SOLUTION: a coin is tossed until for the first time the same result occurs in twice succession.Find the probability that an even number of tosses is required.

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Question 737203: a coin is tossed until for the first time the same result occurs in twice succession.Find the probability that an even number of tosses is required.
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
To be successful you either have to get

HH or HTHH or HTHTHH or HTHTHTHH

or get

TT or THTT or THTHTT or THTHTHTT


So it's the probability of starting with a head and
ending with two heads + the probability of starting 
with a tail and ending with two tails.

The desired probability is the sum of sums of these 
two infinite geometric series:

1.   P(HH)+P(HTHH)+P(HTHTHH)+··· =  
1%2F2%5E2%22%22%2B%22%221%2F2%5E4%22%22%2B%22%221%2F2%5E6+··· =
1%2F4%22%22%2B%22%221%2F16%22%22%2B%22%221%2F64+···   


2.   P(TT)+P(THTT)+P(THTHTT)+··· =  
 1%2F2%5E2%22%22%2B%22%221%2F2%5E4%22%22%2B%22%221%2F2%5E6+··· =
1%2F4%22%22%2B%22%221%2F16%22%22%2B%22%221%2F64+···   

Each of those has a1=1%2F4 and r =1%2F4

S∞ = a%5B1%5D%2F%281-r%29 = %281%2F4%29%2F%281-1%2F4%29 = %281%2F4%29%2F%283%2F4%29 = %281%2F4%29%2A%284%2F3%29 = 1%2F3

So the desired probability is twice that or 2%2F3

Edwin