Blackjack, or twenty-one as it is frequently called, is a...

Blackjack, or twenty-one as it is frequently called, is a popular gambling game played in Las Vegas casinos. A player is dealt two cards. Face cards, queens, and kings) and tens have a point value of $10 .$ Aces have a point value of 1 or $11 .$ A 52 -card deck contains 16 cards with a point value of 10 jacks, queens, kings, and tens) and four aces. a. What is the probability that both cards dealt are aces or 10 -point cards? b. What is the probability that both of the cards are aces? c. What is the probability that both of the cards have a point value of 10$?$ d. A blackjack is a 10 -point card and an ace for a value of $21 .$ Use your answers to parts $($ a), $(b),$ and $(c)$ to determine the probability that a player is dealt blackjack. (Hint: Part (d) is not a hypergeometric problem. Develop your own logical relationship as to how the hypergeometric probabilities from parts $(a),(b),$ and $(c)$ can be combined to answer this question.)

Blackjack, or twenty-one as it is frequently called, is a popular gambling game played in Las Vegas casinos. A player is dealt two cards. Face cards, queens, and kings) and tens have a point value of $10 .$ Aces have a point value of 1 or $11 .$ A 52 -card deck contains 16 cards with a point value of 10 jacks, queens, kings, and tens) and four aces.
a. What is the probability that both cards dealt are aces or 10 -point cards?
b. What is the probability that both of the cards are aces?
c. What is the probability that both of the cards have a point value of 10$?$
d. A blackjack is a 10 -point card and an ace for a value of $21 .$ Use your answers to parts $($ a), $(b),$ and $(c)$ to determine the probability that a player is dealt blackjack. (Hint: Part (d) is not a hypergeometric problem. Develop your own logical relationship as to how the hypergeometric probabilities from parts $(a),(b),$ and $(c)$ can be combined to answer this question.)

Key Concepts

Combinatorial Analysis

This concept involves counting the number of ways events can occur, typically using combinations or permutations. In probability problems, especially those with a finite sample space, combinatorial analysis helps determine the total number of possible outcomes as well as the number of favorable outcomes.

Hypergeometric Distribution

The hypergeometric distribution models the probability of drawing a specific number of successes from a finite population without replacement. It is essential for scenarios where the probability changes as items are removed from the sample space, such as dealing cards from a deck.

Sampling Without Replacement

This principle refers to selecting items from a finite set where each item cannot be chosen more than once. In probability calculations, sampling without replacement alters the probability of subsequent events since the total number of outcomes decreases with each draw.

Conditional Probability

Conditional probability is used to calculate the probability of an event occurring given that another event has already occurred. In sequential events like successive card draws, the outcome of the first draw influences the probabilities in subsequent draws, making this concept critical for accurate calculations.

Transcript

00:01 In problem number 49, we're asked to analyze a standard deck of cards, and we're asked to find the probability for different scenarios.

00:08 So the first scenario is that the probability that both cards are aces or ten point cards.

00:14 And in the prompt, it told you that there were 16 ten point cards, and then there are four aces.

00:23 And if you didn't know, there are 52 cards total in a deck.

00:28 So keep that in the back of your mind.

00:31 All right, we're going to need to know that.

00:32 So there's a couple ways to go about this.

00:34 There's the formula way and then there's kind of a logical way.

00:37 So i'll show you the formula way.

00:39 So here there are 20, you can think of like favorable.

00:42 There are 20 that you want or the probability that both cards or aces or 10.

00:47 So 20 and it says that we're selecting two.

00:50 So both cards.

00:51 So we're choosing two.

00:54 And then there are 32 non 10 point or aces.

00:59 So that's the 23, 4, 5, 6, 7, 8, 9 for all 4.

01:03 Suits there are 32 of them and we're choosing zero of those and then that's divided by 52 choose two now a way to double check to make sure that you do the formula right 20 plus 32 is 52 and then two plus zero is two okay so then you can use the formula or i like to use the calculator personally so if you go to 20 and then if you have a t -i -84 you go to math and then you go to probability and it's that ncr.

01:33 So 20 choose 2 and then times 32 math choose 0 and then that's divided by 52 math choose 2.

01:57 And that gives us about 0 .1433 let's say.

02:08 And that is the correct answer.

02:10 Now also if you think of it, you know, logically we're assuming this is without replacement.

02:15 So you can kind of say like or.

02:18 There are 20 possible choices out of the 52 for your first selection.

02:22 And then whenever you draw that 10 or ace or king or queen or, you know, of any suit, then there are 19 left over out of a total of 21.

02:29 And whenever you do that, you take 20 times 19 divided by 52 times, i'm sorry, not 21, that should say 51.

02:37 51.

02:39 That gives you the same thing.

02:40 So basically, you know, exactly the same way, just kind of the different way to think about it.

02:46 So then the second part of this, it says what's, find the probability that both cards are aces.

02:52 All right...

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