Will you play the "Red and Black" card game?

boing
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“Read” and black? Or is this part of a joke concerning a sunburned zebra?

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First thought.

I won’t take any bet anyone proposes, so no.

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My statistics is fuzzy, but the standard deviation appears to be 2.2 hands, so the bet is definitely worth it.

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Awesome. Let’s play. Send me a dollar and we can get started.

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Except that the solution doesn’t require statistics.

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No stats involved. No matter what, you have 52 cards in a deck, 26 R, 26 B. If you pull B/B for a pair, at some point you’ll have to pull a corresponding R/R, leaving the two players even. The R/B pairs will just winnow down the deck.

For the stats folks: if the rules were changed so that if 26 R/Bs are indeed pulled (leaving both players with none), the challenger offers a “grand prize” payout of $15, would you take that bet?

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Thanks for the fix. Always nice to know that the staff is paying attention.

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“But, son, do not accept this bet, because as sure as you stand there, you’re going to wind up with an ear full of cider.”

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Dude, you got there before I did!

Also, never play cards with a man named Doc.

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Edit: of course @SteveBosman is correct. Damn my midnight itchy trigger finger. Corrections made below.

There are 10914854252082341062587699469448622479085492240384000000000000 out of 80658175170943878571660636856403766975289505440883277824000000000000 ways of doing this. So a probability of 8388608/61989816618513 (approx 0.000000135322355470475) assuming all shuffles are equally likely.

So, not unless the jackpot was about $7,389,762.

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How about if the jokers were left in? If I buy an ‘ordinary deck of cards’, it usually has a couple. And an incomprehensible Bridge scoring chart.

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You are missing a lot of combinations, e.g.

Red-Black-Black-Red-Black-Red-Red-Black…

would just be one starting order satisfying the 26 R/B

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Bart: Pick the red, get ahead. Pick the black, set you back.
Principal Skinner: Hmm. I don’t recall authorizing this booth.
Bart: Good-bye, gentlemen.
disappears in smoke

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