Probability help with deck of cards am I going in the right direction?

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calcnoob145

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No jokers in the decks

Say a person has to choose between two decks. One deck however has 51 cards because the King of Diamonds has been taken out of that deck. This person however does not know which deck has just 51 in it. So they choose one of the decks. Then this person is dealt a card from that deck.

What would the probability be that this person gets an Ace of Spades?



So here is where I am going if I was going to solve this problem.

Ok so if both decks had 52, then it doesn't really matter which deck that person chooses,
the probability would be (1/52) since there is only one card that is an Ace of Spades out of the entire 52 cards.




But since one deck has 51 and this person doesn't know which deck is one less, then the probability is (103/2652) that the card they are dealt with is an Ace of Spades.

Ok here is my reasoning. I got this because when that person is dealt the card after they chose the deck, the probability it is an Ace of Spade is 1/52 if he chooses the deck with 52 cards or 1/51 if he chooses the deck with the 51 cards. So I added (1/52)+(1/51)=(103/2652).



Do I have this concept understood or am I off.

Does the person not knowing which deck is which affect the probability. Say if they did know which deck is which, say Deck 1 has 52 and Deck 2 has 51, and they chose Deck 2 knowing that it has the King of Diamond card removed, then would their probability of getting an Ace of Spade equal (1/51).
 
Last edited:
calcnoob145 said:
No jokers in the decks

Say a person has to choose between two decks. One deck however has 51 cards because the King of Diamonds has been taken out of that deck. This person however does not know which deck has just 51 in it. So they choose one of the decks. Then this person is dealt a card from that deck.

What would the probability be that this person gets an Ace of Spades?



So here is where I am going if I was going to solve this problem.

Ok so if both decks had 52, then it doesn't really matter which deck that person chooses,
the probability would be (1/52) since there is only one card that is an Ace of Spades out of the entire 52 cards.




But since one deck has 51 and this person doesn't know which deck is one less, then the probability is (103/2652) that the card they are dealt with is an Ace of Spades.

Ok here is my reasoning. I got this because when that person is dealt the card after they chose the deck, the probability it is an Ace of Spade is 1/52 if he chooses the deck with 52 cards or 1/51 if he chooses the deck with the 51 cards. So I added (1/52)+(1/51)=(103/2652).



Do I have this concept understood or am I off.

Does the person not knowing which deck is which affect the probability. Say if they did know which deck is which, say Deck 1 has 52 and Deck 2 has 51, and they chose Deck 2 knowing that it has the King of Diamond card removed, then would their probability of getting an Ace of Spade equal (1/51).
Click to expand...

A simple way to approach this problem is with a probability tree. You can see an example of a probability tree (for a different problem) here:

http://www.amstat.org/publications/jse/v14n1/larsen_figure1.jpg

The first two branches of your tree would be selecting one of the two decks, .5 probability each.

From the ends of those two branches make two more branches. These represent the drawing of a card. The probabilities are 1/52 (the Ace of spades) and 51/52 (not the Ace of spades); and 1/51 (the Ace of spades) and 50/51 (not the Ace of spades).

So, we can see that there are two paths that result in drawing the Ace of Spades. The probability of following the first path is (.5)(1/52) and the probability along the second path is (.5)(1/51). Add the two probabilities together for your answer.
 
calcnoob145 said:
No jokers in the decks
Say a person has to choose between two decks. One deck however has 51 cards because the King of Diamonds has been taken out of that deck. This person however does not know which deck has just 51 in it. So they choose one of the decks. Then this person is dealt a card from that deck.
What would the probability be that this person gets an Ace of Spades? .
Click to expand...
This is a rather sparse question. So we have the make several assumptions.
Assume that the person is equally likely to choose either deck. Each card is equally likely to be dealt from the chosen deck.

Let \(\displaystyle A\) be the event that the deck of fifty one; \(\displaystyle B\) be the event that the deck of fifty two; and \(\displaystyle S\) be the event that the ace-of-spades is dealt.
\(\displaystyle \begin{align*} \mathcal{P}(S)&=\mathcal{P}(C\cap A)+ \mathcal{P}(S\cap B)\\ &= \mathcal(P)(A)\mathcal{P}(S|A)+ \mathcal(P)(B)\mathcal{P}(S|B)\\&=(0.5)\left( {{1 \over {51}}} \right)+(0.5) \left( {{1 \over {52}}} \right)\end{align*}\)
 
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