Probability of Drawing X of One Suit Before Y of Any Another Suit

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I am looking for a generalized answer to a problem like the following.

What is the probability of drawing S spades before, D diamonds or C clubs or H hearts.

The purpose of this is to be able to determine the win probability for a given suit in the following game.

Remove all 4 aces from the deck of cards and place the horizontally on the table, then draw the top X cards from the deck and align them vertically ( where X+1 will represent the game length). Below is an example where X = 1, and the card drawn from the top of the deck was the 7 of hearts (7H).

      7H
AH AD    AC AS

From here to win the game cards are drawn from the top of the deck and the Ace that corresponds to the suit drawn gets to move forward, when a specific suit gets to the end the game is over.

So for the scenario above if we want to know the probability of diamonds winning we would need to know the probability of drawing 2 diamonds before drawing 2 hearts, or 2 spades, or 2 clubs.

To simplify the game further I started with the base case of X = 0.

AH AD    AC AS

In this scenario the chances of any suite winning would be 25% (12/48)

From here I tried to work out the case for X=1 but other than manually calculating all of the possible scenarios I wasnt able to see how the problem can be generalized for any value of X.

any help would be greatly appreciated.

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I see at least three different possible questions here. Which one did you intend?

  1. In the title and the first line, you're asking: What is the probability of drawing S spades before, D diamonds or C clubs or H hearts. The deck is not mentioned, so I assume you're starting with a regular 52 card deck.

  2. In your explanation of the game, it seems like you instead need: What is the probability of drawing X+1 spades before, X+1 diamonds or X+1 clubs or X+1 hearts. This would be the win probability in your game if we know the value of $X$ but we haven't seen the $X$ initial drawn cards yet. This time it seems like the deck would be 48 cards: the original 52 cards, minus aces.

  3. Another way of interpreting your explanation of the game is: Let's say $X=1$ and the 7H has already been revealed. Now what is the probability of each suit winning? This time the deck has $48-X$ cards: the full deck minus aces and also minus the $X$ cards displayed on the table.

Question (2) is easy: it's fully symmetrical (you could switch the suits without changing anything), so each suit must always have win probability $25\%$ no matter what $X$ we use. If you meant one of the other versions, please clarify in comments and I or another answerer can try to help.