Use a tree diagram to calculate the probability

[alfred2]

Hi everybody! I have to solve this exercise.

We observe a tournament of 2 ^ n players, which consists of n rounds. In each round every 2 players play against each other and only the winner goes to the next round. At the beginning of each round random pairings are defined, all with the same probability.
Each player is assigned a number corresponding to its position in a ranking, ie if [2 ^ n] is the set of all participants, player i is worse than player j if i> j (1st place, 2nd place, ...)
The encounters of a round are played in the following order: of all remaining encounters to be played, the next encounter will count with the 2 worst players. In each of the encounters of 2 players the best of both always wins with a probability p∊(0.1) - regardless of all other ecounters of the tournament

(a) Is n = 2
Draw a tree diagram of the experiment above exposed calculating the probabilities depended on p where:
i) player 1 wins
Solution?:
http://img546.imageshack.us/img546/8072/79065860.jpg
If n = 2 then there are 2 ^ 2 = 4 players. I have to find all possible combinations between players A = 1, B = 2, C = 3 and D = 4 but considering that the round (A versus B) and (C versus D) ist the same as (B, A) and (C, D) or (C, D) and (A, B). So I concluded that there are only three possible events:
(A, B) and (C, D)
(A, C) and (B, D)
(A, D) and (B, C)
However, the statement says "The encounters of a round are played in the following order: of all remaining encounters to be played, the next encounter will count with the 2 worst players."
Does this mean then for n = 2 that the only possible case is ((A, B) and (C, D)) because the two worst are C and D because their corresponding ranking numbers are 3 and 4 respectively? If it is so, would be the tree like this?:
http://img855.imageshack.us/img855/3879/87552251.jpg
Therefore there would be two possible cases where A (Player 1) wins,:
A beats B and C beats D. Then A beats C
A beats B and D beats C. Then A beats D
The problem is that I do not know how to calculate the probabilities of these events.
Can someone help me? Conditional probability should be used?
ii) player 2 plays in the final round with player 2
iii) player 1 plays at some point with player 2

(b) Let n be any number
i) How many random events Ω kε (branches) have?
ii) Calculate the probabilities for the events E ⊆ Ω of (a) for any n

I'll be glad if somebody helps me =)

 

[jvicens]

Hi there! I am happy to help you solve this exercise.

To answer your first question, yes, for n = 2, the only possible case is ((A, B) and (C, D)) because the two worst players, C and D, will always play against each other in the next round. Your tree diagram for n = 2 looks correct.

To calculate the probabilities of the events where A (Player 1) wins, we can use conditional probability. Let's define the following events:
- P(A) = probability that player 1 wins
- P(B) = probability that player 2 wins
- P(C) = probability that player 3 wins
- P(D) = probability that player 4 wins

Since we know that player 1 always beats player 2 with probability p, and player 3 always beats player 4 with probability p, we can say:
- P(A) = p
- P(C) = p

For the first event (A beats B and C beats D), we can use the product rule of probability to calculate the probability of both events happening:
P(A and C) = P(A) * P(C) = p * p = p^2

For the second event (A beats B and D beats C), we can also use the product rule:
P(A and D) = P(A) * P(D) = p * (1-p) = p(1-p)

Therefore, the total probability of player 1 (A) winning is:
P(A wins) = P(A and C) + P(A and D) = p^2 + p(1-p) = p^2 + p - p^2 = p

For the second question, to calculate the probability of player 2 playing in the final round with player 2, we can use the same approach as above. We know that player 2 always beats player 3 with probability p, so:
- P(B) = p
- P(C) = p

Using the product rule, the probability of player 2 (B) playing in the final round with player 2 is:
P(B and C) = P(B) * P(C) = p * p = p^2

For the third question, to calculate the probability of player 1 playing at some point with player 2, we can use the sum rule of probability. This means