## Thursday, December 9, 2010

### Chances Of A Random Encounter

You stop by Starbucks every day after dropping kids at school. You see that you come there randomly between 8 and 9am as sometimes you pause near the school to chat to some other parents, occasionally you go to a CVS first. A few times you run into your boss at Starbucks who is also stopping there between 8 and 9am randomly. You each spend around 10mins standing in line, getting your latte, mixing milk or sugar and zooming off to work. You really do not like these "early morning" encounters that demand a "small talk". You prefer to get caffeinated, clear your mind on the way and slowly transition from the "family" into the "work" mode. You consider switching to a different coffee shop but this one is right on your way. You contemplate: what are the chances of me encountering my boss at this Starbucks location on any given day?

This puzzle is very applicative to other life situations but it is not at all easy. I think we can solve it all together. Everyone who participate will get a puzzle point. First - please submit your individual answers by end of the day on Friday, December 10th. Answers will be hidden till Saturday morning. After the answers are revealed on Saturday, we'll continue brainstorming together.

anne-marie said...

I will think of a binomial distribution with p=1/6, q=5/6 and n=6 during a period of one hour so one day.
P being the probability of meeting the boss and q, not meeting the boss.
There are 60 mn in one hour.
The probability of not meeting the boss during this particular hour would beC(6,0) (1/6) (5/6)^6
the probability of meeting the boss at least one time would be 1- the last result.
This is for one trial, there are five working days so five independent trials so the result should be at the power five.
Now, there are little chance that you encounter the boss several times so we could calculate for y=1, C(6,1) (1/6)^1 (5/6)^5 and the result would be at the power five because there are five independent trials.(five days)

anne-marie said...

I am not sure that my first post went through so I post one more time my view of this problem.
There are 6 times 10mn period in one hour so the probability of meeting the boss is p=1/6 and q, the probability of not meeting the boss is 5/6. I think of a binomial distribution with n=6.
for any given day, the probability of meeting one time the boss would be C(6,1)(1/6)(5/6)^5
there are five working days so five independent trials, the result sould be at the power five if the five days are taken into consideration.

Anonymous said...

how do i give an hidden answer? i will try first this message

Anonymous said...

once every 5 days on average. what ever 10 minutes i pick there is a 1:5 chance that his 10 minutes will overlap. (L)

Tom said...

From Tom, who has never had a latte...but I do drink a little, and I wish I had the money back.

Seems the correct answer will require exact "given's." So let's agree I'm always there sometime between 8 and 9, and out of there by 9am, and so is the boss. And assume each of us is there exactly ten minutes, never more and no less. Never earlier, never later. Every day. And assume we both have absolutely no tendencies toward coming early or late within the hour.

I am there one sixth of the hour, and so is the boss. I am NOT there the other 50 minutes of the hour, nor is he. Without those assumptions I think an exact answer is not possible.

It's a leap, but I suspect the chances of any intercept on any one day are exactly one in six, the same as rolling two dice and getting doubles. (Or rolling one die first, and then the other, and getting a match; same thing.)

There might be something I'm overlooking but I'll stick with the hunch, and I look forward to reading others' ideas on Sunday.

Donna said...

OK....probability was never strong, but here's my thinking. In order for the boss' and my paths to cross, our respective 10 minutes would have to overlap. It doesn't really matter which 10 minutes I am there; on any given day, the probability that my boss will NOT be there the same 10 minutes is 50/60 (since we both go during the same 60 minute interval). So, the probability that our times will overlap would be 1 - 50/60, or 10/60=1/6. This seems too simplistic, though, so I'm sure I'm missing something. I may re-think this.

Anonymous said...

I used a graph to figure out the answer to this question. Along the x axis was your boss's time from 8-9, and along the y axis was your morning. I shaded in all the "safe" times. I found the area of the whole square and the "danger" area to figure out the probability of having a better morning. That was fun!

Solution Chart

Anonymous said...

The Mathematical Way: Let's start off with putting everything in minutes. You go to the coffee shop at a random time within a 60-minute window of time. So does your boss. Pretend it takes one minute to get coffee. Out of the 60 minutes in an hour, the chances of your boss running into you within that one minute are 60*60, or 1 out of 3600. But, as it takes you 10 minutes to get coffee, and there are (at most) 6 10-minute periods, the probability is much higher. Your boss can run into you at any time within those 10 minutes, and you the same. So now the probability is 6*6, or 1 out of 36 times you go there you will run into your boss. :)

-Daniel

Fe said...

1day out of 5 school days and 10minutes out of 60 minutes equals 1 out of 30 chances.

Maria said...

You all have been very brave to tackle this non-trivial puzzle and everyone who dared gets a puzzle point: anne-marie, L, Tom, Donna, Morgan, Daniel, Fe.

I think we should emphasize a few facts before we start discussing solutions:

1)we are looking for a probability of random encounter on any given day, not within a week
2)random encounter means that the 10 min interval you and your boss spend at the coffee shop should not completely overlap but should just intersect for at least 1 min
3)both, you and your boss, can enter the store anytime between 8:00 and 8:59 (60 equally possible options for each)

Some of us understood the puzzle slightly different. But let’s stick with these points now.

I should admit that my statistics is also rusty and after explaining my own solution to this problem (that seems to be very similar to Morgan’s, although hers explanation is much cooler), I welcome your comments. I will also review binomial distribution that anne-marie is referring to and try to get opinion of a professional mathematician to make sure we are not misleading ourselves.

My thoughts: there are three slightly different possible events:

1) I enter between 8:09 and 8:50 (42 out of 60 overall options for me)
2) I enter before 8:09 (9 out of 60 overall options for me)
3) I enter after 8:50 (9 out of 60 overall options for me)

Probabilities of the above events are:
1) P1 = 42/60
2) P2 = 9/60
3) P3=9/60
They add up to 1. Good!

Now, for each of these events let’s consider the probabilities of me encountering my boss and multiply these encounter probabilities by the corresponding probabilities of events.

In the case of event 1, I will be at the shop from time X to time X+9. I will encounter my boss if boss will come anytime between X-9 (overlapping for 1 min when I come) and X+9 (overlapping for 1 min when I leave).
So, there are (X+9)-(X-9)+1=19 minutes when boss may come and we will overlap. Overall, there are 60 possible minutes when boss can come. So, the probability of overlap is 19/60.
We will be multiplying this by P1: 19/60 x P1 = 19/60 x 42/60

In the case of event 2, overlap gets a bit shorter than 19 as boss can’t come before 8:00. It will be 18 minutes if I come at 8:08, 17 minutes if I come at 8:07, … and 10 minutes if I come at 8:00.
So, here we will have (18/60 + 17/60 + … + 10/60) x 1/60.

In the case of event 3, similarly to event 2 overlap gets a bit shorter than 19 as boss can’t come after 8:59. Overlap will be 18 min if I enter at 8:51, 17 min if I enter at 8:52,…, 10 min if I enter at 8:59.
Similarly to the event 2, we will have (18/60 + 17/60 + … + 10/60) x 1/60.

Overall, our probability is: overlap in event 1 + overlap in event 2 + overlap in event 3
19/60 x 42/60 + 2 x (18/60 + 17/60 + … + 10/60) x 1/60) = 0.22 + 0.07 = 0.29 (approx)

Morgan proposed a very similar solution but presented it in an ingenious way – with a graph!

Do you think this makes sense?

An interesting suggestion here is try to come as close as possible to 8:00 or 8:59 as in this case we are minimizing the overlap interval and therefore our chances of encountering our boss.

SteveGoodman18 said...

The answer is 11/36. The problem can be solved with geometric probability. Consider a unit square, representing the 1 hour arrival window for me (x) and the 1 hour arrival window for my boss (y). We will meet if the absolute value of (y-x) is less than or equal to 10. Graphing this region on the unit square yields a diagonal strip whose area is 11/36 the area of the whole square.

anne-marie said...

I thought of random as equal probability.

anne-marie said...

It makes sense.it was interesting!
Thanks!

Maria said...

SteveGoodman18 also gets a puzzle point for submitting his solution before midnight. His solution description is perfectly matching the Morgan's graph. Their explanations are brilliantly simple and allow us all to remember this puzzle and share it with our friends and family this weekend. I will try it om my elementary-school age kids.

Tom said...

Yeah the overlap of even a second (I guess) does mean that my 1/6 is wrong. Good job folks and thanks Maria.