Probability getting King

Card is drawn without replacement from a deck. What is the probability that the 22nd card drawn is a king?

Attempt:

Let Case X represent a scenario where the first 21 cards contain X number of Kings.

Case 0: 48C21 * 4C1

Case 1: 4C1 * 48C20 * 3C1

Case 2: 4C2 * 48C19 * 2C1

Case 3: 4C3 * 48C18 * 1C1

No Restrictions: 52C22

Ans: (Case 0+1+2+3) / No Restrictions

Is this right? Hmm

• Maybe I'm misunderstanding, but why would this be different than the probability that the first card drawn is a king ($4/52 = 1/13$)? – Alexis Olson Sep 7 '16 at 2:47

Yes, you want $$\dfrac{\binom{48}{21}\color{silver}{\binom{4}{0}}}{\binom{52}{21}}\dfrac{\binom{4}{1}}{\binom{31}{1}}+\dfrac{\binom{48}{20}\binom{4}{1}}{\binom{52}{21}}\dfrac{\binom{3}{1}}{\binom{31}{1}}+\dfrac{\binom{48}{19}\binom{4}{2}}{\binom{52}{21}}\dfrac{\binom{2}{1}}{\binom{31}{1}}+\dfrac{\binom{48}{18}\binom{4}{3}}{\binom{52}{21}}\dfrac{\binom{1}{1}}{\binom{31}{1}}= \dfrac{4}{52}$$

However, while correct, you are way overcomplicating this.

Any one of the fifty-two cards in the deck might be in twenty second place with equal probability, and there are four kings in the deck.

The probability is thus...

• Oh yea! true. So its still 1/13 – RStyle Sep 7 '16 at 2:53
• Hi why is it that i just cant do 52 choose 22. for the denominator – RStyle Sep 7 '16 at 3:02
• @RStyle As the 22nd place is special, you are choosing for two categories. The first 21 and the 22nd (and the remainder go to the last 30). $\binom{52}{21}\binom{31}{1}\color{silver}{\binom{30}{30}}$ counts the ways to divide a 52 items in three piles of 21, 1, and 30. – Graham Kemp Sep 7 '16 at 3:08
• "As above, so below." You count ways to select the first 21 cards and then the 22nd in the numerator, and so must do likewise for the denominator. The distinction being that the numerator counts ways to do so to produce the favoured event. – Graham Kemp Sep 7 '16 at 3:20

As others have pointed out there is no need to overcomplicate this. Without any other knowledge (for example what cards have passed before), any card can be in the 22nd position and there are 4 cards that are Kings. So the probability is $4/52 = 1/13$

But let's try to do it your way. That is, define distinct cases where the first 21 cards have 0, 1, 2, 3, kings, and then (the 22nd card) a king is drawn.

My feeling is that you wanted to enumerate all possible draws for each case and then divide the sum of the number of draws with the all the possible ways you can draw 22 cards. (I don't really know what you had in mind, since you dive straight into some formulas. I am just making a guess based on the pattern of mistakes, and what seems intuitive to me.)

But if we want to count all possible ways to draw, we should be working with permutations of cards not combinations of cards. Of course the problem can be solved working with combinations as the accepted answer nicely shows. But thinking in combinations seems harder to me, and you seem to be confused as well by the choice of the denominator. So here's a solution looking at number of different draws, which means using permutations:

Case 0 Kings: (possible draws for 21 cards that are not Kings) $\times$ (possible draws for the 22nd card to be King). $$_{48}P_{21} \cdot\space _{4}P_{1} = \frac{48!}{(48-21)!} \cdot\space 4$$ This is very similar to your first formula, I am just using permutations instead of combinations to count all possible draws.

Case 1 King: (possible draws for 20 non-King cards and one King card) $\times$ (possible draws for the 22nd card to be King). This is a bit trickier. Someone could say that the possible draws for 20 non-King cards and one King card are equal to the possible draws for one King card times the possible draws for 20 non-King cards, so $_{4}P_{1} \cdot\space _{48}P_{20}$ Then we just multiply with the possible draws to get a king at the 22nd card, resulting in a formula very similar to yours $_{4}P_{1} \cdot\space _{48}P_{20} \cdot\space _{3}P_{1}$ (again, we now have permutations instead of combinations)

But this is not correct. This assumes that the King in the 21 cards is in a specific position (say the first). We need to multiply this with the possible ways you can place one card in 21 cards or equivalently pick a card from 21 cards. This is just 21 choose 1. So the correct number of possible draws is: $$_{4}P_{1} \cdot\space _{48}P_{20} \cdot\space _{21}C_{1} \cdot\space_{3}P_{1} = 4 \cdot\space \frac{48!}{(48-20)!} \cdot\space \binom{21}{1} \cdot\space 3$$

Similarly for the remaining two cases.

Case 2 Kings: (possible draws for 19 non-King cards and two King cards) $\times$ (possible draws for the 22nd card to be King). $$_{4}P_{2} \cdot\space _{48}P_{19} \cdot\space _{21}C_{2} \cdot\space_{2}P_{1} = \frac{4!}{(4-2)!} \cdot\space \frac{48!}{(48-19)!} \cdot\space \binom{21}{2} \cdot\space 2$$

Case 3 Kings: (possible draws for 18 non-King cards and three King cards) $\times$ (possible draws for the 22nd card to be King). $$_{4}P_{3} \cdot\space _{48}P_{18} \cdot\space _{21}C_{3} \cdot\space_{1}P_{1} = \frac{4!}{(4-3)!} \cdot\space \frac{48!}{(48-18)!} \cdot\space \binom{21}{3} \cdot\space 1$$

We sum all these numbers of draws and we divide them with the all ways we can draw 22 cards out of a deck of cards: $_{52}P_{22} = \frac{52!}{(52-22)!}$. The result is $\frac{1}{13}$

$$\frac{\frac{48!}{(48-21)!} \cdot 4 + 4 \cdot \frac{48!}{(48-20)!} \cdot \binom{21}{1} \cdot 3 + \frac{4!}{(4-2)!} \cdot \frac{48!}{(48-19)!} \cdot \binom{21}{2} \cdot 2 + \frac{4!}{(4-3)!} \cdot \frac{48!}{(48-18)!} \cdot \binom{21}{3} \cdot 1}{\frac{52!}{(52-22)!}} = \frac{1}{13}$$