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.
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.
Intuitively, in the original 52-deck card example, we could argue that removing a spade card isn't going to make any particular rank any less likely, because a spade can be of any rank at equal probability. If one rank became less likely on removing of a spade, some rank(s) should become more probable.