36
$\begingroup$

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.)

  • 20
    Is that girl really worth it? Just for a date? ;p2011-02-08
  • 1
    On the other hand, I once thought about a variant of this game: suppose we're not really interested in that one specific girl, but only interested in assessing the probability one can pick up a girl. Now, since each time you succeed or fail, we can assume you have learned something, the probabilities will not be constant over time but will have to be updated. That seems like more interesting to model, although still a bit too general. We have to narrow it down to some specific Bernoulli model for individual dates with a set rule for updating its probability.2011-02-08
  • 22
    Since we don't yet know the motivation behind the question, we must add the caveat of "restraining order" at question $N$, where $N$ is a positive integer that is determined empirically, varies slightly from girl to girl and is directly proportional to age. In the U.S., $N$ is approximately $5$, so I'd say if $p$ is less than $20$%, then $5$ times is the expected number of questions. Otherwise, the expected number of questions is $\infty$ _from a jail cell_.2011-02-08
  • 10
    I think she's just saying that to get you to ask her out 100 times. I'd go for someone else if I were you.2011-02-08
  • 0
    If the girl gets to know this, then your probability will probably drop down further to 0%. But still it doesn't mean you will never get a date, though it means your expectation becomes $\infty$2011-02-08
  • 3
    Reminds me of this: http://movieclips.com/LpKN-dumb-and-dumber-movie-theres-a-chance/2011-02-08
  • 0
    I can't see the above from my country, but is it related to: "It's 50% chance you'll follow me home. Either you do or you don't."? Might work on uneducated targets though... ;-)2011-02-11
  • 6
    I have decided to protect this question. In the past 2 hours we have had at least three pointless answers added. Besides, I really wanted to try the protect feature out :-)2011-02-12
  • 0
    @Marcus: that's a shame!2011-02-12
  • 0
    @Moron: Dear Moron, thank you for doing this.2011-02-12
  • 0
    @Akhil: You are welcome :-)2011-02-12
  • 10
    Looking at the number of views this question has got, one learns how important it is to choose the right phrasing for a maths problem.2011-02-12
  • 1
    This reminds me of a test question a friend had: "You flip a fair coin. What is the probability you get tail?" In the margins he wrote "0%, because I'm a math major."2012-07-23

5 Answers 5

26

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 $100$.

  • 9
    Actually, there is a neat little trick you can exploit to avoid the major calculation: interpret the $\sum n (1-p)^{n-1}$ as the derivative of a series of the form $\sum (1-p)^n$ (whilst not forgetting to verify proper convergence). Can save you a lot of effort!2011-02-08
  • 1
    @Gerben: Sure. I was just sticking to solution techniques at the level at which I've typically seen this type of problem—before calculus.2011-02-08
  • 0
    The way I see is: 1% = 1 in 100 = ask her a 100 times. I just hope that this large derivation is to satisfy the mathematician's inner self. I am not a mathematician (but am interested in math, thats why I am here).2011-02-12
  • 0
    This derivation is far too complicated for such a simple problem. See Jonas's answer below for a rigorous but conceptual solution.2011-02-12
  • 0
    @Jim: Note that Jonas's answer references a question where the accepted answer uses the idea that Gerben suggests above—perfectly valid, but requires calculus. There is certainly an intuitive sense to $\frac{1}{p}$ being the expected number of trials to get an outcome that has single-trial probability $p$, but this is a rigorous answer as to why, at a precalculus level, and also demonstrates a useful technique for handling sums of series where the terms are the product of a linear function and a decreasing exponential function.2011-02-12
  • 1
    @Isaac: Well then, see Rawling's answer below. :-) It's also close to rigorous, and the method (which works for many problems of this sort) is simpler.2011-02-14
  • 2
    What I really like is the combination of Isaac's and Rawling's answers, because comparing them allows one to go the opposite direction of Isaac's answer, using a probabilistic interpretation to sum the series $\sum_{n=1}^\infty na^n$ when $0$\sum_{n=1}^\infty na^n=\frac{a}{1-a}\sum_{n=1}^\infty na^{n-1}(1-a)$, and this new sum has the probalistic interpretation shown in Isaac's answer. Using Rawling's method, probability is used to simplify this to $\frac{a}{(1-a)^2}$. (Then the formula works for all $a$ with $|a|<1$ by the identity theorem for analytic functions :).) – 2011-06-27
22

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.

  • 3
    Which gives E = 1/(1-p) (or rather E=1/p with the notation that p is the probability of saying yes, which with p=1% gives E=100). This is a simpler derivation of the same answer Isaac got. (And is rigorous modulo the assumption that E exists.)2011-02-12
  • 0
    @ShreevatsaR: As in, given E < infinity?2011-02-14
  • 0
    Yes, that's what I meant.2011-02-14
4

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.

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 :-) ]

  • 1
    You should be a little more rigorous than that.2012-07-23
1

Use the fact that expectation is linear. A single coin flip has an expected value of $1/100$th of a date. To get an expected value of 1 date, flip the coin 100 times.