Guest Posted July 14, 2009 Report Share Posted July 14, 2009 This is my first post, and I hope you find it challenging 2 people decided to meet at a given location. They setup the following conditions for their meeting: 1. They will show up at the given location sometime between 12 noon and 1 PM 2. When one shows up he/she waits on the second person for up to 20 minutes then if the other person doesn't show up, he/she will leave 3. Wait time is limited to the upper limit, 1 PM, i.e. if one shows up at 12:50 PM, he/she will wait for up to 10 minutes then leave If they both show up, what's the probability for them to meet each other? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 15, 2009 Report Share Posted July 15, 2009 55% This is my first post, and I hope you find it challenging 2 people decided to meet at a given location. They setup the following conditions for their meeting: 1. They will show up at the given location sometime between 12 noon and 1 PM 2. When one shows up he/she waits on the second person for up to 20 minutes then if the other person doesn't show up, he/she will leave 3. Wait time is limited to the upper limit, 1 PM, i.e. if one shows up at 12:50 PM, he/she will wait for up to 10 minutes then leave If they both show up, what's the probability for them to meet each other? Quote Link to comment Share on other sites More sharing options...
0 bushindo Posted July 15, 2009 Report Share Posted July 15, 2009 55% I agree with simulacric_bro Assume that the two people are called A and B. If A arrives between 12:20 to 12:40, there's a (2/3) chance he'd meet A. If he arrives between 12:00 and 12:20, then this chance of meeting B is (x + 20)/60, where x is the number of minutes past 12:00. The case between 12:40 and 1:00 is the mirror image of that between 12:00 and 12:20. So, integrating this, we get (2/3)*(1/2) + (1/3)*(2/3) = .55555 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 15, 2009 Report Share Posted July 15, 2009 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 15, 2009 Report Share Posted July 15, 2009 You guys are good, it is 5/9th, and Rioth's diagram is what would have posted for my solution if no one found it Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted July 15, 2009 Report Share Posted July 15, 2009 Call the friends P1 and P2, where P1 is the first to arrive. P1 arrives between Noon and 1:00 p.m. Call the meeting probability p. If P1 arrives at Noon, p = 1/3. (P2's successful arrival window is 20 of 60 minutes.) p increases linearly with P1's arrival time, reaching unity at 12:40. <p>= 2/3 for that interval. p is unity if P1 arrives after 12:40. <p> = 1 for that interval. Overall then, p = 2/3 x (40/60) + 1 x (20/60) = 4/9 + 1/3 = 7/9 = .77777 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 15, 2009 Report Share Posted July 15, 2009 (edited) Call the friends P1 and P2, where P1 is the first to arrive. P1 arrives between Noon and 1:00 p.m. Call the meeting probability p. If P1 arrives at Noon, p = 1/3. (P2's successful arrival window is 20 of 60 minutes.) p increases linearly with P1's arrival time, reaching unity at 12:40. <p>= 2/3 for that interval. p is unity if P1 arrives after 12:40. <p> = 1 for that interval. Overall then, p = 2/3 x (40/60) + 1 x (20/60) = 4/9 + 1/3 = 7/9 = .77777 You are taking "p" as unity from 12:40 to 1:00 where as "p" is unity only at 12:40 only (at that point only not the whole interval of 12:40 to 1.) Edited July 15, 2009 by DeeGee Quote Link to comment Share on other sites More sharing options...
0 bushindo Posted July 15, 2009 Report Share Posted July 15, 2009 Call the friends P1 and P2, where P1 is the first to arrive. P1 arrives between Noon and 1:00 p.m. Call the meeting probability p. If P1 arrives at Noon, p = 1/3. (P2's successful arrival window is 20 of 60 minutes.) p increases linearly with P1's arrival time, reaching unity at 12:40. <p>= 2/3 for that interval. p is unity if P1 arrives after 12:40. <p> = 1 for that interval. Overall then, p = 2/3 x (40/60) + 1 x (20/60) = 4/9 + 1/3 = 7/9 = .77777 There is a flaw because a picture is worth 1000 words [spoiler=]Probability of meeting as a function of P1 arrival time Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 15, 2009 Report Share Posted July 15, 2009 Call the friends P1 and P2, where P1 is the first to arrive. P1 arrives between Noon and 1:00 p.m. Call the meeting probability p. If P1 arrives at Noon, p = 1/3. (P2's successful arrival window is 20 of 60 minutes.) p increases linearly with P1's arrival time, reaching unity at 12:40. <p>= 2/3 for that interval. p is unity if P1 arrives after 12:40. <p> = 1 for that interval. Overall then, p = 2/3 x (40/60) + 1 x (20/60) = 4/9 + 1/3 = 7/9 = .77777 My integration results agreed with with these. The final condition: If they both show up, what's the probability for them to meet each other? which means that p = 1 from 12:40 to 1:00... Quote Link to comment Share on other sites More sharing options...
0 preflop Posted July 15, 2009 Report Share Posted July 15, 2009 while I agree 5/9 is close I think the actual probability is slightly less. I think it is actually .551913 instead of .555556 This is because there are actually 61 points at which a person could arrive from 12 noon to 1PM. So the potential data points would look like this Point 0: Person A arrives at 12:00 there are 20 out of 60 minutes where they could meet. Point 1: Person A arrives at 12:01 there are 21 out of 60 minutes where they could meet. . . Point 20: Person A arrives at 12:20 there are 40 out of 60 minutes where they could meet. . . point 41: Person A arrives at 12:41 there are 39 out of 60 minutes where they could meet. . . point 60: Person A arrives at 1:00 there are 20 out 60 minutes where they could meet. notice i zero based this (I can't help it, I am a programmer). so that is 61 points. hence my numbers add up to 2020/3660 or 101/183 = .551913 let me know if I have gone astray here. thanks. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted July 15, 2009 Report Share Posted July 15, 2009 while I agree 5/9 is close I think the actual probability is slightly less. I think it is actually .551913 instead of .555556 This is because there are actually 61 points at which a person could arrive from 12 noon to 1PM. So the potential data points would look like this Point 0: Person A arrives at 12:00 there are 20 out of 60 minutes where they could meet. Point 1: Person A arrives at 12:01 there are 21 out of 60 minutes where they could meet. . . Point 20: Person A arrives at 12:20 there are 40 out of 60 minutes where they could meet. . . point 41: Person A arrives at 12:41 there are 39 out of 60 minutes where they could meet. . . point 60: Person A arrives at 1:00 there are 20 out 60 minutes where they could meet. notice i zero based this (I can't help it, I am a programmer). so that is 61 points. hence my numbers add up to 2020/3660 or 101/183 = .551913 let me know if I have gone astray here. thanks. 1. Assuming the arrival times are limited to integral numbers of minutes [0-60] after noon. 2. Assuming that both 0 and 60 have the same standing as 1, ..., 59. Assigning arrival time from 1-minute intervals to integers means rounding up or rounding down. So you really can't use both end points. If you stay with integers, use either the semi-closed interval (0-60] or [0-60) Or use the 60 half-integers 0.5, 1.5, ..., 59.5 to avoid bias. Either way, there are only 60 minutes between Noon to 1:00, not 61. But this point is moot if the arrival times are real number of minutes in 0-60 [open or closed]. Quote Link to comment Share on other sites More sharing options...
0 bushindo Posted July 15, 2009 Report Share Posted July 15, 2009 while I agree 5/9 is close I think the actual probability is slightly less. I think it is actually .551913 instead of .555556 This is because there are actually 61 points at which a person could arrive from 12 noon to 1PM. So the potential data points would look like this Point 0: Person A arrives at 12:00 there are 20 out of 60 minutes where they could meet. Point 1: Person A arrives at 12:01 there are 21 out of 60 minutes where they could meet. . . Point 20: Person A arrives at 12:20 there are 40 out of 60 minutes where they could meet. . . point 41: Person A arrives at 12:41 there are 39 out of 60 minutes where they could meet. . . point 60: Person A arrives at 1:00 there are 20 out 60 minutes where they could meet. notice i zero based this (I can't help it, I am a programmer). so that is 61 points. hence my numbers add up to 2020/3660 or 101/183 = .551913 let me know if I have gone astray here. thanks. Suppose person A plans to arrive exactly 12:30, but he is delayed for 39 seconds by some bum asking for change. When he arrived the hour is 12 o'clock 30 minutes and 39 seconds, what is his chance of meeting B then? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 15, 2009 Report Share Posted July 15, 2009 while I agree 5/9 is close I think the actual probability is slightly less. I think it is actually .551913 instead of .555556 This is because there are actually 61 points at which a person could arrive from 12 noon to 1PM. So the potential data points would look like this Point 0: Person A arrives at 12:00 there are 20 out of 60 minutes where they could meet. Point 1: Person A arrives at 12:01 there are 21 out of 60 minutes where they could meet. . . Point 20: Person A arrives at 12:20 there are 40 out of 60 minutes where they could meet. . . point 41: Person A arrives at 12:41 there are 39 out of 60 minutes where they could meet. . . point 60: Person A arrives at 1:00 there are 20 out 60 minutes where they could meet. notice i zero based this (I can't help it, I am a programmer). so that is 61 points. hence my numbers add up to 2020/3660 or 101/183 = .551913 let me know if I have gone astray here. thanks. I don't agree with "there are actually 61 points at which a person could arrive from 12 noon to 1PM." Can't someone arrive at 12:30, 22 secs and 17/100 of a second? The difference between 5/9 and your result .551913 is probably due to the level of granularity of solving the problem Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted July 15, 2009 Report Share Posted July 15, 2009 There is a flaw because a picture is worth 1000 words [spoiler=]Probability of meeting as a function of P1 arrival time Your picture does not apply to the flawed analysis. If it did, the analysis wouldn't be flawed. Now can you spot the flaw? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted July 15, 2009 Report Share Posted July 15, 2009 My integration results agreed with with these. The final condition: which means that p = 1 from 12:40 to 1:00... That is indeed the question: Where is the flaw? Quote Link to comment Share on other sites More sharing options...
0 bushindo Posted July 15, 2009 Report Share Posted July 15, 2009 (edited) Call the friends P1 and P2, where P1 is the first to arrive. P1 arrives between Noon and 1:00 p.m. Call the meeting probability p. If P1 arrives at Noon, p = 1/3. (P2's successful arrival window is 20 of 60 minutes.) p increases linearly with P1's arrival time, reaching unity at 12:40. Here's the flaw <p>= 2/3 for that interval. p is unity if P1 arrives after 12:40. flaw number 2. <p> = 1 for that interval. Overall then, p = 2/3 x (40/60) + 1 x (20/60) = 4/9 + 1/3 = 7/9 = .77777 Here's another attempt that p only increases linearly between 12:00 and 12:20. That is because of the cutoff at 12, restricting B from arriving earlier than noon. Suppose A arrives at time x, x being the minutes after 12:00 and 0<=x <= 20, then there is a chance of (x + 20)/60 that A would meet B. If A arrives between 12:20 and 12:40, then his chance of meeting B is a fixed 2/3, since there's a full 40 minutes window during which B can arrive. Between 12:40 and 1:00 the probability of meeting B has negative slope, for the same reason as the positive slope in the first 20 minutes of the hour. Edited July 15, 2009 by bushindo Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 15, 2009 Report Share Posted July 15, 2009 Here's another attempt that p only increases linearly between 12:00 and 12:20. That is because of the cutoff at 12, restricting B from arriving earlier than noon. Suppose A arrives at time x, x being the minutes after 12:00 and 0<=x <= 20, then there is a chance of (x + 20)/60 that A would meet B. If A arrives between 12:20 and 12:40, then his chance of meeting B is a fixed 2/3, since there's a full 40 minutes window during which B can arrive. Between 12:40 and 1:00 the probability of meeting B has negative slope, for the same reason as the positive slope in the first 20 minutes of the hour. If you let A be the first person to arrive - First one of the flaws is that from the interval [0,2/3) the probability of meeting B after A arrives is p = 1/3 * 1/(1-a) (1/3 of 1 in (1 minus time of A's arrival) - not linear Ok, gotcha, i did the same flaw thinking it was linear... However, my thinking is that the original condition in the problem states that BOTH A and B arrives. To me, that means that if the first person to arrive is at time point 2/3 and beyond [2/3,1], they have 100% chance of meeting after that point. integral(2/3,1,1) + integral(0,2/3,1/3*(1-a)^-1) = 1-2/3 + 1/3 integral(0,2/3)((1-a)^-1) = 1/3 + 1/3(-ln(1/3) - -ln(1)) = 1/3 - 1/3(ln(1/3)) ~= 0.699537 Better? I'm not quite comfortable with this though... Quote Link to comment Share on other sites More sharing options...
0 bushindo Posted July 15, 2009 Report Share Posted July 15, 2009 If you let A be the first person to arrive - First one of the flaws is that from the interval [0,2/3) the probability of meeting B after A arrives is p = 1/3 * 1/(1-a) (1/3 of 1 in (1 minus time of A's arrival) - not linear Ok, gotcha, i did the same flaw thinking it was linear... However, my thinking is that the original condition in the problem states that BOTH A and B arrives. To me, that means that if the first person to arrive is at time point 2/3 and beyond [2/3,1], they have 100% chance of meeting after that point. integral(2/3,1,1) + integral(0,2/3,1/3*(1-a)^-1) = 1-2/3 + 1/3 integral(0,2/3)((1-a)^-1) = 1/3 + 1/3(-ln(1/3) - -ln(1)) = 1/3 - 1/3(ln(1/3)) ~= 0.699537 Better? I'm not quite comfortable with this though... Ah, I see what you're saying. All this time I thought that P1 denotes one person, while it actually denotes the first person to arrive. Mea culpa. In that case If P1 is the probability that the first person arrives. You correctly derived P( P1 meets P2 | P1 arrives at x), which is the chance of P1 meeting P2, given that P1 arrives at time x, where x is a time between noon and 1:00. Then the chance of meeting P overall is P = \int P( P1 meets P2 | P1 arrives at x) * P( P1 arrives at x) dx The flaw here is that the analysis above assumes that P( P1 arrives at x), the chance that the first person arrives at time x, is uniformly distributed. While the arrival time of each person might be uniformly distributed, the arrival time of the first person is NOT. P( P1 arrives at x) belongs to the extreme value distribution, in case you want to see see a graph. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted July 16, 2009 Report Share Posted July 16, 2009 Ah, I see what you're saying. All this time I thought that P1 denotes one person, while it actually denotes the first person to arrive. Mea culpa. In that case If P1 is the probability that the first person arrives. You correctly derived P( P1 meets P2 | P1 arrives at x), which is the chance of P1 meeting P2, given that P1 arrives at time x, where x is a time between noon and 1:00. Then the chance of meeting P overall is P = int P( P1 meets P2 | P1 arrives at x) * P( P1 arrives at x) dx The flaw here is that the analysis above assumes that P( P1 arrives at x), the chance that the first person arrives at time x, is uniformly distributed. While the arrival time of each person might be uniformly distributed, the arrival time of the first person is NOT. P( P1 arrives at x) belongs to the extreme value distribution, in case you want to see see a graph. Yes, that's the flaw. Assuming that P1's arrival throughout the one-hour interval has constant probability is wrong. If the two persons are John and Joe, the assumption is true for Joe: his arrival is equally likely at Noon, 1:00 or anywhere in between. If he arrives at Noon, the probability that he's the first to arrive is unity. But that probability decreases with time, being zero if Joe arrives at 1:00. So the arrival probability for P1, the first person to arrive, is weighted heavily away from the last 20 minutes, where the meeting probability is unity. This error can appear in discrete problems when the enumerated possibilities are not equally probable. Quote Link to comment Share on other sites More sharing options...
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This is my first post, and I hope you find it challenging
2 people decided to meet at a given location. They setup the following conditions for their meeting:
1. They will show up at the given location sometime between 12 noon and 1 PM
2. When one shows up he/she waits on the second person for up to 20 minutes then if the other person doesn't show up, he/she will leave
3. Wait time is limited to the upper limit, 1 PM, i.e. if one shows up at 12:50 PM, he/she will wait for up to 10 minutes then leave
If they both show up, what's the probability for them to meet each other?
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