Combinations and Card Drawing

• member 428835
In summary, the probability of choosing 1 Jack from a card deck and then one heart (no replacement) can be calculated by breaking it into two mutually exclusive cases (the first drawn Jack is also a Heart, and the first drawn Jack is not a Heart) and adding their probabilities together. Alternatively, it can be calculated using a mapping from the original deck to a deck where all suits are rotated by one position, which results in the same probability.
member 428835

Homework Statement

What is the probability of choosing 1 Jack from a card deck and then one heart (no replacement)?

Homework Equations

None come to mind.

The Attempt at a Solution

I was thinking ##(4C1)/(52C1)\cdot(13C1)/(51C1)## but what happens if the jack drawn is a heart? Do I make cases? Then I'm thinking something like ##(1C1)/(52C1)\cdot(12C1)/(51C1)## added to the above probability? Please help.

You cannof simply add the HJ case to what you already had since that includes drawing the HJ.
Yes, you could break it into mutually exclusive cases, but there really is no need. On average, how much is the number of Hearts reduced by the drawing of a J? On average, how much is the number of Spades reduced by the drawing of a J?

haruspex said:
On average, how much is the number of Hearts reduced by the drawing of a J?
1/4
haruspex said:
On average, how much is the number of Spades reduced by the drawing of a J?
1/4

I write 1/4 because the odds of a J being a spade that we draw would be one of four suits, right?

joshmccraney said:
1/4
1/4

I write 1/4 because the odds of a J being a spade that we draw would be one of four suits, right?
Right. So after drawing a J, how does the probability that the next card is a Heart compare with the probability that it is a Spade (or any other specific suit)?

haruspex said:
So after drawing a J, how does the probability that the next card is a Heart compare with the probability that it is a Spade (or any other specific suit)?
It is the same, regardless of the suit. Sorry, I'm trying to think instead of being spoon-fed the answer but I'm not seeing where you're going with this.

joshmccraney said:
It is the same, regardless of the suit. Sorry, I'm trying to think instead of being spoon-fed the answer but I'm not seeing where you're going with this.
Right. And what must those four probabilities (the suit of the second card given that the first is a J) add up to?

The four add to 1. The odds of picking a Jack that is heart, spade, diamond, clover are 1.

joshmccraney said:
So what is the probability that the second card is a H given the first was a J?

##1/4 \cdot 1/4##? I'm lost, sorry.

joshmccraney said:
##1/4 \cdot 1/4##? I'm lost, sorry.

There are two cases: (1) The first drawn Jack is also a Heart; and (2) the first drawn Jack is not a Heart. Is the probability of next drawing a Heart the same in cases (1) and (2)? When you say there is "no replacement", doesn't that mean that the first drawn card is now absent from the deck?

joshmccraney said:
##1/4 \cdot 1/4##? I'm lost, sorry.
I see Ray is taking you along the path of breaking it into cases, but I shall persevere with showing it is unnecessary. Please do not be confused by the two approaches.
Notation: let the events be JH that first card is a J and 2nd is a H, JS that the 1st is a J and 2nd a S etc. Similarly conditionals, H|J etc.
You agreed that P(H|J)=P(S|J)=P(D|J)=P(C|J), and that they add up to 1. So what is each of them numerically?

Ray Vickson said:
There are two cases: (1) The first drawn Jack is also a Heart; and (2) the first drawn Jack is not a Heart. Is the probability of next drawing a Heart the same in cases (1) and (2)?
Nope, they are not the same. So would the probability be $$\frac{1C1}{52C1}\frac{12C1}{51C1} + \frac{3C1}{52C1}\frac{13C1}{51C1}$$
Ray Vickson said:
When you say there is "no replacement", doesn't that mean that the first drawn card is now absent from the deck?
Yep.

haruspex said:
Notation: let the events be JH that first card is a J and 2nd is a H, JS that the 1st is a J and 2nd a S etc. Similarly conditionals, H|J etc.
You agreed that P(H|J)=P(S|J)=P(D|J)=P(C|J), and that they add up to 1. So what is each of them numerically?
P(H|J)=P(S|J)=P(D|J)=P(C|J)=(1C1)/(4C1), since each case is identical!

joshmccraney said:
Nope, they are not the same. So would the probability be $$\frac{1C1}{52C1}\frac{12C1}{51C1} + \frac{3C1}{52C1}\frac{13C1}{51C1}$$
Yes. What does that simplify to?

haruspex said:
Yes. What does that simplify to?
1/52

joshmccraney said:
1/52
So the methods agree, right? Always reassuring.

member 428835
Thanks for the help!

And yes, it's reassuring for sure when different approaches yield the same solution.

joshmccraney said:
And yes, it's reassuring for sure when different approaches yield the same solution.
Fwiw, I've just thought of another interesting way to look at it. Construct a mapping from the actual arrangement of cards in the deck to one in which all the denominations are the same at the various positions (e.g. if the original deck goes Q383A... then so does the mapped deck) but the suits are all rotated around by one position (so if the original deck goes HHSCDS... then the mapped deck goes HSCDS...H.)
The event of drawing a J then a H in the first deck maps to the event of drawing the HJ straight away in the second.
The mapping is clearly 1-1 onto (a bijection).

haruspex said:
Fwiw, I've just thought of another interesting way to look at it. Construct a mapping from the actual arrangement of cards in the deck to one in which all the denominations are the same at the various positions (e.g. if the original deck goes Q383A... then so does the mapped deck) but the suits are all rotated around by one position (so if the original deck goes HHSCDS... then the mapped deck goes HSCDS...H.)
The event of drawing a J then a H in the first deck maps to the event of drawing the HJ straight away in the second.
The mapping is clearly 1-1 onto (a bijection).
That is interesting, but I'm not seeing the relevance. Could you explain please?

joshmccraney said:
That is interesting, but I'm not seeing the relevance. Could you explain please?
It means that the probability of drawing a J then a H is the same as the probability of drawing the HJ in a single draw, so 1/52.

member 428835
haruspex said:
It means that the probability of drawing a J then a H is the same as the probability of drawing the HJ in a single draw, so 1/52.
Could you explain it to me please? I'm really interested in this!

joshmccraney said:
Could you explain it to me please? I'm really interested in this!
Not sure that I can think of any clearer way of putting it.
I set up a 1-1 mapping, call it f, between decks. (By deck, I mean a specific ordering of all 52 cards.)
So if the deck A starts SQ, D7, D9, H2,... then f(A) starts DQ, D7, H9, ... and finishes with a S.
Consider the deck properties
P = that the deck starts with a Jack, followed by a Heart;
Q = that the deck starts with HJ.
P(A) if and only if Q(f(A)).
So the number of decks satisfying P is the same as the number satisfying Q.

member 428835
Thanks!

1. What is a combination in card drawing?

A combination in card drawing refers to a set of cards that are chosen or dealt from a deck according to a specific rule or criteria. The order in which the cards are drawn does not matter in a combination. For example, in a game of poker, a combination could be a hand of five cards with three of a kind.

2. How many combinations are possible in a deck of cards?

In a standard deck of 52 cards, there are 2,598,960 possible combinations. This number can be calculated by taking the total number of cards (52) and calculating the number of unique combinations that can be made, which is 52 choose 5.

3. What is the difference between a combination and a permutation in card drawing?

The main difference between a combination and a permutation in card drawing is the order in which the cards are drawn. In a combination, the order does not matter, whereas in a permutation, the order does matter. For example, in a game of poker, a combination could be a hand of five cards with three of a kind, while a permutation could be a hand of five cards with a specific sequence, like a straight.

4. How can combinations be used to calculate probabilities in card games?

Combinations can be used to calculate probabilities in card games by dividing the number of desired combinations by the total number of possible combinations. For example, in a game of blackjack, the probability of being dealt a blackjack (an ace and a card with a value of 10) would be 4 (number of aces) x 16 (number of 10-value cards) / 2,598,960 (total combinations) = 0.024%. This can help players make informed decisions based on the likelihood of certain combinations.

5. Can combinations be applied to other areas besides card games?

Yes, combinations can be applied to a variety of fields, including mathematics, statistics, and computer science. In mathematics, combinations can be used to calculate the number of ways to choose a group of objects from a larger set. In statistics, combinations can be used to calculate the probability of certain events occurring. In computer science, combinations can be used in algorithms and data structures to solve problems efficiently.

• Precalculus Mathematics Homework Help
Replies
5
Views
2K
• Precalculus Mathematics Homework Help
Replies
2
Views
850
• Precalculus Mathematics Homework Help
Replies
12
Views
1K
• Precalculus Mathematics Homework Help
Replies
2
Views
2K
• Precalculus Mathematics Homework Help
Replies
6
Views
1K
• Precalculus Mathematics Homework Help
Replies
10
Views
3K
• Set Theory, Logic, Probability, Statistics
Replies
9
Views
1K
• Engineering and Comp Sci Homework Help
Replies
11
Views
1K
• Precalculus Mathematics Homework Help
Replies
5
Views
3K
• Set Theory, Logic, Probability, Statistics
Replies
7
Views
1K