Chance of getting a date

A girl you like tells you that there's a $1$% chance she'll go out on a date with you. What is the expected number of times you have to ask her out before you get a date?

(Not a homework problem - I figure it's like flipping a coin with $99$% heads, $1$% tails, stop flipping when you get a tail.)

Solution 1:

Under your assumption that it's like flipping a coin—that is, that each time asking is independent of every other time—the probability that it takes asking exactly $n$ times (first $n-1$ no's, then 1 yes) is $(1-p)^{n-1}p$, where $p=1\%$ is the probability of a yes, so the expected value of the number of times to ask to get to the first yes is $$ \begin{align} \sum_{n=1}^{\infty}n(1-p)^{n-1}p &=& p+2p(1-p)+3p(1-p)^2+4p(1-p)^3+&\cdots\\ &=&p(1-p)+p(1-p)^2+p(1-p)^3+&\cdots\\ &&+p(1-p)^2+p(1-p)^3+&\cdots\\ &&+p(1-p)^3+&\cdots&\\ &&+&\cdots \end{align} $$ $$ \begin{align} =&\sum_{n=1}^{\infty}p(1-p)^{n-1}+\sum_{n=1}^{\infty}p(1-p)(1-p)^{n-1}\\ &+\sum_{n=1}^{\infty}p(1-p)^2(1-p)^{n-1}+\sum_{n=1}^{\infty}p(1-p)^3(1-p)^{n-1}+\cdots\\ =&\frac{p}{1-(1-p)}+\frac{p(1-p)}{1-(1-p)}+\frac{p(1-p)^2}{1-(1-p)}+\frac{p(1-p)^3}{1-(1-p)}+\cdots\\ =&\sum_{n=1}^{\infty}(1-p)^{n-1}\\ =&\frac{1}{1-(1-p)}=\frac{1}{p} \end{align} $$ so for $p=1\%$, the expected number of times is $100$.

Solution 2:

The number of times you'll have to ask her, given you've not asked her yet, is the sum of the following:

• 1 (because you're about to ask her)
• the probability she says no, multiplied by the number of times more you'll expect to ask her if she says no.
• the probability she says yes, multiplied by the number of times more you'll expect to ask her if she says yes.

The last bullet is zero, because if she says yes, you won't have to ask again.

Then note that the number of times more you'll expect to ask her if she says no is the same as the number of times you'd expect to ask in the first place, because—as with flipping a coin—they're all independent.

Thus we have $E = 1 + pE$ where $E$ is the number of times you can expect to have to ask and $p$ is the probability she says no.

Solution 3:

First of all, if a girl tells you that, I figure it's nothing like flipping a coin with 99% heads, 1% tails, and stopping when you get tails.

However, if we accept that model, it's a well-defined problem with a clear answer. The expected number of times you have to ask is just the average number of times you would have to ask if you repeated this process many times. (It's possible you would be really lucky, and she'd agree the first time; it's possible you would be really unlucky, and she'd turn you down 200 times before accepting.) We can apply the same kind of reasoning as in this question. (If people keep having children until they get a boy, how many children will the average family end up with?)

Suppose 1 million guys ask this girl out again and again until she says yes, and her responses really are random, with just a 1% chance of acceptance each time. The 1% chance of acceptance means that about 1 out of every 100 responses was a yes. So once all 1 million guys have gotten their dates, we have 1 million yes responses and about 99 million no responses.

So on average, each guy had to ask 100 times.

See the answer to the linked question for the infinite sum to obtain the same answer.

Solution 4:

1% = 1 in 100  

So you ask her 100 times for one date.

[The second date may take 0, 1, 100 or ∞ times, depending on your first date :-) ]