How do you calculate the odds of simultaneous events?

Question

Given a period of time and a group of people, each of whom has their own button and clicks it one time, how would you calculate the odds that 2 of these clicks would occur at the same time, to the millisecond? What formula should be used?

Answer 1

Your question is actually a cute variation on "The Birthday Problem", as one of the other posters has pointed out. I asked the following very similar question not too long ago:

http://answers.yahoo.com/question/index;_ylt=AvV_XHzHl4fZiIondBTaDDXty6IX?qid=20070613202141AArjTJW

An interesting point shared by both our questions that the Birthday Problem does NOT share is that overlapping areas can change the probability. For example, if two clicks are 1.5 milliseconds apart, then they are not within a millisecond, but there is a 0.5 millisecond overlap. However, as I say in the Additional Details to my question, if the number of people is proportionately small compared to the number of milliseconds, then overlapping areas will not change the odds significantly.

Answer 2

Let n be the number of people with buttons. Let t be the total interval of time (in seconds) in which people may press their button. Assume that each person presses their button exactly once in a randomly-chosen millisecond of the total time interval.

So, the total number of milliseconds is
m = 1000*t.

If one person chooses a certain millisecond to press their button, the probability the next person does NOT choose that millisecond is (m - 1)/m. However, they must therefore have chosen a different millisecond in which to push their button. This means the probability a third person does not push their button in either of these milliseconds is (m - 2)/m. The probability that a fourth person misses the other three is (m - 3)/m. You can see the pattern emerging.

So, if we have 10 people and a total time interval of 1 second (1000 milliseconds), the probability that everyone chooses a different millisecond is:

P = m! / ((m - n)! * m^n)

The probability that everyone does NOT choose a different millisecond is simply 1 minus this probability:

P = 1 - m! / ((m - n)! * m^n)
P = 1 - 1000! / ((1000 - 10)! * 1000^10)
P = 1 - 0.955861
P = 0.044139

If there are 50 people pressing buttons in a 1000-millisecond interval, the probability that two button-pushes happen in the same millisecond is:

P = 1 - m! / ((m - n)! * m^n)
P = 1 - 1000! / ((1000 - 50)! * 1000^50)
P = 1 - 0.287731
P = 0.712269

Here's another use. Assuming a random distribution, the probability that at least two people in a room of 30 people have the same birthday is:

P = 1 - 365! / ((365 - 30)! * 365^30)
P = 1 - 0.293684
P = 0.706316

Higher than you thought, eh?

Answer 3

The probability that any two people would respond in the same interval in time approaches zero as the interval duration approaches zero. I believe that is all that can be said with certainty given the problem statement.

To say any more, you would need to specify an overall time period for the experiment.

Answer 4

that would be very subjective to the case at hand
if i ask the people to click one after another, the odds will be very very low indeed

however, i guess you are talking about a highly theoretical scenario, where the time at which a person clicks is a random variable within a fixed band of time. the concept of confidence intervals applies here.