What is the probability that a hand of five cards will contain at least one king?
Another way to get CPhill's answer:
Still using his idea of subtracting the probability of getting no kings from 1, the total possible probability.
The only way that you can get no kings is:
to get no king for the first card: probability = 48/52
to get no king for the second card: probability = 47/51
to get no king for the third card: probability = 46/50
to get no king for the fourth card: probability = 45/49
and, finally:
to get no king for the fifth, and last, card: probability = 44/48
Multiplying all these probabilities together: 48/52 x 47/51 x 46/50 x 45/49 x 44/ 48 = 0.6588
Subtracting this from 1.000: 1.000 - 0.6588 = 0.341
2 jokers, 4 aces, 4 2's, 4 3's, 4 4's, 4 5's, 4 6's, 4 7's, 4 8's, 4 9's, 4 10's, 4 jacks, 4 queens, and 4 kings
I'm not good with probabiltiy...but in 54 cards....there are 4 kings, so maybe 4/25 because their are 5 cards in the hand....like I said, not good, giving my best try
What's up with all these probability problems?
This equals..... [1 - probability of no kings ]
So, the ways to choose no kings is to just select 5 cards from the remaining 48
This is ginen by C(48,5) = 1712304
And the total way of selecting 5 cards from 52 is C(52,5) = 2598960
So....the probability of at least one king is
1 - 1712304/2598960 = about 34.1%
Another way to get CPhill's answer:
Still using his idea of subtracting the probability of getting no kings from 1, the total possible probability.
The only way that you can get no kings is:
to get no king for the first card: probability = 48/52
to get no king for the second card: probability = 47/51
to get no king for the third card: probability = 46/50
to get no king for the fourth card: probability = 45/49
and, finally:
to get no king for the fifth, and last, card: probability = 44/48
Multiplying all these probabilities together: 48/52 x 47/51 x 46/50 x 45/49 x 44/ 48 = 0.6588
Subtracting this from 1.000: 1.000 - 0.6588 = 0.341