The hardest math problem yet in my GRE review book

July 28, 2010

Last night, Dave and Kathy both arrived at Pizza Palace at two different random times between 10:00 p.m. and midnight. They had agreed to wait exactly 15 minutes for each other to arrive before leaving. What is the probability that Dave and Kathy were together at Pizza Palace last night between 10:00 p.m. and midnight?

I can rock the hell out of a math class, but I’m no math prodigy. Not by a long shot. After looking up the worked answer to the problem above, I wrote “holy shit” in my notes. How was I supposed to get that? I’ll write you a free blog post if you know the answer and can explain how you got it.

Update: We have a winner.

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7 Responses to “The hardest math problem yet in my GRE review book”


  1. I just made a very quick analysis of this and came up with .75, but I have only slight confidence in it because I don’t have much time to think about this right now and because probability can be very tricky sometimes. Unless I’m missing something, you have one number line of 60 minutes and two sliding windows of 15 minutes each. Each of the sliding windows consists of an arrival time and the 15 minutes after that arrival time. The problem is to figure out if there is any overlap of the two sliding 15 minute windows within the 60 minute number line, assuming they are independent events and uniformly distributed. So for example if one is sitting at the end of the line covering the last 15 minutes, then the probability of the other overlapping it is 1-.25 = .75. Then again, I’m no math genius either, so this analysis may just be stupid. I just couldn’t resist the challenge.


  2. Wait, even by my stupid analysis this is backward. It should have been .25 not .75. .75 covers the area of the line that one person might arrive in while the other was sitting in the last .25, so they only overlap 1-.75 = .25. I’ll say .25 … final answer. 🙂


  3. Nevermind, I just realized the window is 2 hours not 1 hour so I’m still wrong. I shouldn’t try to do math problems late at night.

  4. Jenny Goldstein Says:

    My approach:

    The total time is represented by a square with sides of length 120 (2 hrs) and the time that they can’t meet is represented by a square with sides of length 105 (120-15). For both of them, the ratio of time they can’t meet to the total time is 105/120, or 7/8. So you can simplify and say the squares have sides of 8 and 7, respectively. Then the probability that they will meet would be [(8×8)-(7×7)] over the total time, which is 8×8. That comes out to 15/64, which is slightly less than 1/4.

  5. JR Minkel Says:

    My coworker got it by subtracting (7/8)^2 from 1, which makes a lot more sense to me than the way my book explained it. To wit: Let d be the time Dave arrived and k be the time Kathy arrived. Simply graph |d-k| <= 15/64 for a square of side length 2 with its lower left corner at 0,0. This gives a strip going from lower left to upper right, which represents the combinations of times both of them could have shown up in the same 15-minute interval. Or something like that. A math PhD must have written it.

    And yes, in a math PhD's hands, the alphabet is sexist.

  6. Charles Says:

    Pretty tough for a GRE question! I got 15/64 by taking Dave’s arrival and looking at three cases:

    1) he arrives in the first 15 minutes (p = 1/8)
    2) he arrives between 10:15 and 11:45 (p = 6/8)
    3) he arrives in the last 15 minutes (p = 1/8)

    Case 2 is easy to analyze — if he lands there, then there’s a 30 minute window where Kathy can arrive for an overlap (p = 1/4)

    In case 1, if Dave arrives at 10:00, there’s only a 15 minute window for Kathy to arrive so that they overlap(p = 1/8) and if he arrives at 10:15, there’s a 30 minute window (p=1/4) — and the probability goes up linearly with time between 10 and 10:15. Thus, the overall probability of Kathy’s landing in the window given that Dave arrived in the first 15 minutes is the average of these two probabilities: (1/8 + 1/4)/2 = 3/16

    Case 3 is the mirror image of 1: p=3/16

    So, overall, we have a probability of:
    1/8 * 3/16 + 6/8 * 1/4 + 1/8 * 3/16
    = 3/128 + 6/32 + 3/128
    = 15/64

    Not as elegant as the geometric solution, but gets you there.


  7. I like Charles’ logical way of solving this, but I give the geometric solution extra credit for ingenuity! A dunce cap and a good night’s sleep for me.

    Thanks for sharing this, very interesting!


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