tamnicufl

2022-09-19

Tossing a coin. X is the number tosses. What is the probability that X is an odd number?

A "fair" coin is tossed until the first time we get tails. Let X be the number tosses. What is the probability that X is an odd number?

A "fair" coin is tossed until the first time we get tails. Let X be the number tosses. What is the probability that X is an odd number?

Dillon Levy

Beginner2022-09-20Added 12 answers

Step 1

Consider the first toss. We have a $P=\frac{1}{2}$ probability that it's our only toss (and thus an odd number of tosses).

If it's heads, then for our second toss, we have $P=\frac{1}{2}$ that we get a tail, for an overall probability of $P=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$ that we have an even number of tosses.

So if the game ends in the first two tosses, there's a $\frac{\frac{1}{2}}{\frac{3}{4}}=\frac{2}{3}$ chance that we have an odd number of tosses.

Step 2

Now here's the key - if we keep going, then the pattern just repeats. The third and fourth tosses will provide the same results as above. What does that tell you about the overall probability?

Alternately, consider that ending on the first toss is a $\frac{1}{2}$ chance. Ending on the third toss means a result of HHT, which has a $\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}$ chance. What about ending on the fifth toss? The seventh? The pattern may become clear as you list them out.

Consider the first toss. We have a $P=\frac{1}{2}$ probability that it's our only toss (and thus an odd number of tosses).

If it's heads, then for our second toss, we have $P=\frac{1}{2}$ that we get a tail, for an overall probability of $P=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$ that we have an even number of tosses.

So if the game ends in the first two tosses, there's a $\frac{\frac{1}{2}}{\frac{3}{4}}=\frac{2}{3}$ chance that we have an odd number of tosses.

Step 2

Now here's the key - if we keep going, then the pattern just repeats. The third and fourth tosses will provide the same results as above. What does that tell you about the overall probability?

Alternately, consider that ending on the first toss is a $\frac{1}{2}$ chance. Ending on the third toss means a result of HHT, which has a $\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}$ chance. What about ending on the fifth toss? The seventh? The pattern may become clear as you list them out.

Ignacio Casey

Beginner2022-09-21Added 3 answers

Step 1

We can simply sum the probabilities.

Step 2

Let $n\in \mathbb{N}$

$P(X=2n+1)=P(X=1)+P(X=3)+...$

$=\sum _{n=0}^{\mathrm{\infty}}\frac{1}{{2}^{2n+1}}$

$=\frac{1}{2}\ast \sum _{n=0}^{\mathrm{\infty}}\frac{1}{{2}^{2n}}$

$=\frac{1}{2}\ast \frac{4}{3}$

$=\frac{2}{3}$

We can simply sum the probabilities.

Step 2

Let $n\in \mathbb{N}$

$P(X=2n+1)=P(X=1)+P(X=3)+...$

$=\sum _{n=0}^{\mathrm{\infty}}\frac{1}{{2}^{2n+1}}$

$=\frac{1}{2}\ast \sum _{n=0}^{\mathrm{\infty}}\frac{1}{{2}^{2n}}$

$=\frac{1}{2}\ast \frac{4}{3}$

$=\frac{2}{3}$

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