combinatorics conditional probability

Question

Yeechi has a deck of cards consisting of the 2 through 5 of hearts and the 2 through 5 of spades. She deals two cards (at random) to each of four players. What is the probability that no player receives a pair?

now the answer given is thus:

We see that player #1 does not get a pair with probability $\frac67$ (since for any first card that player #1 receives, 6 of the 7 remaining cards do not match it). Suppose without loss of generality that player #1 receives a 2 and a 3, and that a 4 is dealt to player #2. (Think about why we can make these assumptions.) The other 4 gets dealt to a different player with probability $\frac45$, since there are 5 slots remaining for the 5 cards left in the deck. Now the deck is left with a 2, and 3, and the pair of 5’s, and the probability that the pair doesn’t get dealt to the remaining person is $\frac56$. Therefore, the probability of no pair is $\frac67 \cdot \frac45 \cdot \frac56 = \frac47$

my question is how do we find the the probabilities--$\frac67$,$\frac45$,$\frac56$?any hint would be of utmost help!

Answer 1

First, we require that player 1 not receive a pair. Let them receive their first card. It can be anything, but without loss of generality suppose that it were the 2 of hearts. Their second card is selected from the remaining seven cards available and will not pair with the 2 of hearts they already have with probability $\dfrac{6}{7}$ since there are six cards which are not 2's remaining. Without loss of generality, suppose that second card they receive happens to be the 3 of hearts.

Our hypothetical scenario currently looks like: $(2\heartsuit~~3\heartsuit)~(\square~~\square)~(\square~~\square)~(\square~~\square)$ where the squares represent unknowns/undecideds.

Now... among the remaining six cards, one of them is the 4 of hearts. One of the remaining three players will necessarily be given this card. Without loss of generality, let it be the second player who receives it and also without loss of generality let it be the first card they received. Our hypothetical scenario looks like $(2\heartsuit~~3\heartsuit)~(4\heartsuit~~\square)~(\square~~\square)~(\square~~\square)$

Now, the remaining 4 is equally likely to go into any of the remaining positions. There are $5$ positions still available that we have not assigned a card to and each of these positions are equally likely to receive it. In order to continue to avoid anyone getting a pair it will go into one of the positions held by a player other than the second player with probability $\dfrac{4}{5}$. Without loss of generality, suppose it went to the third player in their first position. Our hypothetical scenario now looks like $(2\heartsuit~~3\heartsuit)~(4\heartsuit~~\square)~(4\spadesuit~~\square)~(\square~~\square)$

At this point, there are four open positions left and the only way in which we can form a pair in someone's hand is if the fives happen to both be dealt to the fourth player. There are $\binom{4}{2}=6$ ways in which we can select both positions used for the fives simultaneously, and each of these will be equally likely to occur (given what has already occurred). Only one of these ways to simultaneously select the positions for the fives will be such that they both go to the fourth player. Avoiding this then occurs with probability $1-\dfrac{1}{6}=\dfrac{5}{6}$.

Multiplying gives the stated result.

$\dfrac{6}{7}\times \dfrac{4}{5}\times \dfrac{5}{6} = \dfrac{4}{7}$

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As for a different approach, you could have used inclusion-exclusion principle. Let $A$ be the event that the first player received a pair, $B$ that the second player received a pair etc... Finding the probability that at least one player did get a pair would be

$$\Pr(A\cup B\cup C\cup D) = \Pr(A)+\Pr(B)+\Pr(C)+\Pr(D)-\Pr(A\cap B)-\Pr(A\cap C)-\dots + \Pr(A\cap B\cap C)+\dots - \Pr(A\cap B\cap C\cap D)$$

$$\binom{4}{1}\times \dfrac{1}{7} - \binom{4}{2}\times \dfrac{1}{7\times 5} + \binom{4}{3}\times \dfrac{1}{7\times 5\times 3} - \dfrac{1}{7\times 5\times 3\times 1} = \dfrac{3}{7}$$

Find the answer to the probability that noone received a pair by subtracting this away from $1$, giving $\dfrac{4}{7}$

Answer 2

for player 1--the chance of not getting a pair is $\frac{{C^8_1}\cdot{C^6_1}}{P^8_2}$

for player 2 the chance of getting a pair is $\frac{{C^6_1}\cdot(0+{C^1_1})}{P^6_2}$ --here we choose form 2 unpaired and 2 paired cards.if we choose the 1st card from the unpaired cards the chance of getting a pair is 0.for the case where we choose from the 2 sets of paired cards the facourable casees are $C^6_1.C^1_1$ i.e$\frac15$--so chance of not getting a pair is $\frac45$ now how do we get to value for the 3rd player?