Aces distribution between 4 card players

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I recently was playing a card game with a few people and we got to talking about probabilities. I intuitively claimed that it's much more likely that (a) 1 of 4 players gets all 4 aces than (b) all players get 1 ace each. But after some discussion it became clear that for most people (b) seems more likely than (a).

We assume that a full card deck is distributed between the players.

I'm not very good with probability, but from elementary considerations we have the following:

$$p_a=\frac{1}{4^4} \cdot 4=\frac{1}{4^3}$$

This is because each player drawing each ace is an independent event, while the players themselves are indistinguishable, so there's $4$ possible cases satisfying the condition.

$$p_b=\frac{1}{4^4}$$

And here there's only one possible case, so the probability is much smaller.

Is my reasoning correct?

And, going beyond mathematics, do most non-mathematicians expect a "fair" distribution to be more probable than an "unfair" one? Is it a common enough mistake?

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You calculated $p_a$ correctly, but your result of $p_b$ is much smaller than its actual value.

Let the four players be $A, B, C, D$. Distributing the four aces independently to one of the four players, we obtain the equiprobable sample space $$ \Omega = \{A, B, C, D\}^4 $$ where, say, $(A, B, C, C)$ means that the four aces went to $A, B, C,$ and $C$, respectively.

Therefore, \begin{align*} p_b &= \frac{\text{number of permutations of $(A, B, C, D)$}}{\text{total number of possibilities}} \\ &= \frac{4!}{4^4} = \frac{3}{32} = 9.375 \%, \end{align*} whereas $p_a = 1/64 = 1.5625 \%$.

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For the “each player gets one Ace each” case, you undercounted by a factor of $4!$

This is because player 1 can get any of the 4 aces; player 2 can get any of the 3 remaining aces, and so on.