Friday evening puzzle
When I met Lia’s mom (see a couple of posts back) last evening, Lia mentioned to her mom and her sister about my FB posts. And quickly added that she totally skips my puzzles. Which gave me an idea to form a puzzle including Lia and see what she does about it 🙂
Lia and I, both Saturday morning runners at Fowler Park, decide to run together tomorrow morning. We agree to meet up between 6am and 7am. Both of us are definitely going to arrive in that timeframe. But, since we were likely to reach there at different times, we agreed on the rule that upon reaching the trailhead, if one does not find the other waiting there, he/she will wait exactly for 20 minutes for the other person before proceeding to run by himself/herself. Of course, in the middle of the wait, if it becomes 7 am already, the person just goes ahead for the run alone, (realizing that the other person came earlier and must have left after waiting for 20 minutes.)
Question is what is the probability that we will actually get to run together?
Tagging Lia 🙂
(1/60)*21/60 + (1/60)*22/60 + (1/60)*23/60 + … + (1/60)* 40/60 + 20*(1/60)*41/60 + (1/60)*40/60 + (1/60)*39/60 + (1/60)*38/60 + … + (1/60)*20/60?
Each should bring their phones, it’s the simplest solutions that are the best.
David, i doubt math teachers would accept that.
My question is, how much of this question did Lia read before she stopped reading? Would love to see the formula for that. But I do enjoy the puzzles, don’t get many correct.
Prasenjit, I think I followed your logic. There are a couple of points. First, the approach is assuming that there are discrete points when we arrive – 6:52 or 6:53 etc. But it will be a continuous function- meaning we can arrive at 6:52:07. Etc. Your thought process should still work. Also, I do not think it ever goes to 41/60 (in either discrete or continuous mode – the max would be 40/60)
You’re absolutely right, Rajib. Although arrival times would, in reality, be continuous, I felt assuming them to be discrete, while being an approximation, would make the puzzle easier to attempt on a smartphone. Also I was afraid continuous arrival times would involve formulating and solving an integral (which I’d last done in 1982 ☺). So to approximate ‘on the side of caution’, I started with Lia (for example) arriving at any moment during the 1st minute and you at any moment during the 1st-21st minutes to be able to run with her. Which made me take (1/60)*21/60 rather than (1/60)*20/60 as the first term in the sequence. As I went along the sequence, Lia could arrive at any moment during the 21st minute and you at any moment during the 1st-41st minutes to be able to run with her. Which made me take (1/60)*41/60 rather than (1/60)* 40/60 as the highest term. Of course, with continuous arrival times, not only would the highest term be (1/60)*40/60, even the starting term would be (1/60)*20/60.
David, Knowing Lia, moment she saw the word puzzle, she stopped 🙂
My answer is 0.5
7:00 to 7:20 = 1/2
7:21 to 7:40 = 2/3
7:41 to 8:00 = 1/3
Avg of these 3 intervals is 1/2
Mukund,
It will be a little more than half
The solution:
Building up on Prasenjit’s approach, note that if I showed up at 6:00, then Lia and I get to run together if she shows up in the next 20 minutes. Since it is equally likely when she will show up, the probability is 20/60. If I showed up at 6:10 though, Lia could show up anytime between 6:00 to 6:30 – 30 minute window and we would run together (prob is 30/60). If I showed up at 6:15, then she would have had a 35 minute window.
Now, not that from 6:20 thru 6:40, she has a 40 minute window (prob is 40/60). And then after 6:40, it starts becoming the reverse of the 6:00-6:20 window.
So, there are 3 windows – 6:00-6:20, 6:20-6:40 and 6:40-7:00.
6:00-6:20 window:
Since every moment is equally likely and the “window of opportunity” grows linearly every moment, the average of all those probabilities will be that of the mid point – 6:10. And that probability is 30/60 = 1/2
6:20-6:40 window:
The prob for each moment is 40/60 = 2/3
6:40-7:00 window:
It is similar to the first window. The average of all the prob in that window will be the same as for 6:50 which is 30/60 = 1/2.
Now, I am equally likely to come in any of those windows (1/3)
So, the total prob of running together is 1/3*1/2 + 1/3*2/3 + 1/3*1/2 = 5/9.
The answer should be 5/9.
I should mention that Karthik solved the problem.
The solution:
Building up on Prasenjit’s approach, note that if I showed up at 6:00, then Lia and I get to run together if she shows up in the next 20 minutes. Since it is equally likely when she will show up, the probability is 20/60. If I showed up at 6:10 though, Lia could show up anytime between 6:00 to 6:30 – 30 minute window and we would run together (prob is 30/60). If I showed up at 6:15, then she would have had a 35 minute window.
Now, not that from 6:20 thru 6:40, she has a 40 minute window (prob is 40/60). And then after 6:40, it starts becoming the reverse of the 6:00-6:20 window.
So, there are 3 windows – 6:00-6:20, 6:20-6:40 and 6:40-7:00.
6:00-6:20 window:
Since every moment is equally likely and the “window of opportunity” grows linearly every moment, the average of all those probabilities will be that of the mid point – 6:10. And that probability is 30/60 = 1/2
6:20-6:40 window:
The prob for each moment is 40/60 = 2/3
6:40-7:00 window:
It is similar to the first window. The average of all the prob in that window will be the same as for 6:50 which is 30/60 = 1/2.
Now, I am equally likely to come in any of those windows (1/3)
So, the total prob of running together is 1/3*1/2 + 1/3*2/3 + 1/3*1/2 = 5/9.
The answer should be 5/9.
I should mention that Karthik Mani solved the problem.