Learning to be Terse

Random Probabilities: How long do you eat?

Posted in Campus, People, Science by Croor Singh on September 21, 2010

[The calculations I've written down here are wrong. Go here for the correct answer and a discussion.]

I have this friend who, for the most part, did no maths in her undergrad. So although she’s quite the nerd, she’s hopeless at maths… so far. We’ll see if that doesn’t change.

Both of us eat with people from our own labs, and those schedules seem to be self-determined and uncorrelated (except for beginning within the two hours of lunch – there’s a subtlety involved here, which I’ll come back to). I asked her, as part of a deal, to predict what the chances that she’d run into me were, if she wanted to hand over her hard-drive for the TV shows she wanted (House and TBBT, if you’re curious. I said she was a nerd!). If the problem seems like it’ll be uselessly simple, I obviously didn’t write this for you. Also, do read through. There’s more than one version.

Let’s say we’ll both be at the mess for 20 minutes. What are the odds that our lunch times will overlap? If you’ve done the arithmetic in your head, did you get one-sixth? (I said two-hour lunch break, remember?)

Or more accurately than last time, let’s say I eat for 15 minutes, while she eats for 25. What are the odds now? Would you be surprised if I told you they are the same as before? (Do you know how to get this? I’ve written it down in white at the end, just in case.)

Now, we’ve assumed in the above that the mess will let you keep eating for as long as you might take to finish if you made it to lunch before the two-hour mark. What if the mess requires that you stop eating at 2pm – end effects, if you want? It’s straightforward to account for those, now that I’ve told you.

One last case: what happens if, like in the bus-stop case, the time I take to eat lunch varies within certain bounds? How about if that’s true for both of us? I haven’t done the arithmetic for this (mostly because I just made this part up), but I’d imagine that it’s doable. (I should also think the answer won’t change, if the mean-times don’t. I’m not sure.)

Okay, here’s the answer to the unequal eating times case (I hope you tried doing it before looking here):

There are only two possibilities regarding who’ll be at the mess first (the same-time case is just a degeneracy of these), and both are equally probable. If I get to the mess first, she has 15 minutes to get to the mess before she won’t meet me. If she gets to the mess first, I have 25 minutes before I won’t meet her. Mutatis Mutandis for ‘m’ and ‘n’ minutes respectively. It’s the average of ‘m’ and ‘n’ divided by the two hours of the lunch break that’s the answer.

[Ananth pointed out a flaw in the above answer, in the comments here. There are end-corrections even when people are allowed to keep eating after the 2 hour mark. If you take those into account, the probability is lesser than (m+n)/2T. The correction can actually be written down algebraically in terms of 'm' and 'n'. For 15 and 25 minutes, respectively, the probability of interest is 0.152, for example, and not 0.1667.]

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7 Responses

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  1. Anant said, on September 22, 2010 at 3:00 am

    Not quite. If you came first and yet you came at the very last minute, then there is a very small duration of time in which she has to be there. The probability should be less than 1/6 or (m+n)/120.

  2. Anant said, on September 22, 2010 at 3:01 am

    I’m assuming, of course, that she wants to eat if she comes to the mess.

  3. Croor Singh said, on September 22, 2010 at 9:43 am

    That’s a set of events – (WWBS – what would Balki say?) – of zero measure [/balki].

    I’ve conveniently cheated by showing calculations as if ‘m’ and ‘n’ are integers, but then assuming that they’re really not. So what you’re saying is that if I come to the mess at 1:59:59.xyz,there’d be no time left. That’s a zero probability event, right?

  4. Ananth said, on September 22, 2010 at 10:28 am

    No nothing about m and n being integers. I was talking about this statement.

    If I get to the mess first, she has 15 minutes to get to the mess before she won’t meet me.

    This is not true if you go there after t+1:45, is it?

  5. Croor Singh said, on September 22, 2010 at 11:02 am

    Um. I didn’t think of that. Maths in undergrad – only as good as no maths in undergrad.

    I’ve added the correction to the answer.

  6. Anant said, on September 23, 2010 at 6:14 am

    There is a bigger mistake in your calculation actually. The probability for 20min-20min should be close to 1/3 (11/36 to be precise) and not 1/6. When you assumed that she had 15 minutes for her to get to the mess after you came, the denominator to calculate the probability should the rest of the time (120-croor’s t)(since you already assumed that you came first). The average denominator will be 60 in that case and not 120. Hence 1/3. Hope I made sense.

  7. Croor Singh said, on September 23, 2010 at 10:59 am

    @Lakkay, That’s a very interesting point. It’s wrong, but the point is very interesting.

    When I say the probability that I get there before her, or vice versa, is one half each, and she has to get there in the (120-t) minutes after me, you also have to put in the probability that she hasn’t already come to the mess before me. The (120-t) should go away.

    {UPDATE: It’s actually much simpler than the above. (120-t) doesn’t have to come in the denominator, ever. The probability is still that she gets there in the final ‘k’ minutes, which is k/120. The fact that she hasn’t come to the mess before the final ‘k’ minutes is already included in the probability being one half.}

    The end correction (after Ananth with the ‘h’) still has to be included. The answer for the equal time problem (20 minutes each) will be 0.15277, instead of one-sixth (or 0.1519 for the 15-25 minutes case).


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