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InertialObservr's profile
〈 Berger | Dillon 〉
〈 Berger | Dillon 〉
〈 Berger | Dillon 〉
@InertialObservr

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〈 Berger | Dillon 〉

@InertialObservr

PhD student of Theoretical Particle Physics @UCIrvine l @NSF Fellow l Physics & Math Animations l Patreon: https://www.patreon.com/inertialobserver …

DC → CA
youtube.com/c/InertialObse…
Joined August 2015

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    1. Greg Egan‏ @gregeganSF 3 Dec 2019
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      You have a pack of 52 playing cards. • Shuffle the cards. • Pick a card at random. • Note what it is. • Return it to the deck. • If you’ve now seen every card, you’re done. • Otherwise, repeat. What is the expected number of cards picked, rounded to the nearest integer?

      60 replies 89 retweets 262 likes
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    2. John Carlos Baez‏ @johncarlosbaez 4 Dec 2019
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      Replying to @gregeganSF

      It's annoying how hard this is for me. After some despair I decide to think about what happens when I've seen 51 of the cards and am waiting for the last. I keep picking cards; each time with probability 51/52 I fail to get the one I want. The expected time to success? (1/n)

      3 replies 0 retweets 38 likes
    3. John Carlos Baez‏ @johncarlosbaez 4 Dec 2019
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      Replying to @johncarlosbaez @gregeganSF

      The probability that I succeed on the nth try is the probability I fail (n-1) times and then succeed: (1/52)(51/52)^{n-1} = 51/52^n. Sanity check: these probabilities sum to 1. So the expected time to success is the sum of n (51/52)^n from n=1 to infinity. (2/n)

      1 reply 0 retweets 17 likes
    4. John Carlos Baez‏ @johncarlosbaez 4 Dec 2019
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      Replying to @johncarlosbaez @gregeganSF

      The sum of nx^n from 1 to infinity is x times the sum of nx^(n-1) from 1 to infinity. So it's x times the derivative of 1/(1-x), i.e. x/(1-x)^2. Taking x = 51/52 I get (51/52) / (1/52)^2 = 51 times 52 (3/n)

      1 reply 0 retweets 12 likes
    5. John Carlos Baez‏ @johncarlosbaez 4 Dec 2019
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      Replying to @johncarlosbaez @gregeganSF

      So if I haven't screwed up the expected number of tries to get the *last* card is 51 times 52 = 2652. I should be able to compute the expected number of tries to get the second to last card, etc., and add these up to get the answer. Bed-time. (4/n)

      8 replies 0 retweets 14 likes
      〈 Berger | Dillon 〉‏ @InertialObservr 4 Dec 2019
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      Replying to @johncarlosbaez @gregeganSF

      (4/Zzz)

      12:37 AM - 4 Dec 2019
      • 9 Likes
      • Termy Cornall eripsa have fun Ну, погоди! claudio 🇪🇺 👨🏻‍💻⚡@🇮🇳 Vincent R.B. Blazy Ryan Reece Tim Hosgood
      0 replies 0 retweets 9 likes

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