# Independence of removing a spade and picking an ace.

In this video, around 8:00, it is stated that if a spade is removed from a deck of cards, the probability of picking an ace stays the same.

I don't understand why. I tried to simplify with three cards: a red ace, a black ace, and a black ten. (After removing a black card, what is the probability of picking an ace? Before that, it's 2/3.)

1) Remove (black) ten. Two aces remain -> probability 1
2) Remove (black) ace. One ace remains -> probability 0.5


So how does that add up to 2/3? Half the time I get an ace, and half the time I have a 50% chance of getting one. Feels like 0.75, although I am somewhat new to probability.

• The trick is that a removing a spade will not change the chance of any of the other cards being picked since it is equally probable for all cards to be a spade (for every card, there is one that is a spade), so as the man in the video says, it does not add information to the problem. – Ruben Nov 17 '15 at 10:06
• @Ruben I get it—kind of. – 355durch113 Nov 17 '15 at 10:36

Your intuition is correct, but it doesn't match the original example because there are more aces than there are of other cards, initially. Let's even the odds by having there be four cards: a black ace, a black ten, a red ace and a red ten. Initially the probability of picking an ace is $1/2$.
We remove one black card. This means we have $1/3$ of picking an ace if we removed the black ace, and $2/3$ if we removed the black ten. $\frac{1}{3} \times \frac{1}{2} + \frac{2}{3} \times \frac{1}{2} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6} = \frac{1}{2}$, our original probability.
• Oh, thanks. With the formula $\frac{1}{13} \cdot \frac{3}{51} + \frac{12}{13} \cdot \frac{4}{51} = \frac{1}{13}$, but I still have to get my head around it intuitively. – 355durch113 Nov 17 '15 at 10:29