My Math Forum A coin is tossed five times

 September 11th, 2011, 01:05 PM #1 Newbie   Joined: Sep 2011 Posts: 1 Thanks: 0 A coin is tossed five times A coin is tossed five times. By counting the elements in the following events, determine the probability of each event. a. Heads never occurs twice in a row. First I thought it could be: H T H T H or T H T H T but still thinking more this does not seem right.
 September 11th, 2011, 01:41 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: A coin is tossed five times Remember that tails are allowed to appear more than once in a row.
September 12th, 2011, 08:21 AM   #3
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Re: A coin is tossed five times

Hello, nomss!

Quote:
 A coin is tossed five times. By counting the elements in the following events, determine the probability of each event. [color=beige]. . [/color]a. Heads never occurs twice in a row.

I found no formula for this problem.
I made an exhaustive list; there are four cases.

$\begin{array}{cccccc}\text{0 Heads} && T\:T\:T\:T\;T && \text{1 way} \end{array}$

$\text{1 Head}\;\;\begin{array}{c}H\:T\:T\:T\;T \\ \\ \\ T\:H\:T\;T\;T \\ \\ \\ T\:T\:H\:T\:T \\ \\ \\ T\:T\:T\;H\:T \\ \\ \\ T\:T\:T\:T\:H \end{array}\;\;\text{5 ways}$

$\text{2 Heads}\;\;\begin{array}{c}H\:T\:H\:T\:T \\ \\ \\ H\:T\:T\:H\:T \\ \\ \\ H\:T\:T\:T\;H \\ \\ \\ T\:H\:T\:H\:T \\ \\ \\ T\:H\:T\:T\:H \\ \\ \\ T\:T\;H\:T\:H \end{array}\;\;\text{6 ways}$

$\text{3 Heads}\;\;H\:T\:H\:T\:H \;\;\text{1 way}$

$\text{Hence, there are }13\text{ ways with non-consecutive Heads.}$

$\text{There are: }\,2^5= 32\text{ possible outcomes.}$

$\text{Therefore: }\:P(\text{non-consecutive Heads}) \:=\:\frac{13}{32}$

September 12th, 2011, 11:19 AM   #4
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Re: A coin is tossed five times

Quote:
 Originally Posted by soroban I found no formula for this problem.
One way of thinking about it: it's the same as the number of ways to flip six coins such that no two heads are adjacent and the last coin is tails. This is the same as the number of strings of length 6 composed of HT and T. So for length n there are
$\sum_{k=0}^{(n+1)/2}{n-k+1\choose k}.$
For five it's easy to take the sum and get 13.

Another way of thinking about it: there's one way to do it for length 0 and 2 ways for length 1. For each longer length you can do it in as many ways as the last two added together: the last one followed by tails, or the second-to-last one followed by tails then heads. Thus the answer is the (n+2)-th Fibonacci number.

Alternately: You're trying to match zero or more tails, followed by a sequence of (heads followed by one or more tails), followed by an optional heads. For those familiar with regular expressions, it can be expressed as
Code:
T*(HT+)*H?
The number of solutions can be expressed as a matrix exponentiation problem: in this case,
Code:
M=[1,1,0;1,0,1;0,0,2]^5;M[1,1]+M[1,2]

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# a coin is tossed 5 times what is the probability that tail appears more than three times

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