3 of a kind/ A pair: Say there are 3 Aces 3 Kings and 3 Queens...

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Probable

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Hi!
Pardon my ignorance. I am not good at math. Say there are 3 Aces 3 Kings and 3 Queens. You shuffle them and place them down on the table. You are allowed to turn over 3 cards. What is the probabilty of getting 3 of a kind? Is it 33 percent. What is the probability of getting a pair? Cheers for any explanation. Chris
 
Probable said:
Hi!
Pardon my ignorance. I am not good at math. Say there are 3 Aces 3 Kings and 3 Queens. You shuffle them and place them down on the table. You are allowed to turn over 3 cards. What is the probabilty of getting 3 of a kind? Is it 33 percent. What is the probability of getting a pair? Cheers for any explanation. Chris
Click to expand...

You say you are not good at math; you don't say why you are asking. Are you taking a class in probability, or just curious for some reason?
 
Probable said:
Hi!
Pardon my ignorance. I am not good at math. Say there are 3 Aces 3 Kings and 3 Queens. You shuffle them and place them down on the table. You are allowed to turn over 3 cards. What is the probabilty of getting 3 of a kind? Is it 33 percent. What is the probability of getting a pair? Cheers for any explanation. Chris
Click to expand...

I'll assume that you want to learn a little about probability, and perhaps have learned some, so a hint will be appropriate. Neither of these questions is very difficult, though neither is as obvious as your guess of 1/3.

To find P(3 of a kind), you can calculate (number of ways to choose 3 cards of the same type)/(number of ways to choose 3 of 9 cards). This is easy if you know about combinations. Give it a try.

To find P(at least two cards alike), you can do something similar, but it is easiest if you first think about P(no two cards alike).
 
Probable said:
Hi!
Pardon my ignorance. I am not good at math. Say there are 3 Aces 3 Kings and 3 Queens. You shuffle them and place them down on the table. You are allowed to turn over 3 cards. What is the probabilty of getting 3 of a kind? Is it 33 percent. What is the probability of getting a pair? Cheers for any explanation. Chris
Click to expand...
You really don't care what the first card is: any type will do. Let's identify the type that you turn over as the desired type. So there are eight cards left to choose from, and only two are of the desired type. So the probability that you will pick one of the desired type on your second pick is

\(\displaystyle \dfrac{2}{8} = \dfrac{1}{4}.\)

Now you have seven cards left, and only one is of the desired type so the probability of success on the third card, given success on the second card, is

\(\displaystyle \dfrac{1}{7}.\)

Thus the probability of three of a kind is

\(\displaystyle \dfrac{1}{4} * \dfrac{1}{7} = \dfrac{1}{28} \approx 3.6\%.\) Not so good.

As for an exact pair, you once again do not care what type turns up on the first card. Again, let's call that the desired type. We know that the probability of getting the desired type on on the second card is 1/4 from the prior analysis (two of the desired type in the eight cards remaining). If that happens we have seven cards left, six of which are of the undesired type. But now we want the third card to be of an undesired type (because we want a pair, not a triplet.) So that probability is 6/7. However, the probability of getting an undesired type on the second card is

\(\displaystyle \dfrac{6}{8} = \dfrac{3}{4}.\)

But now we have two desired types, that of the first card and that of the second card. Out of seven remaining cards, four are of a desirable type. So the overall probability of a pair is

\(\displaystyle \left ( \dfrac{1}{4} * \dfrac{6}{7} \right ) + \left ( \dfrac{3}{4} * \dfrac{4}{7} \right ) = \dfrac{6 + 12}{28} = \dfrac{18}{28} \approx 64.3\%.\)

Those don't add up to 100% because we have ignored the possibility of one of each type.

Again, we do not care what type we get on the first card. There are six non-matching cards of the eight remaining. So the probability of getting a non-match on the second card is 6/8 = 3/4. There are three cards of the third type left out of the seven remaining. So the probability of one of each type

\(\displaystyle \dfrac{3}{4} * \dfrac{3}{7} = \dfrac{9}{28} \approx 32.1\%.\)

\(\displaystyle \dfrac{1}{28} + \dfrac{18}{28} + \dfrac{9}{28} = \dfrac{28}{28} = 100\%.\)

The math is not hard. It is making sure that you count all the possibilities that can become hard.
 
Dr.Peterson said:
I'll assume that you want to learn a little about probability, and perhaps have learned some, so a hint will be appropriate. Neither of these questions is very difficult, though neither is as obvious as your guess of 1/3.

To find P(3 of a kind), you can calculate (number of ways to choose 3 cards of the same type)/(number of ways to choose 3 of 9 cards). This is easy if you know about combinations. Give it a try.

To find P(at least two cards alike), you can do something similar, but it is easiest if you first think about P(no two cards alike).
Click to expand...

Since you say this is not a homework problem, and others have shown the answer already, I'll show what I meant here.

The number of ways to choose 3 of 9 cards is represented by 9C3, or \(\displaystyle \binom{9}{3}\), which can be calculated by a simple formula, or by a button on many calculators. In this case, the value is \(\displaystyle \frac{9\cdot8\cdot7}{3\cdot2\cdot1} = 84\).

The number of ways to choose 3 cards of the same type is 3: they can be the three aces, or kings, or queens.

So the probability of three of a kind is 3/84 = 1/28.

The probability that at least two cards are alike (which is different from exactly two cards alike -- this means "a pair or better") is 1 minus the probability that none are alike. The denominator is the same; the numerator is the number of ways to choose one ace, one king, and one queen, which is \(\displaystyle 3\cdot3\cdot3 = 27\). So the probability is 1 - 27/84 = 57/84 = 19/28.
 
There are a total of 9 cards, 3 queens, 3 kings and 3 aces.

The probability the first card turned over is an ace is 3/9= 1/3. There are then 8 cards left, 2 of them aces. The probability the second card turned over is also an ace is 2/8= 1/4. There are then 7 cards left, 1 of them an ace. The probability the third card turned over is the last ace is 1/7. The probability the three cards turned over are all aces is (1/3)(1/4)(1/7)= 1/84.


It should be clear that the probabilities of "three queens" or "three kings" are also 1/84. The probability that the three cards are the same is 1/84+ 1/84+ 1/84= 3(1/84)= 1/28.
 
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