It's all in the math

*McDowell states "if a small group of cards is known to contain an ace, simply knowing its suit increases the probability of one or more additional random aces in the group." The effect is not subtle, resulting in over a twofold increase in the probability of at least one additional random ace in a group of four cards.
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*I'm aware of the effect of conditional probabilities. That is, I understand that an answer of "yes" to, "Is there an ace of clubs in these four cards?" results in a greater number of expected aces than an answer of "yes" to, "Is there any ace in these four cards?" *

First of all, your example is not right. The questions should be "Are there more aces knowing that the ace of clubs is already in the group?" and "Are there more aces knowing at least one ace is in the group?"

*However, in the case described by McDowell, the only difference is that the player remembers which ace is expected. It's the same ace, and the same group of four cards, whether the ace's suit is remembered or not. In other words, one would not expect the group of four cards to change spontaneously if the player suddenly forgets or remembers the suit of the predicted ace. *

The short answer is that knowing the ace is in clubs reduces the total number of combinations to about 1/4 while only reducing the number of combinations which two aces to about 1/2.

Here's the math

For the case where the Ace of Clubs is known to be in the among the four cards, the remaining 51 cards are divided into a group of 3 aces and 48 non-aces. Following are the number of combination for having the AC + 0, 1, 2, or 3 aces. (C is combination function, COMBIN(n,r) in Excel).

AC + 0 A = C(3,0)*C(48,3) = 17,296

AC + 1 A = C(3,1)*C(48,2) = 3,384

AC + 2 A = C(3,2)*C(48,1) = 144

AC + 3 A = C(3,3)*C(48,0) = 1

Altogether, the number of combinations is 20,825 of which 3,529 have at least one more ace: 16.95%.

For the case where at least one ace is known, but the suit is unknown, the 52 cards are divided into a group of 4 aces and 48 non-aces. Following are the number of combination for having the 1, 2, 3, or 4 aces.

1 A = C(4,1)*C(48,4) = 69,184

2 A = C(4,2)*C(48,3) = 6,768

3 A = C(4,3)*C(48,2) = 192

4 A = C(4,4)*C(48,1) = 1

Altogether, the number of combinations is 76,145 of which 6,961 have at least one more ace: 9.14%.

This can be a difficult concept for some people. It falls under the Principle of Restricted Choice. Once you know one of the cards is the Ace of Clubs, that restricts the choices of the other cards.

Here is a similar problem (how to win a lot of money from your friends): You flip two coins (presumably fair), but don't reveal the results. If one of the coins is heads, say so, *and even show it**. (If both are tails, simply reveal both coins and flip again.) Now give someone 6/5 odds that the other coin is heads. Most people will swear up and down the others is 50/50. Other will say is 1/4 (the probability of two heads from two coins). The correct answer is 1/3. Once a coin is revealed to be heads, the only equally probable choices for the two coins is HH, HT, and TH, where HH is the only one where but coins are heads.
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