The Quantum Pontiff has posted a cool probability question which is guaranteed to torment your brain. Here it is.

The Quantum Pontiff has posted a cool probability question which is guaranteed to torment your brain. Here it is.

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I’m pretty much satisfied with CRM-114’s explanation of why there are equal odds for either hand. I was thinking the same thing: What are the odds of at least one ace being drawn in a 4-card hand from a 51 card deck that is missing one ace? Which ace is irrelevant. A specific ace is missing, but only explicit in the 2nd case. It is only the card value that is important, however; the suit is a red herring.

Of course, I will reserve the right to shame-facedly admit my error once I can look closer at this at home.

And less than a minute afterwards, I think that the order of dealing may make a difference, i.e. it is not the same as a 4-card hand from a 51 card deck.

This is why statistics is so much fun!

1/3 vs 1/2 for 2 coins, 4/7 vs 3/4 for 3 coins, 1/5 vs. 1/3 for 2 triangles. Ace of Diamonds wins, with 99.9% confidence.

Also, this is why quantum mech is so fun (and why it can be a pain in the ass, as I can’t do qm with pencil and paper, unless I have my big books of integrals laying around).