given that the first two cards are kings, what is the probability that two of the next three are also kings

Question

a poker player is being dealt 5 cards from a standard deck. given that the first two cards are kings, what is the probability that two of the next three are also kings

Answer 1

These discrete probability problems are tricky, and it's very easy to get confused. That being said, I'm afraid I have to disagree with the previous answer.

 

Three more cards are going to be dealt, and the question is asking the for the probability that any two of them are a king. Since there are 50 cards left in the deck, there are

 

50_nC_r3=(50!)/(3!(50-3)!) possible combinations of 3 cards that may form the rest of the hand. This number simplifies to

(50!)/(3!47!)=(50*49*48)/(3*2)=50*49*8. I won't bother with simplifying it further at this point.

 

That is the number of possible 3-card combinations that could be dealt. Now we need the number of 3-card combinations that contain two kings.

 

We count this by counting the number of possible cards that could be the third card. There are 48 non-king cards, so that is how many combinations are possible. The probability is therefore

 

48/(8*49*50)=3/(49*25)=3/1225=.00245 after rounding.

 

If you want to calculate the probability using the order the cards are dealt in, then you would start by using the process the previous tutor demonstrated. You need to also calculate the probability that the first card is a King, the second card is something else and the third card is a king, as well as the probability that the first card is something else and the second and third cards are the kings.

 

So we would have

 

P(First two cards are Kings)+P(First card and third card are kings)+P(second and third cards are Kings)=

(2/50)*(1/49)+(2/50)*(48/49)*(1/48)+(48/50)*(2/49)*(1/48)

In the second term, you have the probability of 2/50 for the first factor of getting a king with 50 cards still in the deck. Then you have the probability of getting something other than a king with 49 cards still in the deck, finally the probability of getting the last king when 48 cards are in the deck. The third factor comes from the same thought process. It simplifies to

 

(1/25)*(1/49)+(1/25)*(1/49)+(1/25)*(1/49)=3/(49*25), the same result I got with the non-ordered method. In each term, a 48 in the numerator canceled a 48 in the denominator and a 2 in the numerator turned a 50 in the denominator into a 25. Hopefully it makes sense.

 

 

Answer 2

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