# What is the probability that they meet?

1. Sep 30, 2016

### jaus tail

1. The problem statement, all variables and given/known data
A and B decide to meet between 1pm and 2pm on a given day.
Whoever arrives first will not wait for the other for more than 15 minutes. What's the probability that they will meet that day?

2. Relevant equations
Probability of event occurring = (number of favorable chances)/(total number of possible chances)

3. The attempt at a solution
Honestly I didn't know how to do it. The book has something like this:
Let's take A arrives before B.
The probability that A arrives during the first 3/4 hour is 3/4.
He'll then wait 1/4 hour.

The probability that A arrives during the last 1/4 hour is 1/4, and then (on average he'll wait) 1/8 hour.(This i understood. Like if A arrives at 1:45, he'll wait for maximum 15 minutes, but if A comes at 2:00, he won't wait, so average wait time is extremes/2 is (15+0)/2 is 7.5 minutes is 1/8 hours)

So altogether A will wait (3/4)*(1/4 ) + (1/4)(1/8) = 7/32(i guess this is minutes. Like time for which A waits). So the probability that B arrives while A is waiting is 7/32.
I didn't understand this. What is this probability? This is A waiting time. B can come anytime between 1 and 2.

Similarly if B arrives before A we have P(A and B) meet is 7/32. So altogether, the probability of them meeting is 7/32 + 7/32 = 7/16.

2. Sep 30, 2016

### Buzz Bloom

Hi jaus:

Perhaps this will help. If X is the time one person arrives, and Y is the time the other person will arrive, how do your express the relationship P(X,Y) that expresses the probability that they meet on a given day? You should assume that the time of arrival distribution is flat between 1pm and 2pm.

Regards,
Buzz

3. Oct 1, 2016

### jaus tail

Hi Buzz,
Thanks for the reply.
In your question, probability that they will meet on a given day, would be if they both arrive at same time. Or one must arrive within waiting time.
So it's like if they are waiting for 10 minutes. And X arrives at 1:30.
Then Y can arrive from 1:20 to 1:40. So there is a bandwidth of 20 minutes for Y out of total 60 minutes.
Y must come within this 20 minutes. P(Y arrives in this time) is 20/60 is 1:3.
Now how do i proceed?

4. Oct 1, 2016

### Buzz Bloom

Hi jaus:

Can you write down an expression P(X,Y) that represents that the events X and Y occur within the required time limit for a meeting to occur? Once you have that expression, how would you calculate the total probability P that a meeting will occur on a given day in terms of P(X,Y)?

Regards,
Buzz

5. Oct 1, 2016

### jaus tail

Well here goes..

X and Y have to meet between 1 and 2.
Suppose X arrives at 1:30. Probability of this is 1/60. Since there are sixty minutes between 1 and 2.
Y must arrive between 1:15 and 1:45. Probability of this happening is 30/60, is 1/2.
So P(X,Y meet) is (1/60) * (1/2) if X arrives at 1:30.

Now this is true for all X arriving between 1:15 to 1:45. That is 30 separate events.
So Probability they meet is 30 * 1/60 * 1/2
This is 0.25.

This is probability if X arrives between 1:15 and 1:45.

Suppose X arrives at 1:00. Y has to arrive from 1 to 1:15. P(Y arriving in this time) is 15/60.
Likewise
X arrive at 1:01, 1:02, 1:03....1:14 .....Probability of each is 1/60. Total 15 events(from 1:00 to 1:14).
P(Y) comes within meeting gap is 16/60, 17/60, 18/60...29/60
Probability X and Y meet in this case is (1/60)*(330/60) is 0.091678

Likewise if X comes at 2:00
Y has come from 1:46 to 2:00. P(Y arriving in this time) is 15/60.
Probability they meet is again 15*(1/60)*(330/60)
Another 0.091678

Total probability is 0.091678 + 0.091678 + 0.25 is 0.43333.
I don't know if my approach is right.

Actual answer is 0.4375.

Last edited: Oct 1, 2016
6. Oct 1, 2016

### Ray Vickson

There will be a slight difference between the answers you get with discrete arrival times (t= 0,1,2,..., 60) and continuous arrival times (0 ≤ t ≤ 60). The continuous arrival-time case gives the answer of 7/16 = 0.4375. Can you work that out?

Also: are you sure about your 1/60 basic arrival probability? In your discrete model there are 61 arrival times, which are all the integers from 0 t0 60. Also, if Mr. 1 arrives at t = 0, they meet if Mr. 2 arrives at any of the times 0,1, 2,...,15, of which there are 16 possibilities; so the probability they meet would be 16/61, not 15/60

BTW: sometimes it is easier to work out the complementary probability---in this case the probability they do not meet. I think in this particular case the complementary probability is a lot easier to compute. In this case the continuous-time version is much easier than the discrete-time version.

7. Oct 1, 2016

### haruspex

This problem can be easier to handle graphically. Draw a rectangle with axes representing the arrival times. Sketch the area which represents meeting.

8. Oct 3, 2016

### jaus tail

There was one solution on internet using graphs.

Using continuous time variable:
t1 can be 0 to 60.
t2 can be 0 to 60.
Mod of t1 - t2 must be less than equal to 15.
Now how to proceed..?

9. Oct 3, 2016

### Ray Vickson

Draw the region in (t1,t2)-space. Basically, they meet if the point (t1,t2) near enough to the line t1=t2, and they fail to meet if the point (t1,t2) is too far away from that line.

10. Oct 4, 2016

### jaus tail

I plotted this graph.

Okay so mod of t1 - t2 must be less than 15.
I don't know what to do next. How is t1 = t2 line helpful.
Equation can be t1-t2 = 15, so t1 = t2 +15, this gives one line
and t2-t1 = 15. So t2 = t1+15. This gives other line. But where do these lines lead me?

So now how do i find probability. How is graphs helpful?

The equations are t1 - t2 < = 15, so t1 = t2 + 15. 15 is waiting time which can vary from 0 to 15. So this line is t1 = t2 + 15, to t1 <= t2.
Likewise t2 - t2 < =15, waiting time varies from 0 to 15, so region is t2 <= t1+15, to t2 <= t1.
So i got region where they meet that is the middle region between the lines.

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Last edited: Oct 4, 2016
11. Oct 4, 2016

### Ray Vickson

In post #5 you went at the problem by summing over various integer-valued points that satisfied the "meet" condition. Now you have an actual drawing showing you directly which points satisfy the "meet" condition. Surely you can see how that is helpful?

Anyway, in the continuous time analysis, you integrate instead of summing, but to do that you need a probability distribution over the points (t1,t2). What distribution do you think you should use? It should match the type of probability distribution you used when t1 and t2 were integer-valued instead of continuous.

Also: looking at your drawing, the "meet" region has a more complicated shape than the "don't meet" region. Which would be the easier region to integrate over? Which calculation would be easier?

Last edited: Oct 4, 2016
12. Oct 6, 2016

### jaus tail

I thought about it. But i don't know how to go about it. Area seems the only solution, but i dont know why.
Area of region between lines can be
area of entire region (60 X 60) - (area of upper left triangle + area of lower right triangle)
and that is 60 X 60 - (1/2 X 45 X 45 + same as left term)
so that's like 3600 - 2025
so that's 1575 second square.

Entire area is 3600 sec square.

So probability can be 1575 / 3600 = 0.4375 = 7/16.

Okay so i got answer. But i dont understand why must i take area.

13. Oct 6, 2016

### Ray Vickson

Area is relevant because you are (without really saying so) assuming a uniform probability density over the square $S =[0,60] \times [0,60]$. In this case the probability that a point $(t_1,t_2)$ falls in some region $A$ in $S$ is just the ratio of the area of $A$ to the area of $S$.

Last edited: Oct 6, 2016
14. Oct 8, 2016

### jaus tail

Thanks for all the help.:)

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