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The game is that I and my opponent keep on tossing a fair coin alternatively until a tails shows up. When first tails is seen the game stops and whoever got that tails wins the game.

I calculated the expected number of coin tosses to end the game and it comes out to be 2. But the question is whether the below statement is true?

The probability of the above game ending within finite number of tosses is equal to 1.

Please provide your answer and the arguments supporting the answer.

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    how did u get expected number of coin tosses to end the game =2 – 2017-01-27
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    Probability of the game ending in k steps is $(\frac12)^k$. Expected number of tosses to end the game is $\sum_{i=1}^n i(\frac12)^i$. It is an arithmetico-geometric series. Solve it and you will get 2 – 2017-01-27
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    We can derive that the probability of the game ending within k number of tosses is equal to the value of binary number 0.11... to k decimal places. In this sense is it correct to say that the probability of the game finishing within finite number of tosses is equal to the value of binary number 0.11... to finite number of decimal places? So if latter is not equal to 1 then former is also not equal to 1. Of course latter is not equal to 1 as 0.11.. up to infinitely many decimal places is equal to 1 not up to finite number of decimal places. – 2017-01-27
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    That is correct. – 2017-01-27
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    Hence you accept that the probability of the game ending in finite number of tosses is not equal to 1. Right? – 2017-01-27
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    The probability of the game ending within a given finite number of tosses is not equal to $1$. The probability of the game ending within any finite number of tosses is equal to $1$. – 2017-01-27
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    There is no given finite number of tosses. Neither it says any finite number of tosses. The statement is if the probability of the game ending in finite number of tosses is 1. – 2017-01-27
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    Oops, misunderstood what you were asking. – 2017-01-27

2 Answers 2

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HINT Note that if $X$ is the number of tosses until the end, $1 \le X < \infty$ and $$ \mathbb{P}[X = n] = \frac{1}{2^n}, $$ and so what is $$ \mathbb{P}\left[\bigcup_{n \in \mathbb{N}}\{X = n\} \right]? $$

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    how this can be derived? can you give any reference? – 2017-01-27
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    $\mathbb{P}\left[\bigcup_{i=1}^\infty\{X = i\} \right]$ = 1 – 2017-01-27
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If $n$ is the number of tosses until the end, then you'll end up with $n-1$ heads followed by $1$ tails at the end, so, if $n=3$, it'll look like HHT. But the probability of a given trial with $n$ specific H/T appearing is $$P(X=n)=(\frac12)(\frac12)...(\frac12),$$ $n$ times, since $$P(H)=P(T)=\frac12$$ But note that the probability of the game ending within $n$ tosses will be the sum of probabilities $$P(X\leq n)=P(X=1)+P(X=2)+...+P(X=n)$$ Rewriting this with summation notation, we get $$P(X \leq n)=\sum_{i=k}^n{P(X=k)}$$ $$P(X \leq n)=\sum_{i=k}^n{(\frac12)^k}$$ Where $n$ is any finite number. We need to find $$P(X\leq n<\infty)=\sum_{i=k}^n{(\frac12)^k}=1-(\frac12)^k \neq 1$$ So the statement is false.

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    You derivation and final expression is hundred percent accurate. But the question was whether the below statement is true. The probability of the above game ending within finite number of tosses is equal to 1. – 2017-01-27
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    Was just looking up how to write limits in LaTeX – 2017-01-27
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    If it is indeed equal to 1 then we can also say that binary number 0.11.. to finite number of decimal places is equal to 1. Because the probability of the game finishing within k tosses is exactly equal to binary number 0.11.. to k decimal places. What is your comment about this? – 2017-01-27
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    Yes. I am aware of that. So is the statement true or false? – 2017-01-27
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    So the value of binary number 0.11... to finite number of decimal places is 1? – 2017-01-27
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    You agreed that the probability of game ending within k steps is 0.11... to k decimal places. Hence probability of game ending in finite steps is 0.11.. to finite decimal places (just replace k with finite). You say that probability is 1 (by saying the statement is true, not saying that supremum of probability is 1, also the question is not about supremum it is about value). But then you say 0.11.. to finite decimal places is not 1. Isn't that discrepancy? – 2017-01-27
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    I understand that the supremum of the probability of the game ending in finite steps is 1. But the question is if the probability of the game ending in finite steps is 1. Same as the supremum of 0.11.. to finite decimal places is 1 but the value of 0.11.. to finite decimal places is not 1. Aren't these two quantities equivalent and equal? You accept one but not the other. Seems illogical to me. – 2017-01-27
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    I have misunderstood what you meant in the question. I apologize. The probability that the game will end is 1. The game will eventually end, in a finite number of steps. If you were asking the actual value, the statement is false. I changed the solution. – 2017-01-27