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.)

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    $\begingroup$ Is that girl really worth it? Just for a date? ;p $\endgroup$ Feb 8 '11 at 13:13
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    $\begingroup$ 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. $\endgroup$
    – user02138
    Feb 8 '11 at 14:34
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    $\begingroup$ 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. $\endgroup$
    – Zarrax
    Feb 8 '11 at 14:49
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    $\begingroup$ 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 :-) $\endgroup$
    – Aryabhata
    Feb 12 '11 at 0:35
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    $\begingroup$ 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. $\endgroup$ Feb 12 '11 at 11:45

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 $100$.

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    $\begingroup$ 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! $\endgroup$
    – Gerben
    Feb 8 '11 at 23:24
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    $\begingroup$ @Gerben: Sure. I was just sticking to solution techniques at the level at which I've typically seen this type of problem—before calculus. $\endgroup$
    – Isaac
    Feb 8 '11 at 23:58
  • $\begingroup$ 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). $\endgroup$
    – Nivas
    Feb 12 '11 at 7:59
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    $\begingroup$ @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. $\endgroup$ Feb 14 '11 at 9:08
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    $\begingroup$ 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 $0<a<1$. Note that $\sum_{n=1}^\infty na^n=\frac{a}{1-a}\sum_{n=1}^\infty na^{n-1}(1-a)$, 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 :).) $\endgroup$ Jun 27 '11 at 9:29

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.

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    $\begingroup$ 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.) $\endgroup$ Feb 12 '11 at 11:39
  • $\begingroup$ @ShreevatsaR: As in, given E < infinity? $\endgroup$
    – Rawling
    Feb 14 '11 at 8:51
  • $\begingroup$ Yes, that's what I meant. $\endgroup$ Feb 14 '11 at 9:08

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.

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 :-) ]

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    $\begingroup$ You should be a little more rigorous than that. $\endgroup$
    – Potato
    Jul 23 '12 at 1:54

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.

  • $\begingroup$ I don't see how this allows one to conclude. $\endgroup$
    – Asinomas
    Aug 3 at 1:04
  • $\begingroup$ I think there is some stuff that is missing because if for example the probability drops down to $1/200$ after the $100$'th date then this answer would not be affected but the real answer would. $\endgroup$
    – Asinomas
    Aug 3 at 1:30

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