# non-fair coin

• Sep 24th 2008, 09:00 PM
lord12
non-fair coin
Suppose that a non-fair coin with $\displaystyle P(\text{Heads}) = 0.6$ is tossed repeatedly.

(a) What is the probability that the first time we get "Heads" happens on an even numbered toss?

(b) What is the probability that the first "Heads" occurs on an odd-numbered toss?

(c) Explain why the answer (b) is larger than the answer in (a). Does it depend on $\displaystyle P(\text{Heads})$?

So let $\displaystyle P_{n}$ denote the probability that the first head appears on the $\displaystyle n$th toss. Then $\displaystyle P_{1} = 0.6, \ P_{2} = (0.6)(0.4), \ P(3) = (0.4)^{2}(0.6) \dots$

In general, $\displaystyle P_{n} = \frac{3}{5}\left(\frac{2}{5}\right)^{n-1}$. So for part (a) it would be $\displaystyle P_{2} + P_{4}+P_{6}+ \ldots + P_{2n}$ where $\displaystyle n \to \infty$.

Similarly, for part (b) it would be $\displaystyle P_{1}+P_{3} + P_{5} + \ldots + P_{2n+1}$ where $\displaystyle n \to \infty$.

So $\displaystyle P_{2n} = \frac{3}{5} \left(\frac{2}{5} \right)^{2n-1}$ and $\displaystyle P_{2n+1} = \frac{3}{5} \left(\frac{2}{5} \right)^{2n} = \frac{3}{5} \left(\frac{4}{25} \right)^{n}$.

Then $\displaystyle P_{2n+1} = \sum_{n=1}^{\infty} \frac{12}{125} \left(\frac{4}{25} \right)^{n-1} = \frac{\frac{12}{125}}{\frac{21}{25}} \approx 0.1142$. Then $\displaystyle P_{2n} = 1-0.1142 = 0.8858$?

It seems like $\displaystyle P_{2n+1} < P_{2n}$.
• Sep 24th 2008, 11:15 PM
CaptainBlack
Quote:

Originally Posted by lord12
Suppose that a non-fair coin with $\displaystyle P(\text{Heads}) = 0.6$ is tossed repeatedly.

(a) What is the probability that the first time we get "Heads" happens on an even numbered toss?

(b) What is the probability that the first "Heads" occurs on an odd-numbered toss?

(c) Explain why the answer (b) is larger than the answer in (a). Does it depend on $\displaystyle P(\text{Heads})$?

So let $\displaystyle P_{n}$ denote the probability that the first head appears on the $\displaystyle n$th toss. Then $\displaystyle P_{1} = 0.6, \ P_{2} = (0.6)(0.4), \ P(3) = (0.4)^{2}(0.6) \dots$

In general, $\displaystyle P_{n} = \frac{3}{5}\left(\frac{2}{5}\right)^{n-1}$. So for part (a) it would be $\displaystyle P_{2} + P_{4}+P_{6}+ \ldots + P_{2n}$ where $\displaystyle n \to \infty$.

Similarly, for part (b) it would be $\displaystyle P_{1}+P_{3} + P_{5} + \ldots + P_{2n+1}$ where $\displaystyle n \to \infty$.

So $\displaystyle P_{2n} = \frac{3}{5} \left(\frac{2}{5} \right)^{2n-1}$ and $\displaystyle P_{2n+1} = \frac{3}{5} \left(\frac{2}{5} \right)^{2n} = \frac{3}{5} \left(\frac{4}{25} \right)^{n}$.

Then $\displaystyle P_{2n+1} = \sum_{n=1}^{\infty} \frac{12}{125} \left(\frac{4}{25} \right)^{n-1} = \frac{\frac{12}{125}}{\frac{21}{25}} \approx 0.1142$. Then $\displaystyle P_{2n} = 1-0.1142 = 0.8858$?

It seems like $\displaystyle P_{2n+1} < P_{2n}$.

You seem to have lost $\displaystyle P_1$ somewhere.

$\displaystyle P_{odd}=0.6 + 0.4^2 0.6 + ... =0.6 (1 + 0.4^2 + 0.4^4 ...)=0.6 \frac{1}{1-0.4^2} \approx 0.714..$

RonL