MHB Solve Blackjack Probability: No Dealer/Player Blackjacks

[Yankel]

Hello

I have this simple question:

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?

In order to solve the problem, I need the probability of both me and the dealer GETTING a blackjack. The probability is:

\[\frac{4}{52}\cdot \frac{4}{51}\cdot \frac{16}{50}\cdot \frac{15}{49}\cdot 3\]

and I don't understand where the 3 comes from.

Can you assist please ?

Thank you !

 

[Jameson]

I have to run but I want to comment that with card problems I like using combinations, because this takes care of the ordering issue inherently. If you count the cards like you are doing you will have to account for the order at the end.

For example the probability of one player getting blackjack is:

$$\frac{\binom{4}{1} \binom{4}{1}}{\binom{52}{2}}$$

Using the same logic, what is the probability of both you and the dealer getting blackjack? Let's see if we get the same answer two different ways. :)

 

[Yankel]

how can this be the probability of getting a blackjack ? You need 1 card out of 4 aces and 1 card out of 16 possibilities: 4 K's, 4 Q's 4 J's and 4 10's

 

[Jameson]

how can this be the probability of getting a blackjack ? You need 1 card out of 4 aces and 1 card out of 16 possibilities: 4 K's, 4 Q's 4 J's and 4 10's

By your description I thought you were looking for just the AJ combos. If you are using A10, AJ, AQ, or AK then you are correct it would be:

$$\displaystyle \frac{\binom{4}{1} \binom{16}{1}}{\binom{52}{2}}$$

As for your main question, you are looking for $P(B_{p}' \cap P_{d}')$, where $B_{p}$ is the event you receive a blackjack and $B_{d}$ is the event the dealer receives a blackjack. To find this I would use this fact: $P(B_{p} \cup P_{d})'=P(B_{p}' \cap P_{d}')$. If you find the probability that either you or the dealer receives a blackjack, then you can just take the complement for the final answer. Have you made any progress on this question? :)

 

[Opalg]

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?

In order to solve the problem, I need the probability of both me and the dealer GETTING a blackjack. The probability is:

\[\frac{4}{52}\cdot \frac{4}{51}\cdot \frac{16}{50}\cdot \frac{15}{49}\cdot 3\]

and I don't understand where the 3 comes from.

I would prefer to write this as \[\frac{4}{52}\cdot \frac{3}{51}\cdot \frac{16}{50}\cdot \frac{15}{49}\cdot 4\] (with the $3$ and the $4$ interchanged). The reason that the $4$ goes down to $3$ in the second fraction is the same as the reason that the $16$ goes down to $15$ in the fourth fraction, namely that after the first ace is drawn there are only $3$ remaining aces to choose from.

So the real question is: what is the extra $4$ at the end doing? The answer to that is that there are four possible orders in which the cards $AX$ for the two players can be drawn (where $X$ denotes $10$, $J$, $Q$ or $K$). these orders are $$AX \quad AX,$$ $$AX \quad XA,$$ $$XA \quad AX,$$ $$XA \quad XA.$$

 

[Jameson]

As always I enjoy readings Opalg's methods of these calculations because we approach these problems differently. :) Just to show the two methods work the same:

$$\frac{4}{52}\cdot \frac{3}{51}\cdot \frac{16}{50}\cdot \frac{15}{49}\cdot 4 = 0.001773017 $$
$$ \frac{\binom{4}{1}\binom{16}{1}\binom{3}{1}\binom{15}{1}}{\binom{52}{2}\binom{50}{2}} = 0.001773017$$