My Math Forum probability of multiple events with multiple trials

 January 27th, 2010, 08:54 AM #1 Newbie   Joined: Jan 2010 Posts: 2 Thanks: 0 probability of multiple events with multiple trials I know the chances of getting 5 heads in a row by flipping a coin is 0.03125. But how would I calculate the chances of me doing that if I had 100 flips to try to do it. Help would be appreciated. Thank you in advance!
 January 27th, 2010, 10:26 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: probability of multiple events with multiple trials If you must "start again" every five flips (so TTHHH--HHTTT doesn't count) then the answer is 1-(1-0.03125)^20 = 0.47005071531687648 or 47 %. If you don't group them this way, the problem is much harder to solve exactly. The answer is then 1026935919671913581551557828400 divided by 1267650600228229401496703205376 = 0.81010959919635805 or 81 %.
January 27th, 2010, 01:37 PM   #3
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Re: probability of multiple events with multiple trials

Quote:
 Originally Posted by aswoods If you must "start again" every five flips (so TTHHH--HHTTT doesn't count) then the answer is 1-(1-0.03125)^20 = 0.47005071531687648 or 47 %. If you don't group them this way, the problem is much harder to solve exactly. The answer is then 1026935919671913581551557828400 divided by 1267650600228229401496703205376 = 0.81010959919635805 or 81 %.
Thanks, can you tell me how you did the second one, or give me a link to a website with the information?

 January 29th, 2010, 10:57 PM #4 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: probability of multiple events with multiple trials Consider the first five throws. Write heads as 1 and tails as 0, and you get the 32 binary numbers from 00000 to 11111. Now separate these binary strings into six categories: (i) ends with 0, (ii) ends with 01, (iii) ends with 011, (iv) ends with 0111, (v) ends with 01111, and (vi) contains 11111. If you add a new bit (0 or 1) to a binary string, then it will change category, unless it is in (i) and you added 0, or in (vi), which is an absorbing category. For five bits, the number of members in each category is 16, 8, 4, 2, 1, 1. For six bits the distribution is 31, 16, 8, 4, 2, 3. If you examine how the progression works you will soon discover the recurrence relation for category (vi): $a_n= 2a_{n-1}+2^{n-6}-a_{n-6} \text{ for } n\ge 6$ where n is the number of bits, and the starting values (for n=5, etc) are 1, 3, 8, 20, 48, 112, all previous values being zero. Then the exact answer for your problem is $a_{100}2^{-100}$.
 January 30th, 2010, 02:42 PM #5 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 937 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: probability of multiple events with multiple trials Very nicely analyzed.

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