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

[flip101]

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...l_ 13 cards in a deck _ Problem Set 3 _ 6.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?

 

[Dr.Peterson]

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...l_ 13 cards in a deck _ Problem Set 3 _ 6.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?

Your explanation is convincing. Why do you say it's "numerically not correct"? That's the same numerical answer I get when I evaluate either of their forms, and it can be easily shown that it is equivalent to either of them:

C(48,12)/C(52,12)*4/40 = (48*47*...*37)/(12*11*...*1) * (12*11*...*1)/(52*51*...*41) * 4/40
= (48*47*...*37)/(52*51*...*41) * 4/10 = (48*47*...*37*4)/(52*51*...*41*40)
​

I think you went wrong in checking your answer against theirs, not in your actual work.

But this is a horrible problem, as it asks for "THE correct expression", when they themselves point out that there are more than one correct expression! It would make a lot more sense to let you enter your own expression, and for them to check that it is equivalent. Good software can do that.

 

[pka]

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...l_ 13 cards in a deck _ Problem Set 3 _ 6.pdf

The probability that the first card is a king is the same as the thirteenth card is a king, \(\displaystyle \dfrac{1}{13}=0.07692\)

This is very difficult for many to accept.But if you are at a college with a good mathematics library you should be able to find the Springer texts in Statistics, Probability by Jim Pitiman (Jim teaches at Berkeley). In the addition I have in chapter 3 there is an optional section called Symmetry. In that section there is a complete discussion of this question.

 

[flip101]

I think you went wrong in checking your answer against theirs, not in your actual work.

Yes i think this was it too. I was so entangled in the explanation of the problem that i scrolled back and forth between pages and notes that i had the answer of the first question in mind. Actually in my practice of questions before this, when i corrected my own homework most of the time i didn't read properly. This time i didn't notice it though

The probability that the first card is a king is the same as the thirteenth card is a king, \(\displaystyle \dfrac{1}{13}=0.07692\)

Ok thank you for your advices but this is about question 1, i asked about question 2.