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Q. A standard deck of playing cards has 52 cards. There are four suit (clubs, diamonds, hearts and spades), each of which has thirteen numbered cards ( $2, \ldots . ., 9,10$, Jack, Queen, King, Ace)
In a game of card, each card is worth an amount of points. Each numbered card is worth its number (e.g. a 5 is worth 5 points); the Jack, Queen and King are each worth 10 points ; and the Ace is worth your choice of "either 1 point or 11 points". The object of the game is to have more points in your set of cards than your opponent without going over 21.Any set of cards with sum greater than 21 automatically loses. Here's how the game is played. You and your opponent are each dealt two cards. Usually the first card for each player is dealt face down, and the second card for each player is dealt face up.After the initial cards are dealt, the first player has the option of asking for another card or not taking any cards. The first player can keep asking for more cards until either he or she goes over 21 , in which case the player loses, or stops at some number less than or equal to 21 . When the first player stops at some number less than or equal to 21 , the second player then can take more cards until matching or exceeding the first player's number without going over 21 , in which case the second player wins, or until going over 21 , in which case the first player wins.
We are going to simplify the game a little and assume that all cards are dealt face up, so that all cards are visible. Assume your opponent is dealt cards and plays first.
Suppose your opponent is dealt a King and a 10, and you are dealt a Queen and a 9. Being smart, your opponent does not take any more cards and stays at 20 . The chance that you will win if you are allowed to take as many cards as you need, is

Probability - Part 2

Solution:

$P($ win $)=P($ any 2 's or Ace $)=\frac{4}{48}+\frac{4}{48}=\frac{1}{6}$