TOSS Even

Toss a fair coin until 2 consecutive heads or 2 consecutive tails appear.

What is the probability that an even number of tosses will be required?

1 4 \dfrac{1}{4} 1 3 \dfrac{1}{3} 1 2 \dfrac{1}{2} 2 3 \dfrac{2}{3} 3 4 \dfrac{3}{4}

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2 solutions

Michael Mendrin
Jul 22, 2018

After the coin is thrown an odd number of times alternating between heads and tails, the probability of the last throw being either heads or tails is even. Then the following could occur (first coin being at the end of an odd number of throws)

a : H , T , T a: {H,T,T}
b : H , H b: {H,H}
c : T , H , H c: {T,H,H}
d : T , T d: {T,T}

with events b , d b, d being twice as likely as events a , b a, b , so the probability of an even number of throws is 2 3 \dfrac{2}{3}

Samrit Pramanik
Jul 22, 2018

The Sample Space is Ω = { HH,HTHH,HTHTHH, } \Omega=\big\{\text{HH,HTHH,HTHTHH,}\ldots\big\} or Ω = { TT,THTT,THTHTT, } \Omega=\big\{\text{TT,THTT,THTHTT,}\ldots\big\} .

So, the required probability is 2 [ ( 1 2 ) 2 + ( 1 2 ) 4 + ( 1 2 ) 6 + ] 2\left[\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^4+\left(\dfrac{1}{2}\right)^6+ \ldots\right] = 2 1 4 1 1 1 4 =2 \cdot \dfrac{1}{4} \cdot \dfrac{1}{1-\frac{1}{4}} = 1 2 4 3 =\dfrac{1}{2} \cdot \dfrac{4}{3} = 2 3 =\boxed{\dfrac{2}{3}}

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