Probability of picking a card one out of 52 times.

Question

Let's say we have a standard deck of 52 cards.

What would be the probability of choosing the 2 of diamonds?
Obviously, it would be $\frac{1}{52}$.
If we were to randomly choose another card from the deck, the probability of choosing the 2 of diamonds would be $\frac{2}{52}, or \frac{1}{26}$.

If we keep going on with this (while replacing each picked card), the probability of picking the 2 of diamonds will increase to $\frac{52}{52}$, or 100%.

Logically, we know that it is possible to pick 52 cards from a deck and put them back, and yet still not pick the 2 of diamonds.

So, practically speaking, what would be the probability of picking the 2 of diamonds at least once when picking a card from the deck 52 times, and how would I figure this out?

Answer 1

Same answer as Brian Fitzpatrick, but with IMHO a simpler explanation. The probability that you do not pick the $\diamondsuit2$ on the first draw is $\frac{51}{52}$. The same goes for all $52$ draws. These events are independent, so the probability that you never get the $\diamondsuit2$ in $52$ draws is $(\frac{51}{52})^{52}$, and the probability that you do get it at least once is $$1-\Bigl(\frac{51}{52}\Bigr)^{52}\ .$$

Answer 2

Let $\Bbb P(n)$ be the probability that the first time you draw the $2\diamondsuit$ is your $n$th draw. Then \begin{align*} \Bbb P(1) &= \frac{1}{52} & \Bbb P(2) &= \frac{51}{52}\frac{1}{52} & \Bbb P(3) &= \left(\frac{51}{52}\right)^2\frac{1}{52} & \Bbb P(n) &= \left(\frac{51}{52}\right)^{n-1}\frac{1}{52} \end{align*} Now, the probability that you have drawn the $2\diamondsuit$ at least once after $N$ draws is $$ \sum_{k=1}^N\Bbb P(k) = \sum_{k=1}^N\left(\frac{51}{52}\right)^{k-1}\frac{1}{52} = \frac{1}{52}\sum_{k=1}^N\left(\frac{51}{52}\right)^{k-1} = \frac{1}{52}\frac{1-\left(\frac{51}{52}\right)^{N}}{1-\left(\frac{51}{52}\right)} = 1-\left(\frac{51}{52}\right)^{N} $$ Here, we have used the formula for partial geometric sums.

In particular, the probability that you have drawn the $2\diamondsuit$ at least once after $52$ draws is $$ 1-\left(\frac{51}{52}\right)^{52}\approx 63.568\% $$ It's hard to call your card!

Answer 3

As you mentioned, the probability that the first card you take is the 2 of diamonds is $1/52$. However, after you put the card back and randomly choose another, the probability of drawing a two of diamonds is still $1/52$. In this scenario, the cards I draw in the past do not affect the probabilities of what I draw in the future.

However, you ask a different question: after 52 draws, what is the probability that you draw a 2 of diamonds at least once? A simple way to think about this problem is to consider the probability that this event does not occur. Naturally, the sum of the probabilities that an event does occur and that it does not occur equals 1. So upon calculating one of the two, we can easily determine the other.

What is the probability that you never draw a 2 of diamonds after 52 draws? For each draw, the probability that you don't draw a 2 of diamonds is 51/52. Since each draw is independent from the others, the probability of never drawing a 2 of diamonds is $(51/52)^{52}$.

Since the sums of the probabilities of an event occurring and it not occurring is 1, the probability of drawing at least one 2 of diamonds is $1-(51/52)^{52}$.

Hope this helps!