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

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

I bet 90% of Veeky Forums gets this wrong

sigh....

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

this was painful to read

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

1 in 10

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

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 Veeky Forums)

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

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

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

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

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

0.09

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

?

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.

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

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

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.

Bayes go to bed

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.

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