# Is the probability of below game of coin tosses ending within finite number of coin tosses equal to 1?

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.

• how did u get expected number of coin tosses to end the game =2 – Kiran Jan 27 '17 at 20:50
• 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 – Manish Tiwari Jan 27 '17 at 22:56
• 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. – Manish Tiwari Jan 27 '17 at 23:00
• That is correct. – woogie Jan 27 '17 at 23:03
• Hence you accept that the probability of the game ending in finite number of tosses is not equal to 1. Right? – Manish Tiwari Jan 27 '17 at 23:07

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 $$\mathbb{P}\left[\bigcup_{n \in \mathbb{N}}\{X = n\} \right]?$$
• $\mathbb{P}\left[\bigcup_{i=1}^\infty\{X = i\} \right]$ = 1 – Manish Tiwari Jan 27 '17 at 22:46
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.