Probability of at least one player winning all the matches, and another losing all the matches, they play

Question

Question:

5 players of equal strength play one game with each other. $P(A)=$ probability that at least one player wins all matches he (they) play; $P(B)=$ probability that at least one player loses all matches he (they) play.

Find $P(A), P(B)$ and $P(A\cap B)$.

My attempt:

I have two insights:

1. Each player has a probability of winning = probability of losing = $0.5$ (assume no draw can occur), since they are of equal strength
2. Total number of matches = $^5C_2=10$.

Let us compute $P(A)$. Assuming the players to be distinct, we select the winner player as $^5C_1$. But the question reads: "at least one player". This means there's more than one "winner player".

So, instead I may have to do: $P(A)=1-\text{no player wins all the matches he (they) play}$. But, I don't have any clue on how to compute "no player wins all the matches he (they) play" either.

I am unable to proceed further. This is a high school question. I believe a simple method should exist, but I am unable to find it.

Answer 1

At most one player can win all their matches (since every other player would lose their match against such a player).

At most one player can lose all their matches (since every other player would win their match against such a player).

It follows that $$P(A) = P(B) = {\small{\binom{5}{1}}}\left({\small{\frac{1}{2}}}\right)^4$$

Let $W$ be the event that player $1$ wins all of his or her matches, and let $L$ be the event that player $2$ loses all of his or her matches.

Then $$P(W|L) = \frac{P(W \cap L)}{P(L)}$$ so $P(W\cap L) = P(L)P(W|L)$.

We have $P(L) = \left({\large{\frac{1}{2}}}\right)^4$, and $P(W|L) = \left({\large{\frac{1}{2}}}\right)^3$, hence $P(W\cap L) = \left({\large{\frac{1}{2}}}\right)^7$

Finally, since players $1$ and $2$ could be any choice of two players, we get $$P(AB)={\small{\binom{5}{1}}}{\small{\binom{4}{1}}}\left({\small{\frac{1}{2}}}\right)^7$$

Answer 2

Playing against eachother it cannot happen that $2$ players win all matches.

If $W_i$ denotes the probability that player $i$ wins all $4$ matches then the probability that at least one player wins all matches equals:

$$P(A)=P(W_1)+P(W_2)+P(W_3)+P(W_4)+P(W_5)=5\times2^{-4}$$

Likewise we find that $P(B)=5\cdot2^{-4}$ (at most one player can loose all matches).

If $L_i$ denotes the probability that player $i$ looses all $4$ matches then:

$$P(A\cap B)=\sum_{i\neq j}P(W_i\cap L_j)=5\times 4\times2^{-7}$$

($7$ games are decisive for the occurrence of event $W_i\cap L_j$ with $i,j\in\{1,2,3,4,5\}$ and $i\neq j$)

Answer 3

Let players be 1,2,3,4,5

Total number of matches played =$\binom{5}{2}$ =={(12)(13)(14)(15)(23)(24)(25)(34)(35)(45)}

Every match has 2 possibilities where either one player win or the other

Therefor our sample space is $2^{10}$

P(A)= probability that at least one player wins all matches he (they) play

If one player has won all his matches than every other player must have atleast lose one match, that means there is atmost one player that won all his match.

We can choose that player by $\binom{5}{1}$ and assign his win to all the matches For example let the player 1 be winner than all the outcomes of matches {(12)(13)(14)(15)} are already been determined. The outcomes of the rest of the match is given by $2^6$

Probaility = $\binom{5}{1}\frac{2^6}{2^{10}}$

Same argument for P(B)

For P(A∩B) we choose winner and loser by $\binom{5}{1}\binom{4}{1}$ and assign outcome to them for example let 1 be winner and 2 be loser then outcome of{(12)(13)(14)(15)(23)(24)(25)} are already been determined. The rest can be determined by $2^3$

Probability =$\binom{5}{1}\binom{4}{1}\frac{2^3}{2^{10}} $