Doing union of probabilities by taking product of individual probabilities

Question

I was solving this problem:

Suppose that you are playing blackjack against a dealer. In a freshly shuffled deck, what is the probability that neither you nor the dealer is dealt a blackjack?

Now a blackjack means: one of the cards is an ace and the other one is either a ten, a jack, a queen or a king.

Before this problem I solved bunch of problems which involved calculating probability of set of related events. It involved finding probabilities of individual events and taking product (if they are dependent) or sum (if they are independent). Following similar logic I started solving above problem in this way:

• Let us denote event of getting ace by A
• Let us denote event of getting "either a ten, a jack, a queen or a king" by S
• Let us denote event of getting neither ace nor a ten, a jack, a queen or a king by N
• Thus AAAA means all cards obtained by me are aces and a dealer and have probability $\frac{4}{52}\times\frac{3}{51}\times\frac{2}{50}\times\frac{1}{49}$
• AASA means I received two ace, dealer received one ace and one card which is neither ace, ten, jack, queen or king and have probability $\frac{4}{52}\times\frac{3}{51}\times\frac{2}{50}\times\frac{32}{49}$
• I continued this was for a while and got lost in forming the final equation.
• I tried forming fraction multiplications for other probabilities for different allowed combinations like (first two letter corresponds to cards obtained by me, while last two that of dealer)
  • SSSS
  • SSSN
    SSNS (I am still confused if I should consider these two separately)
  • SNSS
    NSSS (I am still confused if I should consider these two separately)
  • NNSS
  • SSNN
  • NNNN
  • And many more combinations are possible involving A.

After getting lost in forming equation, I felt that I might have to solve this problem other way round, that is through complementation. I had to find probabilities where either me or dealer or both got blackjack and subtract that from 1. So I realised I have to find union. But then I followed approach similar to above to find the union, trying to find probabilities of following events:

• I get blackjack while dealer does not:
  ASNN
  ASSS
  ASAA
  ASSN
  ASNS
  ASAN ASNA
• Similarly I thought I have to find probabilities where I did not get blackjack but the dealer and when both of us got blackjack.
• But after a thinking a bit of above I realized that this is going to be rather long and I must be missing something. So I checked the answer and it was given as simple as follows:

Let A be the event that you are dealt a blackjack, and let B be the event that the dealer is dealt a blackjack. Then the probability that you or the dealer have a blackjack is P(A ∪ B), computed using the Inclusion-Exclusion Identity as follows:
$P(A)=P(B)=\frac{2.4.16}{52.51}$
$P(A\cap B)=\frac{4.4.16.3.15}{52.51.50.49}$
$P(A\cup B)=P(A)+P(B)-P(A\cap B) = 0.0948$ So desire probability $= 1-0.0948$

I was severely disturbed by reading the simple answer. I have following doubts:

1. Is my approach absolutely wrong?
2. If not, how can I arrive to solution using my approach?
3. How can I avoid using non standard approach as mine and follow well know standard approach to solve problem? Is this possible only by practice?

I know these are rather not so neat question. But somewhat abstract. The original problem was very easy. Any view / advice of experienced probability problems solver will be helpful!!!

Answer 1

Your approach is not wrong.   You are evaluating $\mathsf P(A^\complement\cap B^\complement)$ by enumerating all the possible outcomes for the event and their weights.   That is okay.   It is just tedious.

Anyway, your method for evaluation agrees with the textbook.   Just look at $\mathsf P(A\cap B)$, which is by you method the measure of $\rm\{ASAS, ASSA, SAAS, SASA\}$ and so giving the textbooks answer of $4{\cdot}\tfrac{4}{52}\cdotp\tfrac{16}{51}\cdotp\tfrac{3}{50}\cdotp\tfrac{15}{49}$.   There is no problem with what you are doing.   It will work; it just needs a bit more effort.

The textbook is just using the Principle of Inclusion-Exclusion (PIE) to reduce the workload.   $\mathsf P(A)$, $\mathsf P(B)$, and $\mathsf P(A\cap B)$ are easier to evaluate directly than their complements' probabilities, but will indirectly tell you what you need to find.

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So, today's lession is: Mathematicians are lazy, they always look for the smarter way to do their work.

You can too $\ddot\smile$