Calculating Probabilities of a Table-Tennis Tournament

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Homework Help Overview

The discussion revolves around calculating probabilities in a table-tennis tournament involving four players (A, B, C, and D) competing in a knockout format. The problem includes determining the likelihood of specific outcomes, such as player A winning the tournament, players C and D meeting in the final, and players B and C meeting at some stage of the tournament.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the probabilities associated with different match pairings and outcomes, questioning how to calculate the chances of player A winning two matches, the conditions under which C and D can meet in the final, and the scenarios where B and C might face each other. There is discussion about using tree diagrams and the independence of matches.

Discussion Status

The discussion is active, with participants providing various calculations and questioning each other's reasoning. Some participants have offered insights into the probabilities of specific matchups and the implications of different pairing arrangements, while others have raised concerns about the correctness of certain calculations.

Contextual Notes

Participants are navigating the complexities of probability in a tournament setting, with specific attention to the independence of matches and the conditions required for certain players to meet. There is an emphasis on ensuring that calculations reflect the rules of the tournament and the probabilities of winning based on player matchups.

Natasha1
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Could anyone please help me with making a start... and a finish to this little exercise. Many thanks Nat.

Four people A, B, C and D are to play in a small table-tennis tournament played on a simple knock-out basis: their names are drawn at random to play in two pairs, then the two winners play in the final. The probability that a player beats a player with a later letter is 2/3. All matches are independent. Find the probabilities that:

1) A wins the tournament

2) C and D meet in the final

3) B and C meet at some stage

Must I use a tree diagram?
 
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1) A must win two fights, irrespective of the pairing arrangements. What is the probability of that?

2) This does depend on the probability that C and D does NOT meet each other in the first fight, and that BOTH win their first fight (irrespective of who beats A or B)
 
arildno said:
1) A must win two fights, irrespective of the pairing arrangements. What is the probability of that?

2) This does depend on the probability that C and D does NOT meet each other in the first fight, and that BOTH win their first fight (irrespective of who beats A or B)


Is 1) p(A) = 2/3 * 2/3 = 4/9
 
Correct! :smile:
 
arildno said:
1) A must win two fights, irrespective of the pairing arrangements. What is the probability of that?

2) This does depend on the probability that C and D does NOT meet each other in the first fight, and that BOTH win their first fight (irrespective of who beats A or B)

Is 2) p(C & D to meet in final) = 2/3 * 2/3 * 2/3 * 2/3 = 16/81
 
No.
1. There is 2/3 chance they'll not meet in the first round. 1/3 of those times, C will go to the finals, and 1/3 of those times again D will also go to the finals to meet C.
 
arildno said:
No.
1. There is 2/3 chance they'll not meet in the first round. 1/3 of those times, C will go to the finals, and 1/3 of those times again D will also go to the finals to meet C.

I see so...

P(C&D meet in final) = 2/3 * 1/3 * 1/3 = 2/27
 
3*3*3 equals 27 last time I checked..
 
arildno said:
3*3*3 equals 27 last time I checked..

lol, i did spot it straight away and corrected it :shy:

Right then...

for number 3 :rolleyes:
 
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  • #10
Now, for 3) calculate separately the probabilities for the disjoint events a) B&C meets in the first round b) B&C meet in the finals.

Either a) or b) may happen..
 
  • #11
arildno said:
Now, for 3) calculate separately the probabilities for the disjoint events a) B&C meets in the first round b) B&C meet in the finals.

Either a) or b) may happen..

a) p(B&C meets in the first round) = 1/3

b) p(B&C meet in the finals) = 2/27

So the p(B&C meet at some stage) = 1/3 * 2/27 = 2/81
 
  • #12
Incorrect!
a) is correct, but why do you think b) is correct??

Furthermore, you are to have probabilities of disjoint events EITHER of which occurring will mean that b&c meets at some stage. Should you multiply the probabilities together in that case?
 
Last edited:
  • #13
arildno said:
Incorrect!
a) is correct, but why do you think b) is correct??

Furthermore, you are to have probabilities of disjoint events EITHER of which will mean that b&c meets at some stage. Should you multiply the probabilities together in that case?


Ok so

a) p(B&C meets in the first round) = 1/3

b) p(B&C meet in the finals) = ?
Well I thought this was correct as for question b) the p(C&D to meet in final is 2/27 so why would p(B&C meet in final) would be different?

So the p(B&C meet at some stage) = 1/3 + ? = ?
 
  • #14
b) will be different in that in the first case, for example, C must always beat a BETTER player in order to go to the finals.
There is 1/3 chance that the initial pairing is (A&B, C&D) What is the chance that this was the set up AND that both B and C wins (and thus proceed to the finals to meet each other)?

Furthermore, it is a 1/3 chance that the initial pairing was (A&C,B&D) What is the chance that this was the set up AND that both B and C wins
(and thus proceed to the finals to meet each other)?
 
  • #15
arildno said:
b) will be different in that in the first case, for example, C must always beat a BETTER player in order to go to the finals.
There is 1/3 chance that the initial pairing is (A&B, C&D) What is the chance that this was the set up AND that both B and C wins (and thus proceed to the finals to meet each other)?

Furthermore, it is a 1/3 chance that the initial pairing was (A&C,B&D) What is the chance that this was the set up AND that both B and C wins
(and thus proceed to the finals to meet each other)?

b) P = 2/3 * 1/3 * 1/3 * 1/3 = 2/81

So p(B&C meet at some stage) = 1/3 + 2/81 = 29/81
 
  • #16
No, no, no!
Think again please.
 
  • #17
arildno said:
b) will be different in that in the first case, for example, C must always beat a BETTER player in order to go to the finals.
There is 1/3 chance that the initial pairing is (A&B, C&D) What is the chance that this was the set up AND that both B and C wins (and thus proceed to the finals to meet each other)?

Furthermore, it is a 1/3 chance that the initial pairing was (A&C,B&D) What is the chance that this was the set up AND that both B and C wins
(and thus proceed to the finals to meet each other)?

Let me see...


There is 1/3 chance that the initial pairing is (A&B, C&D) What is the chance that this was the set up AND that both B and C wins (and thus proceed to the finals to meet each other)? this gives 1/3 * 1/3 * 1/3 = 1/27

Furthermore, it is a 1/3 chance that the initial pairing was (A&C,B&D) What is the chance that this was the set up AND that both B and C wins
(and thus proceed to the finals to meet each other)? this gives 1/3 * 1/3 * 2/3 = 2/27
 
  • #18
Almost correct, but the probability for the first case is also 2/27, since C beats D 2/3 of the time.

Thus, the TOTAL probability of B&C meeting in the finals is 2/27+2/27=4/27

And the total probability of B&C meeting at some stage is therefore...?
 
  • #19
arildno said:
Almost correct, but the probability for the first case is also 2/27, since C beats D 2/3 of the time.

Thus, the TOTAL probability of B&C meeting in the finals is 2/27+2/27=4/27

And the total probability of B&C meeting at some stage is therefore...?

p(B&C meeting at some stage) = 1/3 + 4/27 = 13/27
 
  • #20
You're done! :smile:
 

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