# Thread: Probability of drawing a king on the 13th draw from a carddeck

1. ## Probability of drawing a king on the 13th draw from a carddeck

Hello, i'm trying to solve question 2 of problem 2 by myself. The question + answer + explanation can be found in the link below.

https://github.com/theGreenJedi/The-...03%20_%206.pdf

I will use the ncr() function of SpeedCrunch to denote combinations.

My reasoning was as follows: when 4 kings can not be chosen there are ncr(48;12) ways to pick the first 12 cards. When any card can be chosen there are ncr(52;12) ways to pick the first 12 cards. So the probability that the first 12 cards are not kings is ncr(48;12) / ncr(52;12)

Then there are (52 - 12) cards left of which 4 are kings, so the probability that the next card is king is 4 / 40. So my answer is: ( ncr(48;12) / ncr(52;12) ) * 4/40 = 0,0337575 But this is numerically not the correct answer. Where did i go wrong in my reasoning?

2. ## Re: Probability of drawing a king on the 13th draw from a carddeck

This problem is solved. The given expression equates to the same numerical answer.