“Read” and black? Or is this part of a joke concerning a sunburned zebra?
I won’t take any bet anyone proposes, so no.
My statistics is fuzzy, but the standard deviation appears to be 2.2 hands, so the bet is definitely worth it.
Awesome. Let’s play. Send me a dollar and we can get started.
Except that the solution doesn’t require statistics.
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?
Thanks for the fix. Always nice to know that the staff is paying attention.
“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.”
Dude, you got there before I did!
Also, never play cards with a man named Doc.
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.
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.
You are missing a lot of combinations, e.g.
would just be one starting order satisfying the 26 R/B
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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