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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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  • $\begingroup$ how did u get expected number of coin tosses to end the game =2 $\endgroup$ – Kiran Jan 27 '17 at 20:50
  • $\begingroup$ 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 $\endgroup$ – Manish Tiwari Jan 27 '17 at 22:56
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    $\begingroup$ 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. $\endgroup$ – Manish Tiwari Jan 27 '17 at 23:00
  • $\begingroup$ That is correct. $\endgroup$ – woogie Jan 27 '17 at 23:03
  • $\begingroup$ Hence you accept that the probability of the game ending in finite number of tosses is not equal to 1. Right? $\endgroup$ – Manish Tiwari Jan 27 '17 at 23:07
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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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  • $\begingroup$ how this can be derived? can you give any reference? $\endgroup$ – Kiran Jan 27 '17 at 21:01
  • $\begingroup$ $\mathbb{P}\left[\bigcup_{i=1}^\infty\{X = i\} \right]$ = 1 $\endgroup$ – Manish Tiwari Jan 27 '17 at 22:46
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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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  • $\begingroup$ 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. $\endgroup$ – Manish Tiwari Jan 27 '17 at 22:49
  • $\begingroup$ Was just looking up how to write limits in LaTeX $\endgroup$ – woogie Jan 27 '17 at 22:54
  • $\begingroup$ 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? $\endgroup$ – Manish Tiwari Jan 27 '17 at 23:05
  • $\begingroup$ Yes. I am aware of that. So is the statement true or false? $\endgroup$ – Manish Tiwari Jan 27 '17 at 23:19
  • $\begingroup$ So the value of binary number 0.11... to finite number of decimal places is 1? $\endgroup$ – Manish Tiwari Jan 27 '17 at 23:30

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