Greetings at a Party
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
10 people are gathering in a room. Each person is meeting every other person. There are three types of greetings: a quick hello, which will take 40 seconds; a short chat, which will take 1 and a half minutes; or a longer greeting, which will take 2 minutes. Each greeting is completely random, and duration depends on the longer person. What is the probability that this gathering will take 1/2 an hour or less?
Assume no time loss between meetings with different people.
Answer
1/3^90
= 1/(8.73*10^42) !!!
All greetings need to be made, so firstly, there are 45 greetings:
9+8+7+6+5+4+3+2+1 = 45
If everyone has a short greeting:
45*40 = 1800 seconds = 30 minutes, which is ok, so if everyone made that short greeting, the probability would be 1/3^90 (Each greeting, each time depends on both people involved taking the short meeting).
Hide
Comments
Beelzebub
Apr 12, 2002
 Each person has to meet 9 people. While he is meeting somebody, everybody else is can also meet other people in the group, so even if everybody has a long meeting, it should only take 18 minutes. Your answer assumes that while one meeting is going on, nobody else is meeting. 
emmm
Jul 01, 2002
 hmmm. still a good teaser, regardless. maybe the poster is a microsoft programmer ;) 
Badge1
Mar 20, 2003
 I think this is much more complex than that because there will always be some waiting time. Even if only one person could ever do the full two minute greeting, the person he was greeting would be out of sync with everyone else, and would have to wait. I think the only way to solve this is by computer. 
jimbo
Mar 22, 2003
 I agree with both comments. If the question was what is the probability the greetings would take 18 min or less, the answer would stand because every greeting would be the same length. Once the meeting times can be different, it would appear to be a much more difficult problem. 
(user deleted)
Nov 28, 2008
 The first comment is right. The worst case is that in each round the time for every meeting is two minutes, and the total never exceeds 18 minutes. 
javaguru
Dec 24, 2008
 It looks like other people pointed out a couple of the issues I had: the simultaneous meetings and the subtly complex scheduling issue.
However, due to the scheduling issue, the maximum meeting time is greater than 18 minutes. In fact, there are scenarios where the meeting even lasts longer than 30 minutes. Using the participants A, B, C, D, E, F, G, H, I and J, I'll illustrate this with two different scenarios.
SCENARIO 1:
1) Of the first five conversations, 4 of them last 1.5 minutes and one between A and B lasts 2 minutes.
2) At the end of 1.5 minutes the 8 people C to J in the shorter conversations then pair up with new partners and have 4 2minute conversations.
3) Meanwhile, at the 2minute mark A and B finish their conversation, but have to wait for one and a half minutes before starting their second meeting since no other participants are available.
4) At the 3.5minute mark five new conversations begin. The conversation between A and C lasts 2 minutes and the other four last 1.5 minutes. One of the participants in the 2minute conversation was in the 2minute conversation at the beginning of the meeting.
5) Again, at the 5minute mark, 4 new 2minute conversations start between B, D, E, F, G, H and I.
6) At the 5.5minute mark A and C finish their conversation, but have to wait 1.5 minutes until anyone else is available.
At the 7minute mark A is beginning her third conversation. At this point, even with no further scheduling delays, if A's remaining 7 conversations take 2 minutes each, the meeting will take 21 minutes to complete. It's possible to make the meeting even longer using the staggered schedule, but eventually you will run out of pairings to tie up all of the other participants that A needs to greet. In theory, using this approach the 9 meetings for participants B through J can alternate between 2 and 1.5minute greetings for a total of 5 2minute and 4 1.5 minute greetings, at which point after 14 minutes A would just be starting her fifth greeting, giving a theoretical maximum of 24 minutes. However, since one of B to J is tied up with A for two consecutive greetings on each cycle, I think it's highly unlikely that you can find a sequence of meetings to stretch the timeline out this far using this approach.
OK, so maybe if greetings only started every two minutes then the 18minute maximum can be preserved. Not only is this regimented approach unnatural in a social gathering, but the maximum meeting time is still longer than 18 minutes or even 24 minutes.
SCENARIO 2:
1) A and B start one of five 2minute conversations.
2) At 2 minutes everyone completes the first conversation. C, D, E, F, G, H, I and J now pair up with new partners within this subgroup. This leaves A and B as the odd couple out with no new partners available.
3) At the 4, 6, 8, 10 and 12minute marks C, D, E, F, G, H, I and J continue to pair up with each other until at the 14minute mark each member of the subgroup has meet each of the other seven members in the subgroup. At the 14minute mark, A and B have still only greeted each other.
4) At the 14, 16, 18, 20, 22, 24 and 26minute marks, A and B greet C, D, E, F, G, H and I.
5) At the 28minute mark, A greets J. Since J is the only participant B hasn't greeted yet, B has to wait until A and J finish.
6) At the 30minute mark, B and J start their 2minute greeting, finally completing the meeting after 32 minutes.
I'm not proofpositive that 32 minutes is the maximum meeting time, but my intuition tells me that it is.

Zag24
Feb 07, 2009
 This is poorly written as well as wrong. The sentence "Each greeting is completely random" means nothing. I think what you meant to say is "The type of greeting each person prefers is consistent for that person, and there is an equal chance, per person, to prefer each of the three types of greeting."
Anyway, the solution is terrible, because (as the first comment points out) you assume that only one greeting can happen at a time, which is stupid. A greets B while C greets D and E greets F, all simultaneous. 
abharath27
May 13, 2013
 Wrong problem.
Please remove this. 
juliano
Jun 19, 2017
 This is wrong. There are 10 people, so there are 10 choose 2 pairs = 45 greetings. In the simplest case, if all greetings were of the same duration, then there would always be 5 simultaneous greetings, such that there would be 45/5 = 9 greeting sessions. At most, each session would be 2 minutes, thus the maximum total time would be 9*2=18 minutes.
The author must have assumed that the greetings were asynchronous, but did not make that clear. 
Back to Top
 

Follow Braingle!