0
$\begingroup$

I have the following problem:

In a card game, 13 cards are given to you out of a deck of 52. This game is being played 1000 times. Identify (with a name and parameter) the random variable representing the first time where the number of aces you receive is at least 1.

So far in class, we have only learned about the Bernoulli, Binomial, Geometric, and Negative Binomial random variables.

Here were my thoughts:

It is tempting to use the Geometric and Negative Binomial random variables since they are about the first or $n$th occurrence of an event, but we can eliminate them since their sample size is infinite (since we play this game only 1000 times). It's not a Bernoulli random variable either because we need to have a notion of when the event occurs, not just whether it occurs or not (we have at least one ace in the first experiment, or the second experiment, and so on). What remains is the Binomial random variable, which count how many times an event occurs. What we want though is the first occurrence of an event, not how many times it occurs. Perhaps there is a way to cleverly express first occurrence as an arrangement of events, but I can't see a way of doing this.

Also, I computed the probability of receiving at least one ace in a hand of 13 cards to be $\displaystyle \frac{4}{52}$ (there are four ways to choose an ace out of 52 cards and the other 12 cards do not matter).

$\endgroup$
  • 2
    $\begingroup$ No, the probability of receiving at least one ace in a hand of 13 cards can be computed with the use of the hypergeometric distribution and the converse probablility: $P(X\geq 1)=1-P(X=0)=1-\frac{{4 \choose 0} \cdot { {48} \choose {13} }}{{ {52} \choose {13} }}\approx 69.62\%$ $\endgroup$ – callculus Oct 11 '18 at 5:23
  • 1
    $\begingroup$ Essentially this is a geometric distribution, except that there is a very small positive probability of the order of $4.1 \times 10^{-518}$ that you never get an ace in any of the $1000$ attempts. So it is a censored geometric distribution, with the parameter @callculus found $\endgroup$ – Henry Oct 11 '18 at 7:33
  • $\begingroup$ I see. Thank you both for your help. $\endgroup$ – Frederic Chopin Oct 11 '18 at 17:57

Your Answer

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

Browse other questions tagged or ask your own question.