Probability Intuition

I was explaining to a friend how to calculate some probabilities having to do with playing cards and I came across a problem that I can understand numerically but not intuitively. I was hoping someone might have some insight.

Situation 1: The probability that, when dealt 2 random cards from a regular deck, you will get 2 aces given that at least 1 of the 2 cards is an Ace.

Situation 2: The probability that, when dealt 2 random cards from a regular deck, you will get 2 aces given that 1 of the 2 cards is the Ace of Spades.

The strange thing is that the probability of Situation 2 is nearly double the probability of Situation 1 despite the fact that Situation 2 is more specific.

Here are my calculations:

Situation 1

Let X = both cards are aces, and let Y = at least 1 of the cards is an ace

\[P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{\frac{\binom{4}{2}}{\binom{52}{2}}}{1-\frac{\binom{48}{2}}{\binom{52}{2}}}\frac{}{}\]

This comes out to 1/33

Situation 2

This is much easier. If one of the cards is the ace of spades then the probability that we have 2 aces is the same as the probability that the other card is an ace, which is 3/51 = 1/17


Can anyone provide an insight into why it would make intuitive sense for situation 2 to be essentially twice as likely as situation 1?

Much appreciated!

thepillow said:

I was explaining to a friend how to calculate some probabilities having to do with playing cards and I came across a problem that I can understand numerically but not intuitively. I was hoping someone might have some insight.

Situation 1: The probability that, when dealt 2 random cards from a regular deck, you will get 2 aces given that at least 1 of the 2 cards is an Ace.

Click to expand...

"Intuitively", there are 52 cards in the deck, four of which are aces. IF you know the first card is an ace, you know there are 51 cards left 3 of which are aces. The probability that the second card is an ace, so the probability that you have two aces given that the first card is an ace, is 3/51. By symmetry, the probability that the first card was an ace, given that the second card was an ace, is exactly the same. So the probability that both cards are aces is 6/51.

Situation 2: The probability that, when dealt 2 random cards from a regular deck, you will get 2 aces given that 1 of the 2 cards is the Ace of Spades.

The strange thing is that the probability of Situation 2 is nearly double the probability of Situation 1 despite the fact that Situation 2 is more specific.

Here are my calculations:

Situation 1

Let X = both cards are aces, and let Y = at least 1 of the cards is an ace

\[P(X|Y)=\frac{P(X\cap Y)P(Y)}{P(Y)}\]

Click to expand...

This formula is incorrect. It should be obvious tha thte two "P(Y)" terms wil cancel so you are asserting that "\(\displaystyle P(X|Y)= P(X\cap Y)\)".

\[ =\frac{\frac{\binom{4}{2}}{\binom{52}{2}}}{1-\frac{\binom{48}{2}}{\binom{52}{2}}}\frac{}{}\]

This comes out to 1/33

Situation 2

This is much easier. If one of the cards is the ace of spades then the probability that we have 2 aces is the same as the probability that the other card is an ace, which is 3/51 = 1/17


Can anyone provide an insight into why it would make intuitive sense for situation 2 to be essentially twice as likely as situation 1?

Much appreciated!

Click to expand...

Oops. Yes I made a typo in that formula. Thank you for pointing it out. I definitely didn't mean to have P(Y) also in the numerator. This didn't impact my calculation (the 1/33 is still correct), it was just a typo. I'll change the original post so it's not confusing for someone if they read it.


I also found that simplifying the problem helps give an intuition. Instead of using an entire deck of cards (in which case the sample space is way too big to draw as a picture), I used just 10 cards (including the 4 aces). The probabilities change obviously, but I can actually draw all the possibilities and can see visually why the calculations are correct.