Been there. Many times. https://xkcd.com/2217/ pic.twitter.com/zgs5ctolR0
Neurodiverse trans geek girl. Yosys, RISC-V, SAT/SMT.
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Been there. Many times. https://xkcd.com/2217/ pic.twitter.com/zgs5ctolR0
Ok, but how do you use a 52 card deck to simulate a 53 card deck? Is there a procedure that produces the same probability distribution as drawing a card from a 53 card deck?
And if so, what's the lowest number of shuffles you can do it in
Oh, that's such a great question! There's an obvious solution that has no lowest bound, because it has a chance of "restart from the top". 53 is a prime number, which excludes the other obvious solution that come to mind from the pool of options.
The obvious solution: Just draw a pair of cards from the shuffled 52-cards deck. There are 2652 possible outcomes. Assign each of the 53 "virtual cards" 50 of those outcomes. Re-shuffle and re-try if you drew one of the two remaining pairs.
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