What Are the Odds that Two Random Events Occur at the Same Time?

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Summary:

Odds of two random events occurring on different dates but at the exact same time down to the second.
Suppose someone randomly went to Starbucks Monday thru Saturday between 8:00 AM and 10:AM.

Suppose one Tuesday in May they purchased a drink and the credit card transaction occurred at 9:14:12 AM.

What are the odds that person could randomly return on a Wednesday in July and purchase a drink and the credit card transaction also occurs at exactly 9:14:12 AM?

I'm not a statistics wizard, but from what I've read these are independent, meaning either event can occur without the other needing or having occurred.

Would I be correct in stating the odds are 1 in 7200?

The rationale being 2 hours is 120 minutes or 7200 seconds, so the chance of either transaction occurring randomly at any given time is 1 in 7200 just like in a coin toss there are two sides so odds are are 1 in 2 and for a 6-sided die the odds of rolling a 1 are 1 in 6.

Would I also be correct in multiplying them?

(1 / 7200) * (1 / 7200) = 1.929 x 10 ^ -8

To convert to a whole number and get 1929 I would have to multiply by 100 Billion. Is that right? Would that mean 1929 chances in 100 Billion? Can that be reduced? Like 192 in 10 Billion or 19 in 1 Billion or 1 in 100 Million?

Am I even on the right track?

If there's a different formula that should be used, what would it be?

Thanks in advance for anyone's participation.
 

Answers and Replies

  • #2
phinds
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First, just to be clear, you are using the term "event" as an English language term, not the scientific version of the term. That's clear from your post and there will be no misunderstanding but I want to be sure you are aware of it because the subject line of your post (on a science forum) immediately led me to think you had things wrong.

I don't think your simplistic use of time intervals to come up with a number is really helpful. Life is WAY more complicted that that. If you Do want to use time intervals, then I would say your number is way off. People are creatures of habit so the odds of someone doing the same thing at the same time of day is quite high, although taking it down to the second would certainly make the odds lower.
 
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  • #3
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Also there are hours when star bucks could be closed and hours where there’s no way you’d go and that would reduce the odds some.

It’s a fuzzy problem and depending on how you calculate it you get wildly different numbers.
 
  • #4
Vanadium 50
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Am I even on the right track?
No.

Apart from all the other caveats you got, your calculation does not match what you wrote. The probability that both transactions occur at the same time is not the same as the probability that both transactions occur at the same time and that time is 9:14:12.
 
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  • #5
phinds
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No.

Apart from all the other caveats you got, your calculation does not match what you wrote. The probability that both transactions occur at the same time is not the same as the probability that both transactions occur at the same time and that time is 9:14:12.
Hm ... not sure that matters. What you are really computing is whether, once the first event occurs, what are the odds that the second event occurs at the same time of day so it shouldn't matter what time the first even occurs.

EDIT: Oh, wait, I think I get where you are coming from now and yes, looked at differently than I did, you have it right. I'm not positive but I think the wording of the problem favors your interpretation.
 
  • #6
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You did specify that the arrival time was random. I assume that you meant that the arrival times are random, uniformly distributed in that time slot and independent of each other. That is a precise way of expressing the problem.

The next thing to decide is whether the time of 9:14:12 was selected before the first visit or whether the first visit determined the time 9:14:12. In the first case, you are right to multiply (1 / 7200) * (1 / 7200). In the second case, you only have one factor, (1 / 7200), since the first visit happened and determined that particular time. The original statement sounds like the first visit sets the time at 9:14:12 and then the problem is to find the probability that the next visit is at the same time. The answer to that would be 1/7200, not (1 / 7200) * (1 / 7200).
 
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  • #7
PeroK
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Another answer is that the probability is 100%. There are only a finite number of time stamps in the interval. Eventually two must be the same. If you keep going to Starbucks often enough it's bound to happen sooner or later.
 
  • #8
256bits
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Isn't there another answer? Or the actual question should be:
What is the probability, given time stamp Xo, that time stamp Xn is the same as X0, with time stamps X1 ... Xn-1 not being the same as X0?
 
  • #9
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All I am saying is what he calculated didn't match what he wrote. It may well be that in addition what he wrote didn't match what he meant.
 
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You did specify that the arrival time was random. I assume that you meant that the arrival times are random, uniformly distributed in that time slot and independent of each other. That is a precise way of expressing the problem.

The next thing to decide is whether the time of 9:14:12 was selected before the first visit or whether the first visit determined the time 9:14:12. In the first case, you are right to multiply (1 / 7200) * (1 / 7200). In the second case, you only have one factor, (1 / 7200), since the first visit happened and determined that particular time. The original statement sounds like the first visit sets the time at 9:14:12 and then the problem is to find the probability that the next visit is at the same time. The answer to that would be 1/7200, not (1 / 7200) * (1 / 7200).
Thank you for your response.

I particularly like the way your phrased or framed the issue more succinctly in a way that I could not. This is not a trivial issue. It's actually a legal matter. It may be several months before a definitive statement can be made and if this thread is still open, I'll update it and if not I'll send you a message.
 
  • #11
mathman
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You got a lot of noise from people who seem to misunderstood the problem. The answer given by FactChecker is correct.
 
  • #12
phinds
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The original statement sounds like the first visit sets the time at 9:14:12 and then the problem is to find the probability that the next visit is at the same time. The answer to that would be 1/7200, not (1 / 7200) * (1 / 7200).
That was my first assumption in post #5 but I re-read it with Vanadium's interpretation and became less convinced. I'm agreeing with you that the OP needs to decide which he is talking about.
 
  • #13
PeroK
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I'm agreeing with you that the OP needs to decide which he is talking about.
Here's my guess:

The OP goes to Starbuck's with unspecified frequency, but sometimes between 8-10 in the morning, with no information on how the visits are distributed. A uniform distribution has been assumed (for no good reason other than it gives an easy mathematical answer).

On two separate occasions, the OP has credit card receipts with the same time (down to the second).

Hypothesis 1: this is so unlikely that it must be a credit card processing error. The OP should get a refund from Starbuck's for the second visit, which is very unlikely to have taken place.

First answer: the odds are 1/7200 that the second visit took place.

Hypothesis 2: sooner or later someone who goes to Starbuck's regularly will have two receipts with exactly the same time (especially as many people will go at approximately the same time every day - quite different from the assumed uniform distribution).

Second answer: there is no evidence from this alone that the second visit never took place.

The critical information required in order to analyse the problem properly:

a) How often does the OP visit the Starbuck's in question?

b) How are those times distributed?
 
  • #14
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The problem with guessing the question is there are a lot of ways to guess.

Let's make a slight change to Hypothesis 1 ("this is so unlikely that it must be a credit card processing error") The OP looks at past receipts - what is the probability that he will have to go at least 60 days back to get a matching time? (That is just under a percent)

Also, who says that just because times are expressed in seconds that the problem has one-second resolution? It could process five times a minute, at 00, 12, 24, etc. That changes all the 1/7200's into 1/600's (and the above answer to just under 10%).

I think we need to wait for the OP to write what he means.
 
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  • #15
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These issues are especially important if the purpose is to use the analysis in a legal context, as @OhioDoug implies. There are a lot of possible technicalities that may change the calculations significantly.
 

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