georggiK
@noidongnguoimatd
82 Following
5 Followers
Casts
Casts + replies
georggiK
@noidongnguoimatd
Politics can be frustrating, but it’s crucial to stay informed and involved. Your vote matters, and your voice can make a difference. Democracy thrives on participation. 🗳️ #CivicDuty #StayInformed
0 reply
0 recast
0 reaction
georggiK
@noidongnguoimatd
Haha, embracing the RBF era sounds like a fun way to handle it. Smiles or no smiles, wrinkles or no wrinkles, you do you!
0 reply
0 recast
0 reaction
georggiK
@noidongnguoimatd
gm to all the Odesza fans! Seeing them for the fifth time tonight at the Gorge Amphitheater in WA. Can't wait! What other outdoor venues are on your bucket list?
0 reply
0 recast
0 reaction
georggiK
@noidongnguoimatd
Been waiting for this moment forever. Can't wait to finally say "I told you so!
0 reply
0 recast
0 reaction
fredwilson
@fredwilson.eth
Hi Casters. For Funky Friday, I'm sharing this excellent playlist of fantastic funky songs from all around the world. There is a tracklist with lots of detail. https://on.soundcloud.com/Nyk5W
4 replies
2 recasts
39 reactions
Zach
@zachterrell
HUBERMAN > HARVARD MAN François, I disagree and here’s why. ON THE SPECIFICS First, on the specifics — I don’t think one can “correct” someone without posting the correct answer. And the equation for the geometric CDF isn’t some random mumbo jumbo. It’s exactly what you need to get the probability of having one conception on or before N=6 cycles with p=0.2 success rate. Moreover, when I first saw the Huberman clip, I thought he was just misspeaking when talking about the expected number of successes in N=6 Bernoulli trials with p=0.2 success rate. In that case E[X] = Np = 1.2. Which does have a real-world interpretation: in an idealized situation, if you had 6 IVF cycles you could have anywhere from 0 to 6 successful pregnancies. On average, you’d have 1.2 successes over 6 trials assuming a 20% independent success rate. Call that the expectation interpretation. Also, even the intuitive-but-wrong approach of just summing probabilities actually works for low values of (np). If you take the
1 reply
1 recast
36 reactions
