0
$\begingroup$

Two cards are drawn at random from a well-shuffled standard deck of 52 cards. If the two cards are the same suit, we stop. If they are from different suits, the cards are returned to the deck, the deck reshuffled, and the process is repeated. What is the probability that it will take n draws until we stop?

$\endgroup$
  • 1
    $\begingroup$ This is just a geometric distribution. What is the chance that in a single attempt, you get two of the same suit? $\frac{12}{51}$ right? Can you explain why? Now... forget the flavor of cards and suits entirely... reword the problem as the following: you have an unfair weighted coin with probability of heads being $\frac{12}{51}$ and probability of tails being $\frac{39}{51}$. What is the probability of flipping $n-1$ tails in a row followed by a head? $\endgroup$ – JMoravitz Nov 18 '16 at 3:13
0
$\begingroup$

Each time you draw two cards from the deck, the probability that the cards are not of the same suit is $39/51$ and the probability that the cards are from the same suit is $12/51$.

Since the processes of drawing two cards are mutually independent, the probability that we stop at the n'th process is $(39/51)^{n-1}(12/51)$.

$\endgroup$
0
$\begingroup$

The probability of having the same suit the first time is:

$$\binom{4}{1}\frac{\binom{13}{2}}{\binom{52}{2}}$$

Now say we didnt get the same suit first time AND=* we get the same suit the second time:

$$\binom{4}{1}\frac{\binom{13}{1}\binom{39}{1}}{\binom{52}{2}}*\binom{4}{1}\frac{\binom{13}{2}}{\binom{52}{2}}$$

If we do this for n tries :

$$\binom{4}{1}\sum_{i=1}^n(\frac{\binom{13}{1}\binom{39}{1}}{\binom{52}{2}})^{i-1}*(\frac{\binom{13}{2}}{\binom{52}{2}})$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.