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   myrmepropagandist (futurebird@sauropods.win)'s status on Friday, 22-Nov-2024 20:20:43 JST myrmepropagandist

   Yes I think fractions cause people as many problems as these other "more advanced" ideas. I've seen students in calc 2 who still had messy ideas about fractions. It's not trivial and just because we teach some of it to 5th graders doesn't mean everyone knows how they work.

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     Michał "rysiek" Woźniak · 🇺🇦 (rysiek@mstdn.social)'s status on Friday, 22-Nov-2024 20:20:42 JST Michał "rysiek" Woźniak · 🇺🇦

     in reply to

     @futurebird I was in a "math/physics/informatics" profiled class, and our math teacher was an absolute legend. Friends who went on to study math easily coasted on what they learned in high school for a year or two.

     That said, conditional probability remains Black Magic to me.

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     Michał "rysiek" Woźniak · 🇺🇦 (rysiek@mstdn.social)'s status on Friday, 22-Nov-2024 20:38:15 JST Michał "rysiek" Woźniak · 🇺🇦

     in reply to

     • Mux2000 (powered down)

     @Mux the main thing to remember is that the Monads have no windows.

     Oh wait, you weren't talking about not the Leibnizian ones. Never mind!

     :blobcatcoffee:

     @futurebird

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     Mux2000 (powered down) (mux@swingset.social)'s status on Friday, 22-Nov-2024 20:38:16 JST Mux2000 (powered down)

     in reply to

     • Michał "rysiek" Woźniak · 🇺🇦

     @rysiek
     Not exactly math, but I could never wrap my brain around monads.
     @futurebird

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     Michał "rysiek" Woźniak · 🇺🇦 (rysiek@mstdn.social)'s status on Friday, 22-Nov-2024 21:23:25 JST Michał "rysiek" Woźniak · 🇺🇦

     in reply to

     @futurebird yeah, that makes a lot of sense.

   • Embed this notice
     myrmepropagandist (futurebird@sauropods.win)'s status on Friday, 22-Nov-2024 21:23:27 JST myrmepropagandist

     in reply to

     • Michał "rysiek" Woźniak · 🇺🇦

     @rysiek

     Part of the reason conditional probability and probability more generally are confusing is that it insists on living on the margin of human language expressed in words and sentences and mathematical representations of that language.

     And our language simply IS NOT precise when talking about cause, effect, probability, dependence and a whole host of topics. It's a big mess.

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     Michał "rysiek" Woźniak · 🇺🇦 (rysiek@mstdn.social)'s status on Saturday, 23-Nov-2024 23:34:41 JST Michał "rysiek" Woźniak · 🇺🇦

     in reply to

     • tuban_muzuru
     • The Sleight Doctor 🃏

     @tuban_muzuru @ApostateEnglishman @futurebird I had asked already to not be mentioned in this thread, thank you.

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     tuban_muzuru (tuban_muzuru@ohai.social)'s status on Saturday, 23-Nov-2024 23:34:44 JST tuban_muzuru

     in reply to

     • Michał "rysiek" Woźniak · 🇺🇦
     • The Sleight Doctor 🃏

     @ApostateEnglishman @rysiek @futurebird

     It hardly matters. I'm violating one of my own rules: never attempt to teach math online, knowing as sure as night follows day, someone will ask your sort of question - about an ace-less deck.

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     The Sleight Doctor 🃏 (apostateenglishman@mastodon.world)'s status on Saturday, 23-Nov-2024 23:34:45 JST The Sleight Doctor 🃏

     in reply to

     • Michał "rysiek" Woźniak · 🇺🇦
     • tuban_muzuru

     @tuban_muzuru @rysiek @futurebird I'm not trying to annoy you! Just pointing out that probability calculations don't work if you're missing crucial info: in this case, that the dealer is crooked. He's only shuffling 49 cards - retaining three of the aces to bottom deal as needed. 🤷♂️

     But even math is tricky! Assuming a *fair* riffle shuffle and any number of *fair* cuts, what's the probability of dealing four cards, one of each suit, first attempt?

     Always 1, if the Gilbreath Principle was used.

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     tuban_muzuru (tuban_muzuru@ohai.social)'s status on Saturday, 23-Nov-2024 23:34:47 JST tuban_muzuru

     in reply to

     • Michał "rysiek" Woźniak · 🇺🇦
     • The Sleight Doctor 🃏

     @ApostateEnglishman @rysiek @futurebird

     Answering you has gotten me in a foul mood.

     Cards are displayed until Ace 1 appears.

     I then ask - how many cards are remaining in the deck.

     THAT NUMBER BECOMES THE DENOMINATOR

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     The Sleight Doctor 🃏 (apostateenglishman@mastodon.world)'s status on Saturday, 23-Nov-2024 23:34:48 JST The Sleight Doctor 🃏

     in reply to

     • Michał "rysiek" Woźniak · 🇺🇦
     • tuban_muzuru

     @tuban_muzuru @rysiek @futurebird Okay help me out here. You somehow do your experiment 100,000 times: deal cards face-up from a face-down shuffled deck, stopping when you get to an ace. You then calculate the probability of the next card also being an ace.

     But the next card is never an ace. All 100,000 iterations of the game, it's an indifferent card. You then learn you could have done it an *infinite* number of times, and still have never seen a second ace. The actual probability is 0.

     Why?

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     tuban_muzuru (tuban_muzuru@ohai.social)'s status on Saturday, 23-Nov-2024 23:34:50 JST tuban_muzuru

     in reply to

     • Michał "rysiek" Woźniak · 🇺🇦
     • The Sleight Doctor 🃏

     @ApostateEnglishman @rysiek @futurebird

     No.

     This isn't about the generation of random numbers and why it's so hard to do it properly.

     It's about the fact that drawing one card from a deck of 52 results in a pile of 51 cards.

     And that's all, bunkie.

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     The Sleight Doctor 🃏 (apostateenglishman@mastodon.world)'s status on Saturday, 23-Nov-2024 23:34:51 JST The Sleight Doctor 🃏

     in reply to

     • Michał "rysiek" Woźniak · 🇺🇦
     • tuban_muzuru

     @tuban_muzuru @rysiek @futurebird Depends on prior knowledge of the deck, because all shuffles (other than a tabled casino wash, but even that can be manipulated to some extent) are non-random.

     Believe it or not, it takes around 2,500 standard overhand shuffles to fully randomize a prearranged deck! A sloppy riffle shuffle is more efficient, achieving this in only seven shuffles - but a perfect, card-for-card interweaving of the packets will restore the deck to original order in eight shuffles.

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     tuban_muzuru (tuban_muzuru@ohai.social)'s status on Saturday, 23-Nov-2024 23:34:52 JST tuban_muzuru

     in reply to

     @futurebird

     1/ Conditional probability explained to my kids:

     Deck of cards. Shuffle. Now draw cards until you get an ace:

     P(Ace) = (Number of favorable outcomes) / (Total number of possible outcomes)

     Let's break it down:

     Favorable outcomes: These are the outcomes we're interested in, which is drawing an Ace. There are 4 Aces in a standard deck.