Probability that one of the players will draw a unique card without replacement There are n=8 cards and p=4 players. 7 of the cards are identical and the 8th is unique. The cards are shuffled randomly and each player draws a card without replacement.
Q_1: What is the probability that one of the players will draw the unique card?
Q_2: What is the probability that the fourth player will draw the unique card?
How does the probability of the previous players affect each subsequent draw? How could one calculate the probability without first calculating the negation? (piecewise or something else?)
I'm trying to account for probability of not chosen for each subsequent draw (draw given not chosen (already) minus draw given chosen by previous player) but my math doesn't seem to add up.
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 A: As hinted by user2661923, we can answer the first 2 questions as follows:
Let's assume that we draw the cards 8 times instead of 4.
And we count only the first 4 draws.
Let's number the draws from 1 to 8.
For each draw (e.g. draw 3), each card has equal chance to be in that draw.
Therefore the unique card has a probability of $\frac{1}{8}$ to be taken in each draw.
Q1: What is the probability that one of the players will draw the unique card?
Answer: This happens when the unique card is among the first 4 cards to be drawn.
The answer is $4 \times \frac{1}{8}= \frac{1}{2}$
Q2: What is the probability that the fourth player will draw the unique card?
Answer: This happens when the unique card is in draw 4.
The answer is $\frac{1}{8}$
A: 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?
Give the one unique card to player four, leaving $7$ cards in the deck. Then, players $1,2$ 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 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
$\frac{210}{1680} = \frac{1}{8} = 0.125$
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 next player will select the unqiue card. In other words, 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?
Simply add the probabilities that players $1,2,3$ and $4$ select the unique card, so that we have $0.125 + 0.125 + 0.125 + 0.125 = 0.5$. You can add in this way because the set of outcomes in which players $1,2,3$ or $4$ get the unique card can be partitioned into four sets, each one consisting of the outcomes in which a particular player gets the unqiue card. Each one of these sets is disjoint (sharing no outcomes) with the others. There is fundamental axiom in probability theory which states that probabilities can be added when this is the case. If you are asked this on a test, then you could answer "well, no more than 1 person can select the unique card at the same time, so adding the probabilities feels like the right thing to do" and you would be absolutely correct because that is what the axiom basically says.
Suggestions for further learning
If you really want a solid grasp on probability theory, then I suggest getting a good foundation in set theory; taking calculus I and II; and then taking a probability course with a university mathematics dept (or walking through a solid textbook on the subject, such as A First Course In Probability by Sheldon Ross 9th ed).
