[Math] Conditional Probability Example using permutations

combinatoricsprobability

You are dealt three cards. The events of interest concern the number of face cards that you are dealt (0, 1, 2, or 3). What is the conditional probability that you are dealt at least 2 face cards given that the last card dealt to you was a face card?

There 12 face face cards (1 jack, queen, king * 4 suits) and 40 non face cards.

Let F= face card, and R = regular or nonface card.

So the question is asking us to find the probability of getting {RFF, FRF, FFF}. I thought I would approach this question using permutations:

$\Pr(\text{@ least 2 face cards | last card has face}) = \Pr(\{RFF, FRF, FFF\})=\cfrac{(40)(12)(11) + (12)(40)(11) + (12)(11)(10)}{(52)(51)(50)} = 0.089593$

but the answer is 0.3882. Can anyone please tell me why this is not working? My understanding is that when you're solving probability you can consider ordering or not consider it. In other words, you could use permutations or combinations.

Thank you in advance.

Here is a tree diagram. Let A, B, and C be events where a face card is obtained.

Tree Diagram

Best Answer

Last or first or second makes no difference. It will feel clearer with first.

We want the probability of at least one face card out of two, from a deck in which one face card is gone.

It is easier to find the probability of no face card in two picks. This is $\frac{40}{51}\cdot\frac{39}{50}$.

So our answer is $1-\frac{40}{51}\cdot\frac{39}{50}$.

Another way: There are fancier ways to proceed. Let $F$ be the event "the third card is a face card," and let $A$ be the event "at least two face cards." We want the conditional probability $\Pr(A|F)$. By the definition of conditional probability, we have $$\Pr(A|F)=\frac{\Pr(A\cap F)}{\Pr(F)}.\tag{$1$}$$ Note that $\Pr(F)=\frac{12}{52}$.

You calculated the top of $(1)$ correctly, and did not divide by the bottom.

Remark: Remember that we are given that the third card is a face card. So the sample space is restricted to those draws in which the third was a face card. What you calculated was the probability that there are at least two face cards and the third is a face card. That is not what the problem asked for, buut once we have found it, the answer is one step away.

Related Solutions

[Math] Need to clarify the “At-least Concept” in Combination.

If you expand your first answer, you have $$ \binom{12}{1} \cdot \binom{40}{1} + \binom{12}{2} \cdot \binom{40}{1}. $$ Notice in the term on the right, you are choosing a total of three cards, which does not count what you want.

A rule of thumb is to turn "at least" problems into several instances of "exactly".

For the problem at hand, "at least one jack/queen/king" translates into "either exactly one jack/queen/king or exactly two jack/queen/king".

The number of ways to get exactly one jack/queen/king and one other card is $$ \binom{12}{1} \cdot \binom{40}{1}. $$ We multiply here because you choose a card from the jack/queen/king pile and a card from the "other" pile.

The number of ways to get exactly two from jack/queen/king is $$ \binom{12}{2}. $$ Since these cases are disjoint (this is important), we simply add the results from the two cases. Thus, the number of ways to get at least one Jack, Queen, or King is $$ \binom{12}{1} \cdot \binom{40}{1} + \binom{12}{2}. $$

[Math] the probability that 5 randomly chosen cards in a deck add up to 40 or greater

Normal approximation:

Consider a single card draw from the deck:

E(X)=7

Var(X)=14

With the sum of 5 cards (drawn without replacement):

$E(X)=7 \times 5$

$\text{Var}(X)=14\times 5\times \frac{52-5}{52-1}$

(the last term being the 'finite population correction', which applies as much to the variance of the sum as it does to the variance of the mean).

The approach to normality in this situation is reasonably rapid.

This suggests the sum on 5 cards might be be roughly approximated by a normal distribution with mean $35$ and variance $70\times\frac{47}{51}\approx 64.5098$. Using a continuity correction this gives an approximate probability of totalling at least 40 on 5 cards of:

$$1-\Phi(\frac{39.5-35}{\sqrt{70\times\frac{_{47}}{^{51}}}})\approx 0.288$$

Simulation (in R) of ten million five-card draws indicates the probability is
about $0.293$ (with s.e. $\approx 1.4\times 10^{-4}$):

 res=replicate(10000000,sum(sample(rep(1:13,4),5,replace=FALSE)));mean(res>=40)  
 [1] 0.2927447

As a check on the earlier calculation of the variance, the standard deviation of of those simulated sums was 8.0316; the previous calculation gives 8.0318.

Edit: Here's a comparison of the empirical cdf of the simulated data with the above normal approximation; they're pretty close:

enter image description here

More extensive simulations are consistent with the other two answers based on complete enumeration:

enter image description here

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[Math] Need to clarify the “At-least Concept” in Combination.

[Math] the probability that 5 randomly chosen cards in a deck add up to 40 or greater

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