The Blackjack

What is the probability of getting a natural or blackjack when a player delt two-card total of 21 from a well-shuffled 5-deck shoe?

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Getting exactly 21 from two cards means you must get an ace (ace can take 1 or 11 at the player’s discretion) and a ten-count card (number 10 or a face card). We will estimate the probability by two different methods.

Method 1

From cards from n-decks, the probability of getting an ace in the first deal followed by a 10 in the second is given by:

\frac{4n}{52n}*\frac{16n}{52n-1}

Similarly, the probability of having a 10-card followed by an ace is:

\frac{16n}{52n}*\frac{4n}{52n-1}

Combining the two, you get the total probability of natural.

P(\text{Natural}) = \frac{4n}{52n}*\frac{16n}{52n-1} + \frac{16n}{52n}*\frac{4n}{52n-1} = \frac{32n}{13(52n-1)}

For a 5-deck shoe, substitute n = 5, and you get 0.0475 or 4.75%.

Method 2

We will use the familiar combinations formula to get the same.

P(\text{Natural}) = \frac{_{4n}C_1*_{16n}C_1}{_{52n}C_2} = \frac{20*80}{33670} = 0.04752 (\text{ for n = 5})

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