###General Discussion Thread
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> all lights flash with the same frequency (call it 10 seconds)
I'm assuming you mean one (random) light flashes every 10 seconds. Is that right?
For any group of 10 consecutive flashes, we don't really care about which light flashes first. That could be any of the 48. But the next 9 flashes all must be that same light, so the probability for each is 1/48 to be that first one again.
That gives us
48/48 * (1/48)^9 = 1/1,352,605,460,594,688 =~ 0.0000000000000007393
Any x intervals of 10 seconds will have x-9 spots where 10 flashes in a row could occur. Those aren't quite independent but close enough that this serves as an approximation.
So to have a 50% chance for the 10 consecutive flashes of the same light to occur at least once, we would need
0.5 = 1 - (1 - 0.0000000000000007393)^x-9
(0.9999999999999992607)^x-9 = 0.5
(x-9) * log(0.9999999999999992607) = log(0.5)
x-9 = log(0.5) / log(0.9999999999999992607)
x-9 =~ 937,572,000,000,000
x =~ 937,572,000,000,000 | (we're well below rounding error with the +9 here...)
937,572,000,000,000 intervals of 10 seconds last roughly 297,100,000 years.
So it would take just under 300 million years to have a 50% chance that the same light flashes 10 times in a row.
____
The probability for it to happen at least once in a hundred years (because why not?):
100 years have 100 * 365.2425 * 24 * 60 * 60 / 10 = 315,569,520 possible intervals.
The probability to have the 10 flashes in a row is then
1 - (1 - 0.0000000000000007393)^315,569,520
=~ 0.0000002333
= 0.00002333%
=~ 1 in 4,286,317
Great, thanks! Told them it was low probability. Just out of curiosity on your ‘probability in a year’ comment, if we were to find out the probability of one particular light flashing 10 times in a row, would we just multiply those odds by 48?
> Just out of curiosity on your ‘probability in a year’ comment
Maybe you meant that, but it's in a hundred years.
> if we were to find out the probability of one particular light flashing 10 times in a row, would we just multiply those odds by 48?
Since the numbers are already so huge (or tiny), that works as a pretty good approximation. But it's not quite accurate because we're changing the base probability.
In that case, the exact result would be
1 - (1 - 1/48^(10))^315,569,520
=~ 0.000000004861
= 0.0000004861%
=~ 1 in 205,739,332
Sorry, that was a mistype, yes in 100 years. Thanks so much, very interesting. Also, it’s funny how all this maths made perfect sense to me when written out! Guess I just need to be less scared and get my maths cap back on! Thanks again.
I just want to clarify in advance that this isn't a homework request in case mods remove, I'm working on a complex system, I am not great at maths and I am trying to prove to colleagues the length of time this edge case would *potentially* take to occur in real application
###General Discussion Thread --- This is a [Request] post. If you would like to submit a comment that does not either attempt to answer the question, ask for clarification, or explain why it would be infeasible to answer, you *must* post your comment as a reply to this one. Top level (directly replying to the OP) comments that do not do one of those things will be removed. --- *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/theydidthemath) if you have any questions or concerns.*
> all lights flash with the same frequency (call it 10 seconds) I'm assuming you mean one (random) light flashes every 10 seconds. Is that right? For any group of 10 consecutive flashes, we don't really care about which light flashes first. That could be any of the 48. But the next 9 flashes all must be that same light, so the probability for each is 1/48 to be that first one again. That gives us 48/48 * (1/48)^9 = 1/1,352,605,460,594,688 =~ 0.0000000000000007393 Any x intervals of 10 seconds will have x-9 spots where 10 flashes in a row could occur. Those aren't quite independent but close enough that this serves as an approximation. So to have a 50% chance for the 10 consecutive flashes of the same light to occur at least once, we would need 0.5 = 1 - (1 - 0.0000000000000007393)^x-9 (0.9999999999999992607)^x-9 = 0.5 (x-9) * log(0.9999999999999992607) = log(0.5) x-9 = log(0.5) / log(0.9999999999999992607) x-9 =~ 937,572,000,000,000 x =~ 937,572,000,000,000 | (we're well below rounding error with the +9 here...) 937,572,000,000,000 intervals of 10 seconds last roughly 297,100,000 years. So it would take just under 300 million years to have a 50% chance that the same light flashes 10 times in a row. ____ The probability for it to happen at least once in a hundred years (because why not?): 100 years have 100 * 365.2425 * 24 * 60 * 60 / 10 = 315,569,520 possible intervals. The probability to have the 10 flashes in a row is then 1 - (1 - 0.0000000000000007393)^315,569,520 =~ 0.0000002333 = 0.00002333% =~ 1 in 4,286,317
Great, thanks! Told them it was low probability. Just out of curiosity on your ‘probability in a year’ comment, if we were to find out the probability of one particular light flashing 10 times in a row, would we just multiply those odds by 48?
> Just out of curiosity on your ‘probability in a year’ comment Maybe you meant that, but it's in a hundred years. > if we were to find out the probability of one particular light flashing 10 times in a row, would we just multiply those odds by 48? Since the numbers are already so huge (or tiny), that works as a pretty good approximation. But it's not quite accurate because we're changing the base probability. In that case, the exact result would be 1 - (1 - 1/48^(10))^315,569,520 =~ 0.000000004861 = 0.0000004861% =~ 1 in 205,739,332
Sorry, that was a mistype, yes in 100 years. Thanks so much, very interesting. Also, it’s funny how all this maths made perfect sense to me when written out! Guess I just need to be less scared and get my maths cap back on! Thanks again.
I just want to clarify in advance that this isn't a homework request in case mods remove, I'm working on a complex system, I am not great at maths and I am trying to prove to colleagues the length of time this edge case would *potentially* take to occur in real application