Combinations and Card Drawing

  • #1
joshmccraney
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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.
 

Answers and Replies

  • #2
haruspex
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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?
 
  • #3
joshmccraney
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On average, how much is the number of Hearts reduced by the drawing of a J?
1/4
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?
 
  • #4
haruspex
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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)?
 
  • #5
joshmccraney
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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.
 
  • #6
haruspex
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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?
 
  • #7
joshmccraney
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The four add to 1. The odds of picking a Jack that is heart, spade, diamond, clover are 1.
 
  • #8
haruspex
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The four add to 1.
So what is the probability that the second card is a H given the first was a J?
 
  • #9
joshmccraney
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##1/4 \cdot 1/4##? I'm lost, sorry.
 
  • #10
Ray Vickson
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##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?
 
  • #11
haruspex
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##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?
 
  • #12
joshmccraney
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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 $$\frac{1C1}{52C1}\frac{12C1}{51C1} + \frac{3C1}{52C1}\frac{13C1}{51C1}$$


When you say there is "no replacement", doesn't that mean that the first drawn card is now absent from the deck?
Yep.

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!
 
  • #13
haruspex
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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?
 
  • #16
joshmccraney
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Thanks for the help!
 
  • #17
joshmccraney
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And yes, it's reassuring for sure when different approaches yield the same solution.
 
  • #18
haruspex
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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).
 
  • #19
joshmccraney
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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?
 
  • #20
haruspex
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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.
 
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  • #21
joshmccraney
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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!
 
  • #22
haruspex
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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.
 
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  • #23
joshmccraney
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Thanks!
 

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