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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: 561 Thanks: 79  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: 561 Thanks: 79  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: 7,593 Thanks: 937 Math Focus: Elementary mathematics and beyond  Quote:
 
April 18th, 2017, 06:27 PM  #7 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,493 Thanks: 752  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: 1,352 Thanks: 463 Math Focus: Yet to find out.  
April 21st, 2017, 02:10 PM  #9  
Senior Member Joined: Oct 2013 From: New York, USA Posts: 561 Thanks: 79  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.  

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