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# An Uncertain Meeting

## Question

There are two friends, who decide to meet at a place between 5 PM to 6 PM everyday. What are the chances that, on a given day, they will be able to meet. Provided-

Case 1: They agreed whoever comes first, will wait for 15 minutes for another friend to arrive.

Case 2: One friend wait for 10 minutes and other for 20 minutes for another friend to arrive.

And which of the above two cases holds better chances of there meeting ?

Edited by ujjagrawal

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You can answer this question if you solve a single problem: What is the probability of meeting if one friend waits 15 + x minutes and the other waits 15 - x minutes?

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There are two friends, who decide to meet at a place between 5 PM to 6 PM everyday. What are the chances that, on a given day, they will be able to meet. Provided-

Case 1: They agreed whoever comes first, will wait for 15 minutes for another friend to arrive.

Case 2: One friend wait for 10 minutes and other for 20 minutes for another friend to arrive.

And which of the above two cases holds better chances of there meeting ?

Assuming the arrival time for the two friends are independently, identically, and uniformly distributed over the hour, then

The probability of meeting can be computed as a sum of simple integrals. The probability of meeting for case 1 is 7/16 = 0.4375, and the probability for the second case is 31/72 = 0.4305556. So case 1 has a slightly higher meeting chance.

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Using assumptions that seem most reasonable to me

1. p = 0.280
2. p = 0.277

The "favorable space" - the meeting time window - is defined: it's between 5:00 and 6:00.

But the "sample space" - the domain of arrival times of the two friends is not.

There seem to be three cases that can be distinguished. The permitted arrival windows are

1. Only from 5:00 to 6:00 [bushindo's assumption]
2. Only times that permit meeting between 5:00 and 6:00 [my assumption]
3. Any time they please. i.e., from midnight one day to midnight the next day.

Then we assume equal arrival probabilities within these windows.

Case 1 uses individual arrival intervals of 15 minutes.

The problem simplifies if we divide the 1-hour meeting time window into 16 15-minute squares.

Coexistence occurs only within an area of 7 in these units.

To create a meeting probability we divide 7 by something.

Under the three assumptions, the permitted arrival times / window sizes / probabilities are:

1. 5:00-6:00 [both persons] / 4x4 = 16 / p = 7/16 = 0.4375
2. 4:45-6:00 [both persons] / 5x5 = 25 / p = 7/25 = 0.280
3. 0:00-24:00 [both persons] / 96x96 = 9216 / p = 7/9216 = 0.000759

Case 2 uses 10 and 20 minutes for individual arrival intervals.

Divide the 1-hour meeting time window into 36 10-minute windows

Coexistence occurs within an "area" of 15 1/2 10-minute windows.

Again, under the three above arrival time assumptions, we get

1. 5:00-6:00 [both persons] / 6x6 = 36 / p = 15.5/36 = 0.431
2. 4:40-6:00 [one] 4:50-6:00 [other] / 8x7 = 56 / p = 15.5/56 = 0.277
3. 0:00-24:00 [both] / 144x144 = 20736 / p = 15.5/20736 = 0.000747

We see for each assumption Case 1 offers the better chance of a meeting.

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Assuming the arrival time for the two friends are independently, identically, and uniformly distributed over the hour, then

The probability of meeting can be computed as a sum of simple integrals. The probability of meeting for case 1 is 7/16 = 0.4375, and the probability for the second case is 31/72 = 0.4305556. So case 1 has a slightly higher meeting chance.

I had got the same answers as you... but would like to see how you had worked it out...
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I had got the same answers as you... but would like to see how you had worked it out...

Here's the integral I used for Case 1,

The first and third terms on the right hand side actually evaluate to the same value, but I kept them distinct since it makes it easier to follow the logic.

I like bonanova's approach much better though.

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Neither would show up at 5:00 since they both can wait 15 minutes. Neither would show at 5:15 since they each would expect the other to show at that time. Ether they both will arrive 5:30 or meet at 5:45 since they would both expect 6:00 to be too late.

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Here's the integral I used for Case 1,

The first and third terms on the right hand side actually evaluate to the same value, but I kept them distinct since it makes it easier to follow the logic.

I like bonanova's approach much better though.

My workings

Case 1: One friend must arrive before the other. The probability that friend 1 arrives during the first 3/4 hour is 3/4. He'll then wait 1/4 hour. The probability that he arrives during the last 1/4 hour is 1/4, and then (on average he'll wait) 1/8 hour. So altogether the friend 1 will wait (3/4)*(1/4 ) + (1/4)(1/8) = 7/32.

So the probability that friend 2 arrives while the friend 1 is waiting is 7/32. Similarly if friend 2 arrives before friend 1. SO altogether, the probability of them meeting is 7/32 + 7/32 = 7/16 = 43.75%.

Case 2: Similarly as case 1.

The probability that friend 2 arrives while the friend 1 is waiting is (2/3)*(1/3) + (1/3)*(1/6) = 4/18 + 1/18 = 5/18 The probability that friend 1 arrives while the friend 2 is waiting is (5/6)*(1/6) + (1/6)*(1/12)= 10/72 + 1/72 = 11/72 Total probability 5/18 + 11/72 = 31/72

Generalizing: Let friend 1 wait w1 and friend 2 wait w2, we get prob of meeting

(1-w1)*(w1) + w1*(w1/2) + (1-w2)*(w2) + w2*(w2/2) = w1(1-w1/2) + w2(1-w2/2)

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