Question

In: Statistics and Probability

In a university games tournament, 64 students are about to participate in a chess knockout competition....

In a university games tournament, 64 students are about to participate in a chess knockout competition.

The first round consists of 32 games, with two students per game.

The 32 winners of the first round get to play in the second round, which consists of 16 games, and so on, until an overall winner is declared in the sixth round. (In the case of a draw on any game, a coin is tossed to determine the winner.)

(a) In how many different ways can the 64 participating students be paired up on the first round? (Do not consider the order in which students can be paired up.)

(b) Suppose that the 64 participating players are of equal ability, and pairing up is purely random on each round. Find the probability that Eva and Brett (two of the 64 students) will get to play each other at some stage during the knockout competition

(very appreciate write in details, thank you very much)

Solutions

Expert Solution

(a)

Number of ways 64 participating students be paired up = ⁶⁴C₂

= 64! / (64-2)! * 2!

= 64! / (62! * 2!)

= (64 * 63) / (2 * 1)

= 2016

(b)

Probability of Eva and Brett (two of the 64 students) will get to play each other in first round = 1 / ⁶⁴C₂

= 1/ 2016

Probability of Eva and Brett will not get to play each other in first round = 1 - 1/2016

Probability of winning any round for a player = 1/2

Probability that both Eva and Brett win 1st round = (1/2) * (1/2) = (1/2)²

Now, in 1st round 32 students will participate.

Probability of Eva and Brett will get to play each other in second round = 1 / ³²C₂

= 1/ 496

Probability of Eva and Brett will not get to play each other in second round = 1 - 1/496

In Similar way we can show that,

Probability that both Eva and Brett win kth round = (1/2²)^k

Probability of Eva and Brett will get to play each other in 1st , 2nd, 3rd, 4th , 5th and 6th round are 1 / ⁶⁴C₂, 1 / ³²C₂,1 / ¹⁶C₂,1 / ⁸C₂,1 / ⁴C₂and 1 / ²C₂respectively.

Probability that Eva and Brett will get to play each other at some stage

= Probability of Eva and Brett will get to play each other in first round +

Probability of Eva and Brett will not get to play each other in first round * Probability of Eva and Brett will get to play each other in second round * Probability that both win 1st round +

Probability of Eva and Brett will not get to play each other in first round * Probability of Eva and Brett will not get to play each other in second round * Probability of Eva and Brett will not get to play each other in third round * Probability of Eva and Brett will not get to play each other in 4th round * Probability of Eva and Brett will get to play each other in 5th round * Probability that both win till 4th round +

Probability of Eva and Brett will not get to play each other in first round * Probability of Eva and Brett will not get to play each other in second round * Probability of Eva and Brett will not get to play each other in third round * Probability of Eva and Brett will not get to play each other in 4th round * Probability of Eva and Brett will not get to play each other in 5th round * Probability of Eva and Brett will get to play each other in 6th round * Probability that both win till 4th round

= 1 / ⁶⁴C₂+ (1 - 1 / ⁶⁴C₂ ) * 1 / ³²C₂ * (1/2²)¹ + (1 - 1 / ⁶⁴C₂ ) * (1 - 1 / ³²C₂) * 1 / ¹⁶C₂ * (1/2²)² +

(1 - 1 / ⁶⁴C₂ ) * (1 - 1 / ³²C₂) * (1 - 1 / ¹⁶C₂ ) * 1 / ⁸C₂ * (1/2²)³ + (1 - 1 / ⁶⁴C₂ ) * (1 - 1 / ³²C₂) * (1 - 1 / ¹⁶C₂ ) * (1 - 1 / ⁸C₂ ) * 1 / ⁴C₂ * (1/2²)⁴

+ (1 - 1 / ⁶⁴C₂ ) * (1 - 1 / ³²C₂) * (1 - 1 / ¹⁶C₂ ) * (1 - 1 / ⁸C₂ ) * (1 - 1 / ⁴C₂ ) * 1 / ²C₂ * (1/2²)⁵

= 1/2016 + (1 - 1/2016) * 1/496 * 1/4 + (1 - 1/2016) * (1 - 1/496) * 1/120 * 1/16 + (1 - 1/2016) * (1 - 1/496) * (1 - 1/120) * 1/28 * 1/64 + (1 - 1/2016) * (1 - 1/496) * (1 - 1/120) * (1 - 1/28) * 1/6 * 1/256 + (1 - 1/2016) * (1 - 1/496) * (1 - 1/120) * (1 - 1/28) * (1 -1/6) * 1 * 1/1024

= 0.003468574513

orchestra answered 2 years ago

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