Probability of being dealt Pinochle if playing 4-person game

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xoticaox

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The card game pinochle is played with two standard decks of cards, but with the 2’s, 3’s, 4’s, 5’s, 6’s, 7’s, and 8’s removed. Thus, there are 48 cards used: 2 cards each of the aces, kings, queens, jacks, tens, and nines. In the four-player version of pinochle, each player is dealt 12 cards.

(a) If you are playing 4-person pinochle, determine the probability that you will be dealt a “pinochle”, which is a Q of hearts and a J of diamonds}.

(b) If you are playing 4-person pinochle, determine the probability that you will be dealt “aces around”, which is an A of spades, an A of hearts, an A of diamonds, and an A of clubs.

Can someone please explain how i can arrive at the solutions of this problem. Thanks
 
Re: Probability of Pinochle

xoticaox said:
(a) If you are playing 4-person pinochle, determine the probability that you will be dealt a “pinochle”, which is a Q of hearts and a J of diamonds}.
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a) Probability of getting one of the four desirable cards = 4/48 (Now there are only two desirable card left)

Probability of getting one of the two desirable cards = 2/47

Now continue....

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.
 
Re: Probability of Pinochle

The card game pinochle is played with two standard decks of cards, but with the 2’s, 3’s, 4’s, 5’s, 6’s, 7’s, and 8’s removed. Thus, there are 48 cards used: 2 cards each of the aces, kings, queens, jacks, tens, and nines. In the four-player version of pinochle, each player is dealt 12 cards.


(b) If you are playing 4-person pinochle, determine the probability that you will be dealt “aces around”, which is an A of spades, an A of hearts, an A of diamonds, and an A of clubs.

Can someone please explain how i can arrive at the solution of this problem. Thanks

Work done: how do i get 4 aces if thre are 2 copies of each ace. I would assume 4/48 x 3/47 x 2/46 x 1/45 for the proabability of getting all aces. Just confused about the 2 copies of cards part.
 
Re: Probability of Pinochle

Did you understand the solution to the part (a)?

What answer did you get for part (a)?

Part(b) is very similar to part (a).
 
Re: Probability of Pinochle

I just multiplied 4/48 x 2/47 to get the probability as being 1/282. Is that correct? And I just told you in the post before how i'd approach b but im still confused can you explain part b please.
 
Re: Probability of Pinochle

xoticaox said:
I just multiplied 4/48 x 2/47 to get the probability as being 1/282. Is that correct?

Yes it is correct -

however did you understand the logic behind those numbers -

more specifically why is it 2/47

(why the numerator is 2 and denominator is 47?

Why does the numerator jump from 4 to 2 - after choosing one -1- card?)

Either you don't understand the logic - or you are not thinking - because your thought process does not follow from process above
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Re: Probability of Pinochle

I do understand why is 4 and 2 in the numerator i just dont understand why that is the only probability dont u have to take into account there are 12 cards dealt??????

Usually one does these things using the combinations formula, nCk. Thus, there are 48C12 hands overall, and this would be a denominator for a probability. There are 2C1 (or 2) ways to get a Qh and 2 ways to get a Jd, and then 44C10 ways to finish the hand, assuming you get only one Qh and one Jd. Add to this the number of ways to get 2 Qh and 1 Jd, the number for 2 Jd and one Qh, and the ways to get 2 Qh, 2 Jd. HOW BOUT THAT??
 
Re: Probability of Pinochle

For part b the chance of getting an ace of hearts is 2/48 to get and ace of diamonds is 2/47 to get an ace of clubs is 2/46 and to get an ace of spades is 2/45 right?
 
xoticaox said:
HOW BOUT THAT??
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How about what? You appear to have copied potentially-useful text from somewhere (the formatting clearly doesn't match your own "voice"), but did you understand what you copied?

Please reply with clarification. Thank you! :D

Eliz.
 
Re: Probability of Pinochle

Subhotosh Khan said:
a) Probability of getting one of the four desirable cards = 4/48 (Now there are only two desirable card left)

Probability of getting one of the two desirable cards = 2/47

Now continue....
Click to expand...

This means the probability would be (4/48) (2/47) whether we're dealing out all 48 cards to two players or 12 players. That just doesn't seem right. The probability should be higher if all 48 cards are dealt to just two players than what it would be if there are 12 players, no? That would be the probability if those were the only two cards dealt to each player.

However, that is not the scenario for this particular problem. In this game, one of the four players HAS to get the first good card, because all cards are dealt. That is a certainty (probability = 1). Then, the probability that it will be YOU is 1/4. And finally, if YOU have the first good card (there are four in the deck), what is the probability that YOU will get one of the two remaining good cards? Keep in mind, you've got 12 chances to get these cards. However, since you already have used up one of those chances in getting the first good card, I think that means you only have 11 chances to get one of the remaining good cards. We aren't just drawing two cards from a deck of 48; we're drawing 12 in a situation where we only care about two of them (for part (a), anyway). Part (a) solution may be closer to (1/4) * (chances of drawing at least one of two desirable cards in a deck of 47, given 11 chances ... I leave this part to you). ???

The combinatorics solution is the way to go in the general case, I think. It's just that there's a small flaw in the logic with what the student wrote. Why would you exclude certain hands from the numerator only to add them back in later? That sort of indicated you don't understand why you're doing this. The same logic would apply for the aces around part, I think.
 
Tsk tsk tsk. You shouldn't be asking that. Think of the code.

Next time, use the internet to check your answer, not outright ask for it.
 
AP high school math is fun as long as your teacher shows up to class and can explain things!
 
xoticaox said:
AP high school math is fun as long as your teacher shows up to class and can explain things!
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Helps if the student also shows up.
 
thechickenknows said:
Tsk tsk tsk. You shouldn't be asking that. Think of the code.

Next time, use the internet to check your answer, not outright ask for it.
Click to expand...

X, are you done with this problem, or do you still need more input from us?

Ted, what farmer forgot to tell the all-knowing chicken to stay in the coop?

Chicken, this IS the internet, or are you lost in a parallel dimension somewhere? If the student shows some initiative, we will help. Read "Read Before Posting" next time, before you post. That is our "code."
 
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