deck of 52 cards, 3 cards are dealt what is the probability that 2 r of one suit and the other is of another?

Question

use combinations

this might help P(2 same and 1 different)

Answer 1

The answers so far are incorrect.
The correct answer is 12168 / 22100 = 234 / 425 ≈ 55%

Instead of working with probabilities the following uses combinitorics (as asked) to count the number of possible hands of each type.
There are C(52,3) = 22100 possible 3-card hands.
The number of these with two of one suit and the other is different is:
4 * C(13,2) * 39 = 12168
(There are four suits to choose for the pair. Then, we choose the combination of the 13 cards in that suit 2 at a time. Then, there are 39 possible cards remaining not in the same suit as the pair.)

In fact, there are only three possibilities:
All three are of different suits.
2 are of the same suit, one different.
All are of the same suit.

# hands with all three different = C(4,3) * 13^3 = 8788
(Choose 3 out of 4 suits, and then 13 choices for each of the suits)
# hands with 2 of one, one different = 12168
# hands all same = 4 * C(13,3) = 1144
(Four choices for the suit, then the combination of 13 taken 3 at a time.)

Note that 8788 + 12168 + 1144 = 22100.

Bramblyspam, the third answerer, has the correct reasoning, however, his arithmetic is flawed. (39*36)/(51*50) = 234/425, the number I have just proven to be correct, and not 221/450 as he stated.

Answer 2

a 1 2 3 4 5 6 7 8 9 10 11 x x
b 1 2 3 4 5 6 7 8 9 10 11 12 13
c 1 2 3 4 5 6 7 8 9 10 11 12 13
d 1 2 3 4 5 6 7 8 9 10 11 12 13

four suits a b c d, 13 per suit. 52 total.

start with first card, does not matter what suit it was,
then you have a 12/51 chance of it being the same suit,
for it to be differen suit it will be 39/50.

1 * (12/(52-1)) * (39/(52-2)) = .1835

Answer 3

The first card is whatever it is.
Suppose the second card is the same suit as the first.
The odds of this happening are (12/51).
In this situation, the third card must be of some different suit.
The odds of that happening are (39/50). That's 1 - (11/50).
Overall probability of this scenario: (12/51) * (39/50)

Suppose the second card is a different suit from the first.
The odds of that happening are (39/51)
In this situation, the third card must be of the same suit as the first or second card.
The odds of that happening are (24/50).
Overall probability of this scenario: (39/51) * (24/50)

So the final answer is: (12/51) * (39/50) + (39/51) * (24/50)
That's (12*39 + 39*24) / (51*50)
That's (39*36) / (51 * 50)
Which equals 221/450, or approximately 0.4911

Alternatively, you can calculate it this way:
(52/52) * (12/51) * (39/50) * 3
The 3 is necessary because there are three different permutations possible: the differently-suited card can be the first card, the second card, or the third card.
This also gives you (39*36)/(51*50), which leads to the same answer as above.

I'll let you figure out which approach makes more sense to you. In any case, the correct answer is 221/450.

Hopefully that helps!

Answer 4

The first card can be any card 1 *the second card with same suit 1/12*the third card of another suit 1/39*how many ways it can be dealt (IE. odd suit first then 2of the same suit)3=14.04%

Answer 5

13 per suit
1/13 first, then 1/12, then 1/39 for the 3rd card of different suit. Combine.