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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?

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  • $\begingroup$ The prob. of picking a particular card at least once in 52 trials is $$1-\left(1-\frac{1}{52}\right)^{52}\approx 0.635$$ $\endgroup$
    – user140541
    Jul 31, 2015 at 5:20

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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 $$1-\Bigl(\frac{51}{52}\Bigr)^{52}\ .$$

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    $\begingroup$ This is true, but this answer shows the probability of picking it only once, but not at least once. $\endgroup$
    – kirtan-shah
    Jul 31, 2015 at 18:04
  • $\begingroup$ I guess this problem is probably right, but I wrote a computer program that gets a 2 almost every 52 times. It even calculated $\frac{correct}{total}$ to be above 100%, almost 110%. The probability, I suspect should be more than 63%. $\endgroup$
    – kirtan-shah
    Jul 31, 2015 at 18:21
  • $\begingroup$ @htmlapps No, this give the probability of picking it at least once. $\endgroup$
    – Brian Fitzpatrick
    Jul 31, 2015 at 21:41
  • $\begingroup$ @htmlapps Does your computer program look for a deuce or specifically the deuce of diamonds? The probability of drawing a deuce at least once in 52 draws is $\approx 98.442\%$ $\endgroup$
    – Brian Fitzpatrick
    Jul 31, 2015 at 21:43
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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!

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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!

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Just my thought. I’d say It depends if the deck gets reshuffled between each draw or the previous cards are placed on the bottom of the pile and the next one drawn.

If the cards are shuffled and a random card out of the deck is drawn. And you always guess the 2♦️ Which is a 1/52 chance probability around 1.92%. If you place that card back in the deck and shuffle its 1/52 2nd time and the 38th time and the 52nd time so the probability of picking the 2 ♦️ Never changes. And you can do it infinite amount of times.

However……….. If you shuffle the deck,draw a random card, you have 1/52 chance 1.92% probability of it being 2♦️

If, when it’s not the 2♦️, you then place that card on the bottom of the pack and draw the next one without shuffling and guess 2♦️ It’s 1/51 chance which is a 1.96% probability.

The probability changes because you know the previous outcome was not the 2♦️

If you continue this way The next time you have 1/50 chance or 2% probability’ eventually you will guess right once even if its not until the 52nd draw when the probability has risen to 100%. But once you have guessed right the chance drops to 0.000% because it wont reappear twice within the said 52 draws. It’s finite.

Thanks in advance for not ripping me a new one just because you disagree.

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