You are dealt 2 cards at random from a standard 52 card deck. What is the probability of getting a blackjack?
Note: A blackjack is a combination of an Ace and a ten point card. All face cards are worth ten points, and any ten card is worth ten points.
+633 Kind of want to do this calculation myself so I'll hop right on board.
The first card we get can be any card in the deck. Let's assume we get the Ace first, or it won't be a blackjack. (This is for purpose of demonstration, and an optimization will be shown later.)
The odds of getting the card we need are 4 in 52, or 1/13.
Now we have 51 cards that we do not know. As long as we don't know what they are, we can say it does not matter if others have been dealt out.
In this case, the odds of getting the 10 card we need, are 16 in 51.
So, the odds of this are 1/13 * 16/51 = 16/663.
Now, let's go over to the inverse: getting the 10 card first. In this case the odds are 4/13, then 4/51.
In this case, we still get 16/663.
... wait a second... they're the same probability?
Well of course! This isn't a permutation based chance, it's a combination based one. Combination based chances will generally be commutable, and can be shortcut in a few ways. I don't know how to write one for this, but because we have 2 permutations ("2 pick 2") of this, we could have just mulitplied the first answer by 2.
Thus, the odds are 32/663.
You have the right answer..., but after reading this meandering swirl of BS, I don’t want to play Blackjack. I want to play Russian Roulette –the original version, where five rounds are loaded and only one chamber is empty. You go first!
I’m 83.3% sure you’ll get a bang out of it.
GA
