Statistically speaking, what are my odds of getting dubs on a

>>8450523
sigh....

fifty fifty either you get dubs or you don't ha ha

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>>8450526
this was painful to read

>>8450539
only half the time. It's either painful or its not. 50 50

1 in 10

>>8450546
Incorrect. Trips, quads, quints, etc. are not dubs

>>8450547
SO actually, you have a higher chance of getting dubs on a board with a lower total post count. (i.e. you won't get octs on /sci/)

>>8450523
I forget dubs rules but if we're talking the last two digits specifically, you first consider the odds of getting a number (1/10) and that number again (1/10), and add up all the different ways you can get that sort of result because they are each independent events that have equal chance of occurring. So if it's 1/100 to get 00, or 11, or 22, that would mean it's a total of 10% chance of getting ANY dubs at the end of a post.

If the dubs can be anywhere in a post, then you have to add more maths for all the different two-digit spots your dubs can occupy

>>8450550
No. The number of posts does not affect the distribution of digits (I assume).

>>8450556
But it does. We have 1/10-1/100-1/1000... up to the number of digits on the board. So it's lower for boards with higher post count

>>8450547
OK I'll bite.

P(singles)=9/10, because for a given value of the last digit, there is a 1/10 chance that the next digit is the same.

Let P(dubs,...) denote the probability of getting at least two repeating digits at the end.

On the one hand,

(1) P(dubs,...)=1-P(singles)=1/10.

On the other,

(2) P(dubs,...)=P(dubs)+P(trips,...).

where P(dubs) is the probability of getting exactly two repeating digits.

But P(trips,...)=P(dubs,...)/10, because if the last two digits match, there's a 1/10 chance that the next will match as well.

Plugging into equation (2),

P(dubs,...)=P(dubs)+P(dubs...)/10

whence

(3) P(dubs,...)=10P(dubs)/9

Plugging this result in to equation (1),

10P(dubs)/9=1/10

whence

P(dubs)=9/100

>>8450559
I assumed they would just add more digits as needed but I could be wrong

0.09

>>8450580
this is if you assume you're drawing numbers at random. only here you do not, so unless the total count is close to a quad you can safely truncate at trips due to the relatively low frequency of posts

>>8450629
?

Of course the numbers aren't truly "random" because they're posted in order, but it still makes sense to talk about the probability, namely the limit of (Number of dubs posts)/(total number of posts) as the total number of posts goes to infinity.

>>8450634
>>8450629

Come to think of it, the argument works as long as you assume that the post numbers have at least 3 digits.

>>8450634
yeah sure only that this does not make much sense for practical application. the chance that I get quints with this post is "zero" since we are nowhere near the next quints. also quads or trips can almost be ruled out in a slow board like this unless the count approaches them. dubs are different because the frequency of posts is large compared to 10/(whatever is the timescale of posting).

also dubs

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>>8450644
Nice digits

You're right about the quints thing, but that's not the question that was asked. If you draw a card at random from a standard deck, what's the probability that it's a king? Now what if I told you that I had already drawn all 4 kings? These are different questions with different answers.

>>8450526
Bayes go to bed

>>8450523
This is basically a binomial problem and does depend on the number of posts. If I haven posts what is the probability of k dubs ( in this case at least 1). The probability of a single post is a dub = 1/10 (10 dubs out of 100 possibilities).

Now the probability of no dubs = (n choose 0)(1/10)^0*(9/10)^n. So the probability of at least one dub = 1- (9/10)^n.

>>8450644
Checking that
So the chances are what then that another person would get dubs