
Advanced Statistics Advanced Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
January 27th, 2010, 09:54 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, 11:26 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 TTHHHHHTTT doesn't count) then the answer is 1(10.03125)^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, 02:37 PM  #3  
Newbie Joined: Jan 2010 Posts: 2 Thanks: 0  Re: probability of multiple events with multiple trials Quote:
 
January 29th, 2010, 11:57 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): 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 . 
January 30th, 2010, 03:42 PM  #5 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 932 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: probability of multiple events with multiple trials
Very nicely analyzed.


Tags 
events, multiple, probability, trials 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Probability of winning the game with multiple variables  egemencoskun  Advanced Statistics  0  February 11th, 2013 12:34 PM 
Probability over multiple tables  joekrebs  Advanced Statistics  1  September 15th, 2011 07:26 PM 
Probability  same answer 4inarow on multiple choice exam  ahhaaa  Probability and Statistics  1  May 18th, 2010 04:23 PM 
Multiple Choice Probability Problem!  krzyrice  Algebra  3  April 19th, 2009 12:55 PM 
Probability Question regarding multiple choice  DW1122777  Algebra  3  January 10th, 2009 05:14 AM 