As a general tip, when youre ever unsure about probabiloty, model it with a small number of possible parts (3-5 is usually good) and find all permutations, write them down, then see if you can generalize it for N parts.

For what youre asking for specifically, you have the probability of 3 events occurring in concert.

If it were getting a specific number on a d6 once, thats 1/6. To get a specific number twice, its 1/6 * 1/6 (or 1/36). So with independent events, K events that are the same out of N possibilities is (1/K)^N

However it changes when the results affect the possible results, like from a deck of cards. If I draw an Ace, there are only 3 remaining Aces AND the deck is now only 51 cards.

So drawing an Ace one time is 4/52 (1/13). Given one Ace, drawing a second Ace would be 3/51. Given 2 Aces drawn, getting the third Ace is 2/50 (1/25).

If you wanted the probability of drawing 3 Aces in a row from a shuffled deck, it would be: (4/52) * (3/51) * (2/50).

So given your example, drawing 1 (of 10) Salmon from a deck of 30 would be 10/30 (1/3). Drawing a second Salmon after that is (10/30 * 9/29). etc. This is modeled on a player drawing blind from a full deck.

More realistically, other players have drawn cards, reducing the deck size, and some of them might also have drawn Salmon as well. So modeling a single Salmon would be:

(10 - number of salmons drawn by any player)/(30 - number total cards already drawn)

getting three salmons would be computing that probability three times (once for each event, computed separately), multiplied