Calculating Probabilities of a Table-Tennis Tournament

[Natasha1]

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?

 

[arildno]

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)

 

[Natasha1]

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

 

[arildno]

Correct!

 

[Natasha1]

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

 

[arildno]

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.

 

[Natasha1]

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

 

[arildno]

3*3*3 equals 27 last time I checked..

 

[Natasha1]

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

 

[arildno]

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..

 

[Natasha1]

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

 

[arildno]

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?

 

[Natasha1]

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 + ? = ?

 

[arildno]

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)?

 

[Natasha1]

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

 

[arildno]

No, no, no!
Think again please.

 

[Natasha1]

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

 

[arildno]

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...?

 

[Natasha1]

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

 

[arildno]

You're done!