>>8056508
>>8056407
Can the first digit be 0? – voldemort Dec 21 '14 at 22:10
@voldemort yes it can be. – Dunka Dec 21 '14 at 22:11
2
Then your solution is correct. – voldemort Dec 21 '14 at 22:14
Check 'em dubs ;) – Hasan Saad Jun 11 '15 at 21:37
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3 Answers
activeoldestvotes
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a, b)
Be X a random variable that counts the repeating numbers from the right. Take into account that when a post is a Quads is not a Trips.
P(X=3)P(X=4)=1×110×110×910=0.9%=1×110×110×110×910=0.09%
That means, you choose the rightests number, then have 110 the next one is the same, and so on, until the fourth or the fifth, which you want it to be different.
c)
Take into account the following. If you have a Trips (xxxxxx111), the next one will come when you sum 111, being xxxxxx222. There are two exceptions, though. First is when you get through the Quads. Where you'll have to sum 222 from one to the next one, since one doesn't count. The other is when you are in xxxxxx999, where you get the next Trips on your next post, xxxxxy000.
Anyway, the mean for that is easier calculated using the probability we just found. If we'd select random number of posts t, by mean we'd find tP(X=3) Trips. Let's find for what number of posts, the expected number of Trips is 1.
1t=tP(X=3)=0.0009t=111.1111
By mean, every 111.1111 posts you'll have a Trips. For Quads it's equivalent.
1t=t′P(X=4)=0.00009t′=1111.111
By mean, every 1111.111 posts you'll have a Quads.
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edited Jun 11 '15 at 10:30
answered Jun 11 '15 at 10:20
Masclins
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A simpler way to answer a and b is to just ignore the rest of the string and consider the final n+1 digits (assuming post numbers in a given thread are effectively random (and thus independent) which for predictive purposes while looking in one thread they are). A. Calling last digits q, r, s, t we have R =/= q: .9 S=q: .1 T = q&s: .1 Multiply ??? Profit B is similar C. Has no finite answer. For any number N there is SOME calculable probability that trips does not occur. Thus for no N can the odds of trips (1- p(no trips)) can never be 1. Analagously, how many times do you have to flip a coin to GUARANTEE heads? Pro tip: you can't.
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answered Jun 11 '15 at 4:05
user247369
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Oh, I should add, my answer that trips is indeterminate is based on the fact that the supply of Ids is inexhaustible. 4chan either will recycle or add a digit. Given that there are multiple threads, there is no way to ensure all combos appear in any given one. You seem to be assuming that numbers are picked randomly with no repeats within a thread and are confined to the set of 9-digit numbers. Under such assumptions, your calculation for (c) is correct, but I don't think that is a good model for 4chan. – user247369 Jun 11 '15 at 4:10
Welcome to math.SE! Before writing, you should learn how to use the text editor of this site. Also, use the "edit" button to add details to your post, not comments. – user228113 Jun 11 '15 at 6:06
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For part C, note that Albert is calculating the expected amount of posts before we see trips. This does not guarantee that trips will show up in that number of posts.
If we want to find the necessary number of posts until we're guaranteed for 3 trips to show up, we can use the Pigeonhole Principle.
Since we're assuming every post ID is unique, and we've discovered there to be 10∗9∗105 post ids that are trips, then there are 109−9000000=991000000 post ids that are not trips. It's possible (though highly unlikely) that all these non-trip post IDs are attributed before any trips show up. But afterwards, every post that shows up will be trips. Thus 991000003 posts are necessary to guarantee 3 trips have happened.
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https://math.stackexchange.com/questions/1077057/probability-of-posting-a-quad-and-trip-on-4chan