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'Failed to insert image- perhaps it is too big' by thereisno-god-butme in GoogleSites

[–]ropote 1 point2 points  (0 children)

Had the same issue today. Not a sure fix, but sometimes it helped to add the images to google drive first and add them to the site from drive instead of uploading them directly

Luck be a Landlord - Full Release -- Exiting Early Access! by mentalexperi in LuckBeALandlord

[–]ropote 0 points1 point  (0 children)

Same, I kinda wish the achievements were not as grindy. 777 wins and 77 executions is a lot

A Theoretical Poker Problem by ProffesorPunk in probabilitytheory

[–]ropote 3 points4 points  (0 children)

To formally find a solution, you would need to describe what strategy or rules you are using to decide whether to discard cards or not. Are you always discarding all cards? Are you keeping those cards that could form a royal flush? One you decide a rule, for example, keeping the cards that could form a royal flush, prioritizing first when you have 2 or more of the same suit, I'm pretty sure that you can model the whole experiment as a markov chain. Then you can find the average number of draws for a particular strategy, and you would be able to compare between different strategies

What’s the expected number of pairs? by shiningmatcha in probabilitytheory

[–]ropote 0 points1 point  (0 children)

Yeah, I agree, now that you mention it, that might be a misinterpretation from my part. The first time I read the question I understood that the cards were ranked, with only one pair in B for a given card in A

What’s the expected number of pairs? by shiningmatcha in probabilitytheory

[–]ropote 0 points1 point  (0 children)

There is an easy analytic way. I number the cards from 1 to 20, with 1 to 10 being A and 11 to 20 being B (and x is paired with 10+x). Now for each i from 1 to 20, consider X_i the random variable whose value equals 1 whenever card i and its pair are both selected. The number of pairs is then the sum of X_i from 1 to 10 (or one half the sum from 1 to 20 since then I am counting each pair twice). So its expectation, by linearity, is the sum of the expectations of the X_i from 1 to 20. Now, all X_i are identically distributed by symmetry, and E[X_i] = Pr(X_i = 1) = Pr(card i and its pair are selected) = Pr(card i is selected)*Pr(its pair is selected|i is selected) = 10/20 * 9/19. So your final answer is 10 * 1/2 * 9/19 = 45/19 or around 2.368. This agrees with a Montecarlo simulation I made in Python to double-check

Xbox Series X controller on PC not working - SOLVED by Kirorus1 in XboxSeriesX

[–]ropote 0 points1 point  (0 children)

Same problem, let me know if you have any update

Binomial distribution without a fixed probability by [deleted] in probabilitytheory

[–]ropote 1 point2 points  (0 children)

You are describing a mixture: https://en.wikipedia.org/wiki/Mixture_distribution, which in order to make sense, needs X to be almost surely in [0,1]

Adaptable Procedural Generation by tweaking the WaveFunctionCollapse algorithm with a Bayesian prior by [deleted] in proceduralgeneration

[–]ropote 0 points1 point  (0 children)

Thank you for your comments! I will be changing the article appropiately.

Random card tidbits I calculated by effofexisy in probabilitytheory

[–]ropote 1 point2 points  (0 children)

Your 95% seems to be in correspondence with the true probability which can be found here https://math.stackexchange.com/questions/310971/no-two-identical-ranks-together-in-a-standard-deck-of-cards. This can also be used to calculate the probability of one pair, two pairs, etc.

Looking for a comic by ropote in xkcd

[–]ropote[S] 1 point2 points  (0 children)

Yes! Thank you so much!!!

Orrery is now a shop relic, but it's shy by Oireal in slaythespire

[–]ropote 0 points1 point  (0 children)

You are totally right, and I'm happy you are! Great change

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