
Probability and Statistics Basic Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
April 16th, 2017, 08:59 PM  #1 
Senior Member Joined: Jul 2016 From: USA Posts: 108 Thanks: 13  Simple/Challenging Coin Probability Question
I have a pretty simple word problem. How many times do I have to flip a coin to have a 90% chance of it landing on heads five times in a row? My answer is approximately 145 times. This should be fun to figure out. I am really looking forward to seeing what answers you guys come up with! 
April 17th, 2017, 10:18 AM  #2 
Senior Member Joined: Jul 2016 From: USA Posts: 108 Thanks: 13 
Here is the solution I came up with: The probability of landing on 5 heads in a row is 0.5^5 So the probability of failure is 10.5^5 So if you flip the coin n times then the probability of failure will be (10.5^5)^n (10.5^5)^n = P Solve for n with algebra using logarithms log(P)/log(10.5^5) = n So P = probability of failure but if we subtract P from 1 then the statement would only be true if P = probability of success log_(10.5^5)(1P)=n Each attempt of success requires 5 flips so we multiply by 5 5*log_(10.5^5)(1P)=n Okay, so here is where things get dirty. As soon as you land on tails, it is an automatic failure for that particular attempt of 5 in a row. It does NOT wait to finish the 5 flips to begin the next attempt of success. So we need to figure out the average number of attempts we are given for some amount of flips. A new attempt begins after we fail (land on tails). So basically the number of attempts is half of the number of flips because the probability of tails (failing) is 1/2. Let's replace that 5 with a 2 and we are finished!!! 2*log(1P)/log(10.5^5) = n That is our final formula. All we need to do is substitute 0.9 as the probability (P) and evaluate. 2*log(10.9)/log(10.5^5) = 145.050677... COOL BEANS! That means if we flip the coin 145 times, we have roughly a 90% chance of landing on 5 heads in a row! And using the formula we can substitute any value of probability for P. 
April 17th, 2017, 02:50 PM  #3 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 602 Thanks: 82  I hate logarithms, so I didn't use them. The value inside the parenthesis is 31/32. I made a spreadsheet where Column A was 31/32 and Column B was the counting numbers, which are n. I calculated Column A to the Column B power and looked at when it got below .1, at which point the probability of at least 5 consecutive heads got above .9. I got an answer of 73, which is about half of your answer.

April 17th, 2017, 06:42 PM  #4 
Senior Member Joined: Jul 2016 From: USA Posts: 108 Thanks: 13  
April 18th, 2017, 02:30 PM  #5 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 602 Thanks: 82  The issue is the probability of 5 consecutive heads compared to the probability of having 5 consecutive of the same side. With five coins, the probability of 5 consecutive heads is 1/32. The probability of 5 consecutive of the same side is 2/32. I don't think the answer for 5 consecutive of the same side would be double the answer for 5 consecutive heads. The first event has double the probability of the second event, but here's an example of how that doesn't make one answer double the other answer. With 4 coins, the probability of at least 2 heads (that do not have to be consecutive) is 11/16. The probability that a coin lands on heads or tails is 1 and is double the probability that it lands on heads. However, the probability of at least 2 heads or at least 2 tails is 1. It can't be double 11/16 because that would make it greater than 1.

April 18th, 2017, 05:42 PM  #6  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond  Quote:
 
April 18th, 2017, 06:27 PM  #7 
Senior Member Joined: Sep 2015 From: USA Posts: 1,977 Thanks: 1026  https://math.stackexchange.com/quest...lengthnconta goes into this problem in excruciating detail. This coin flip problem is completely analgous to the consecutive 1's in a n bit string. For this problem you'd use the formula at the link to determine how many strings of length n have all 1 substrings of only length 4 or less. You would divide that number by $2^n$, i.e. the total number of $n$ bit strings. You would then subtract this from $1$ to obtain the probability you are after. Repeat this with increasing n (in some intelligent way) until the probability becomes 90%. 
April 18th, 2017, 08:47 PM  #8 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,597 Thanks: 546 Math Focus: Yet to find out.  
April 21st, 2017, 02:10 PM  #9  
Senior Member Joined: Oct 2013 From: New York, USA Posts: 602 Thanks: 82  Quote:
I like y = x^2 and y = 3^x because they are easily written in a y = form. A logarithm is like 3 = 10^y (or e^y for natural logarithms). Last edited by EvanJ; April 21st, 2017 at 02:13 PM.  

Tags 
coin, probability, question, simple or challenging 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
need help with umat challenging probability question  magic123  Probability and Statistics  0  July 18th, 2014 02:54 PM 
Probability coin toss question  Skyanne  Probability and Statistics  6  July 27th, 2012 10:54 AM 
SIMPLE COIN Question  helppls123  Advanced Statistics  1  March 25th, 2012 05:59 AM 
challenging probability question  wawar05  Advanced Statistics  7  October 26th, 2011 12:54 PM 
Probability question[ coin flipping ]  tnutty  Probability and Statistics  3  January 27th, 2011 04:32 PM 