There are numerous ways to approach this problem. I will expound upon a more intuitive method below...
Q1: What is the probability that one of the players will draw the unique card?
First, note that each possible outcome of this random experiment is equally likely to occur. This allows us to proceed by simply counting (1) the total number of possible ouctomes $N$ as well as (2) the number of outcomes that satisfy the event $E$ in which "one of the players will draw the unique card."
$N$ may be computed by noting that players $1,2,3$ and $4$ may choose one of $8,7,6$ and $5$ cards, respectively, in that order. No matter which card is drawn by each player, any one of the remaining cards may be drawn by the subsequent player. Hence, by the general rule of counting, $N = 8 \cdot 7 \cdot 6 \cdot 5 = 1680$.
The number of outcomes in $E$ may be determined indirectly by counting the number of outcomes in which no player selects the unique card and then subtracting this quantity from $N$. So, remove the one unique card from the deck, leaving you with $7$ cards. Then, players $1,2,3$ and $4$ may choose one of $7,6,5$ and $4$ cards, respectively, in that order. No matter which card is drawn by each player, any one of the remaining cards may be drawn by the subsequent player. Hence, by the general rule of counting, there are $7 \cdot 6 \cdot 5 \cdot 4 = 840$ outcomes in which no one selects the unique card. In turn, this implies there are $N-840=1680-840=840$ outcomes in which at least one player selects the unqiue card.
Again, since each outcome was equally likely, we compute the probability of $E$ as follows
$P(E) = \frac{840}{1680} = \frac{1}{2} = 0.5$
Q2: What is the probability that the fourth player will draw the unique card?
How doesGive the probability ofone unique card to player four, leaving $7$ cards in the previousdeck. Then, players affect$1,2$ and $3$ may choose one of $7,6$ and $5$ cards, respectively, in that order. No matter which card is drawn by each player, any one of the remaining cards may be drawn by the subsequent draw?player. Hence, by the general rule of counting, there are $7 \cdot 6 \cdot 5 = 210$ outcomes in which player four draws the unique card. We compute the probability as follows
$\frac{210}{1680} = \frac{1}{2} = 0.125$
How does the probability of the previous players affect each subsequent draw?
Note that after each player draws a card, the number of cards left in the deck decreases. So, if the first $r$ players do not select the unique card, then there is a $\frac{1}{8-r}$ chance that the $(r+1)$th player will select the unqiue card. In other wordss, the probability of selecting the unqiue card increases assuming no one else has already selected it.
How could one calculate the probability without first calculating the negation? (piecewise or something else?)