one thing to miss about porn on tumblr is how ridiculous people acted like. there was this one catfish on here who would steal other people’s dick pics and pretend they were all him. he’d post circumcised one day and uncut the next like… sir.

I had a very young kid with a hereditary heart condition come in today to be seen. My colleague excitedly told him that I play “that ‘dungeons and dragons’ game just like you do!”

The kid deadpans me with “I DM for 8 people who keep splitting the party, I think that’s why I’m really here.”

Relatable, kid, relatable.

Marine life specialists noticed a spotted eagle ray mother was having trouble and helped her deliver two baby rays 

(Source)

They’re so cute! Such

:3

There are many benefits to being a marine biologist

tumblr mobile's entire photo editing system is complete trash but it did give way to two iconic images that i will always be remembered by, for better or worse. so. there's that

before and after activating pee pee vision

if I've learned anything from Wheel of Time, it's that forget the politicians - it's going to be slighted academics and a hedge fund manager who are gong to destroy the world

Me: my god.. i did it.. i killed him..!

Angel on my shoulder: we're extremely fortunate. You shot him in the side of the head and you're wearing gloves. Place the gun in his hand and set the house ablaze. Officer Goger's tragic suicide will be the perfect cover story

Devil on my shoulder: Goger was always eating stuffing and spelt wheat and steel cut oats. Bet he'd taste reeeeal good on a spit with an apple in his mouth. Come on, i've seen the way you've looked at him..

My tulpa, a 6'9" DD smokeshow hottie PS1 graphics anthro leopard girl in a lab coat: you must put a baby in me Your Highness, quickly!

Please take this in the most encouraging and constructive way possible: replace the batteries in the CO2 detectors in your home

Girlboy with a he/him pin on one sleeve and a she/her pin on the other. Two people sitting on either side of her aggressively correcting each other on his pronouns

*her pronouns

*his pronouns

Yea hi can I get uhhhhhhhhhhh physical touch and affection combo meal please

This is a real tweet from the National Park Service.

I generally consider myself a fairly grounded person, but when I manage to biff a dice roll with an 85% chance of success seven times in a row, I can't help but start to wonder.

Like, rationally I know it's just because I'm making a huge number of rolls, and the questions "what are the odds of flipping a coin five times and getting five heads?" and "given a thousand sequential coin flips, what are the odds that a run of five consecutive heads appears somewhere in the resulting sequence?" have very different answers, but in the moment the second one feels a whole lot like the first!

OP are you making a million dice rolls? This is super unlucky. I’m blanking on how many rolls you’d need to have even odds of that happening - back in a day or so -

Funnily enough, owing to the tracking features of the dice-rolling software I'm using I can actually give you an exact answer to that: at the time of this posting, I have made that specific 85%-chance-of-success roll 14 987 times.

(I could probably come up with the formula for the odds of a run of seven consecutive failures appearing in 14 987 rolls, but ultimately I'm not sufficiently invested in the answer to feel like working for it.)

I don't know if there's a simple formula. Having thought for a few hours on this, all I can come up with is to do it recursively.

I'm going to model it as a coin that's 85% heads and 15% tails, and you've flipped tails 7 times in a row. Say H=0.85 and T=0.15. The chance of just straight flipping 7 tails in a row is T^7, and out of 15000 flips it could be flips 1 through 7, or 2 through 8, etc. . . but you can't just add up T^7 a bunch of times. That's overcounting - it's possible that you get it on flips 1 through 7 and 2 through 8, and adding separately counts that outcome twice. Or, for that matter, you could get a 10-streak of 1 through 10, or you might get five separate streaks of 7. . . .

(You probably already know all this - this wordy background is for anyone who finds it helpful!)

I think it's easier to calculate the chance of doing 15000 flips and not getting 7 tails in a row. And this is. . . still not straightforward, but I think it can be done recursively.

Math notation ahoy! P(n) will be the chance of not getting 7 tails in a row in n flips.

Ok! Now we're going to figure out P(n+1) based on P(n) and below. In other words, let's say we've recorded 100 coin flips, and got no streaks of 7 tails in a row. Can we use this to figure out P(101)?

This part is a bit fuzzy - I may have some errors in here (please correct me if you find some!) I think that there are two cases that are important:

In case 2, the last flips actually need to be HTTTTTT (1 head followed by 6 tails), or it would not be part of P(100) which represents getting to 100 without getting 7 tails in a row. So the probability of case 2 coming up is P(93)*H*T^6. Since the 94th flip is H, I believe the first 93 flips can be anything (so long as they don't have 7 tails in a row).

In case 1. . . that's the rest of P(100). It's all the ways of getting to 100 flips that aren't in case 2, so the probability of case 1 coming up is P(100) - P(93)*H*T^6.

Time to put the full answer together. All we're missing is the last flip. In case 1, the last flip can be anything, and in case 2, the last flip has to be H. So, all together!

P(101) = (case1 * 1) + (case2 * H) = [ P(100) - P(93)*H*T^6 ] + [ P(93)*H*T^6 * H] = P(100) - (1-H)*P(93)*H*T^6.

One last step. This is the probability of never having 7 tails, so the probability of getting a streak of 7 tails in 101 flips is 1 - P(101).

I believe the general step is the same as the 101st step, so P(n+1) = P(n) - (1-H)*P(n-7)*H*T^6.

---

I messily wrote up this formula in https://www.online-python.com/:

You can just barely see me calling the function with a value of 15000, representing 15000 coin flips. And it came out as:

If my math is right, your chance of getting 7 tails in a row within 15000 flips is a bit over 2%. Fairly unlucky still, but not vanishingly unlikely!

How many flips would it take to have a 50-50 chance of getting 7 tails in a row at some point? I plugged in random values until the output was about 0.5, and 480000 flips is close.

Reblogging this one for being the only attempt that shows its work and doesn't fall into the naïve solution's overcounting trap. Congrats!

There is a simple(r) formula! (At least one that’s non-recursive, and that you can calculate faster.)

For that we need to reach for Markov chains. A Markov chain is a stochastic process (which means a series of random variables, which you can think of as random events) where every step depends only on the one before it.

Again, we will think of the problem as flipping a biased coin a lot of times, with T being the possibility of flipping a tail. If we flip seven tails in a row, we call that a failure. Our Markov chain will have 8 states – 8 possible outcomes for a random event. Now, keep in mind that here “random” doesn’t mean ‘uniform’. Not every event will happen with equal probability. In the extreme, the variable that always outputs the same state is still considered random. (This would coincide with a coin flip where you always flip heads.)

We number the 8 states from 0 to 7, and we think of them as the number of tails we need to flip in a row until the first failure. So if we flip heads, this number is 7. At the first tails it goes down to 6, then 5, etc. If we flip a head – before we reach failure! – it goes back up to 7. If we reach 0, that means that we’ve had a failure, so from that point on, we stay at 0.

This is a Markov chain. This means that if we know what state we’re on, we can tell the probability of arriving on the next state at the next step – without having to know how we got there. For example, if we’re on a 5, then with probability T we go to 4 (5 means we need 5 more tails until failure; if we flip tails, we’ll only need 4 more), with probability 1–T we go to 7, and the probability of going to any other state is 0. The analogue is true for any nonzero state: with probability T we go down one, with probability 1–T we go to 7. From 0, we stay on 0 with probability 1.

We can illustrate this with a graph. Something like this:

(The edges denote the possible transitions, the labels are the probabilities of a given step.)

Another way to summarise the Markov chain is via its transition matrix:

What does this mean? If you take its k’th column (indexing from 0), that is the distribution you get after one step from state k. Taking the k’th column of a matrix is the same as multiplying it by the k’th unit vector. (Vector that’s 0 everywhere and 1 in the k’th place.) The nice thing is that it works for distributions as well: if you multiply it by a vector that denotes a distribution over the states, the result will be the distribution after the next step. This means that the N’th power of this matrix is the distribution after n steps. If you take its last column, that’s the distribution of your N’th step starting from 7, and if you take this vector’s first (zeroth) value, that’s the probability of you having failed.

Now, I did not find a nice easy way to calculate that, you’d have to do the actual matrix multiplication. However, this is computationally a lot cheaper. (For the previous recursive algorithm, you need log(N) multiplications to calculate the final result. With the matrix multiplication method, you need log N matrix multiplications, each of which consists of 4096 regular multiplications.)

Of course the solution is the same (except that I calculated the probability of failure outright, so my number is 1 minus @wobster109​‘s number), and with a number as low as 15000 it won’t make a difference (I believe the recursive algorithm might even be faster), but for larger N’s, the difference grows. (Although I’d be remiss if I didn’t mention that for this algorithm, you use more space: you need to store all log(N) matrices, while the above algorithm only needs to store 7 numbers, regardless of N.)

See, I'm conflicted about this approach. On the one hand, I'm a sucker for any solution that involves expressing the problem as a state machine; on the other hand, matrix transformations make me break out in hives.

(In practice, I'd probably stick with the recursive solution because – as you correctly note – it's likely to be more computationally efficient for any number of dice rolls that a human player could plausibly make.)

The real reason that sex scenes in traditional fantasy erotica tend to involve bathing in improbably convenient crystal-clear pools is that fantasy writers and readers on the whole are the sort of people who won't bat an eye at wizards and dragon, but would be terribly bothered if the text didn't explicitly address the obvious consequences of whipping one's unmentionables out after spending weeks on end traipsing across trackless wilderness in full amour.

If I were an evil wizard, I would simply not use mind control spells which can be dispelled by depleting the victim's health bar.

@bluebandedagate replied:

I mean, I wouldn't tell them it's not that kind of spell. It can be a learning experience.

And then if you want to be real evil you give the mind controlled victim the order to act like they have been freed from the spell after regaining consciousness.

The party thinks they've succeeded and you now have a perfect little spy and ally for the final confrontation.

Quick question: did anyone else experience a sudden tenfold increase in new followers starting around the 24th of June? Are we having another massive bot influx, or is it just me?

In the past 72 hours alone I've been followed by over 1100 default-avatar, zero-post blogs just like these, and literally nobody else seems to be reporting anything like this. Are the pornbots targeting me in particular, or what?

For all the jokers in the notes going "I can't believe you don't know about Reddit" who evidently have no concept of the passage of linear time, a helpful graphic:

"Oh, but our RPG protagonist has to be a teenager, or else they won't appeal to our core 16-to-19-year-old player demographic" literally just make them a DILF.

"But what about straight boys" to the modern heterosexual male gamer the DILF is an aspirational figure. Don't give them what they already are – give them what they want to be.

beaft

HEY. HEY. COLUM WAS NOT IN HARROW'S RIVER BUBBLE DREAM THEATER. COLUM'S SOUL ISN'T IN THE RIVER. WHERE THE FUCK IS COLUM ASHT

Trying to impress cats and failing is universal 

(via)

It’s extremely important that I tell you all that this seal is doing the ‘banana pose,’ something seals do when they are feeling particularly happy and relaxed. This seal, looking directly at the cat, is absolutely overwhelmed with Good Vibes, something we can all related to.

Mad because you don't have bird vision?

Seethe . You will never b them