k runs of length k, we could use chisquare to generate confidence intervals. A really cheap test could actually be to simply count runs R, because then some expression like (NR1)/(N2) should be tdistributed.

20100513, 02:55  #8 
"Lucan"
Dec 2006
England
194A_{16} Posts 

20100513, 04:05  #9  
Dec 2008
Boycotting the Soapbox
2^{4}·3^{2}·5 Posts 
Quote:
https://pantherfile.uwm.edu/ericskey...L29/node2.html If we let R denote the number of runs, H the number of Heads and T the number of Tails, then E[R]=2*H*T/(H+T)+1 and VAR[R]=2*H*T*(2*H*THT)/[(H+T)^2*(H+T1)] and the standardized number of runs is asymptotically normal distributed. In any case, smaller standardized values should be considered 'preferred by henryzz'. I'm too lazy to plug in the actual numbers to give a definite answer. Quote:
Butch Last fiddled with by __HRB__ on 20100513 at 04:16 

20100514, 00:47  #10  
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
Quote:


20100514, 11:21  #11 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
With the greatest (lack of) respect, have any of you
ever considered why nCr = n!/((nr)!*r!) ? Or even glanced at the "coin tossing" thread? David Or heard of the binomial theorem, let alone the binomial distribution? Last fiddled with by davieddy on 20100514 at 11:41 
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