A deck of $4$ cards with values $\{1,2,3,4\}$ on each. A and B pull each one after another a card without putting it back in the deck. The winner is the one who pulls a $sum≥4$ first. A starts the game. What is the probability that A wins?
I tried to observe the cases that A won't win: $\{1,2\};\{2,1\}$ plus when B pull $4$ on his first try. There are $3$ out of $8$ choices $(4×2)\ldots 1-(3/8)=0.625$
In the comments, you write (you should always put your attempt in your original post, BTW):
There are two problems with this:
First, you count the '$(1,2)$' event, the '$(2,1)$' event, and the event that $B$ picks $4$ all as $1$ event. That is, you give them all 'equal weight' and thus consider them equally likely. However, that is not the case:
The '$(1,2)$' event, which is the '$A$ loses because $A$ picks $1$ and then $2$' event, can only happen when $B$ picks the $3$ on the first pick instead of the $4$, because otherwise the game would be over. So, the '$(1,2)$' event really represents only $1$ possible permutation of the $4$ cards: $(1,3,2,4)$. Likewise for the '$(2,1)$' event, which represents the single permutation $(2,3,1,4)$.
On the other hand, the '$B$ picks $4$' event covers many possible permutations, e.g. $(1,4,2,3)$, but also $(1,4,3,2)$ (even though the game ends after the $4$), but also $(2,4,1,3), (2,4,3,1), (3,4,1,2)$, and $(3,4,2,1)$. So, the '$A$ loses because $B$ picks $4$' event represents $6$ possible permutations, not just $1$.
Second, you suggest that there are $8$ choices ... but what are those $8$ choices? There are $24$ permutations of the $4$ cards, so that's what you need to use as the denominator. And, as shown above, there are $8$ possible permutations that leads to $A$ losing. So, the probability that $A$ loses is:
$$\frac{8}{24}=\frac{1}{3}$$