Calculating Probabilities of a Table-Tennis Tournament

In summary: B and C wins? This gives 1/3 * 2/3 * 2/3 * 2/3 = 8/81...thus proceed to the finals to meet each other? This gives 1/3 * 2/3 * 2/3 * 2/3 * 1/2 = 4/81Therefore, p(B&C meet at some stage) = 1/27 + 8/81 + 4/81 = 19/81In summary, in a small table-tennis tournament played on a simple knock-out basis, where four players (A, B, C, and D) are randomly paired and the probability of a player
  • #1
Natasha1
494
9
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?
 
Physics news on Phys.org
  • #2
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)
 
  • #3
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
 
  • #4
Correct! :smile:
 
  • #5
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
 
  • #6
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.
 
  • #7
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
 
  • #8
3*3*3 equals 27 last time I checked..
 
  • #9
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:
 
Last edited:
  • #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:
 

FAQ: Calculating Probabilities of a Table-Tennis Tournament

What is the formula for calculating the probability of winning a table-tennis tournament?

The formula for calculating the probability of winning a table-tennis tournament is the number of possible outcomes where a player wins divided by the total number of possible outcomes. This can be expressed as P(winning) = # of winning outcomes / total # of outcomes.

How do you determine the probability of a specific player winning the tournament?

To determine the probability of a specific player winning the tournament, you need to consider their performance and the level of competition. You can also look at their past record and calculate their win percentage to estimate their probability of winning. However, it's important to keep in mind that probability is not a guarantee of future results and there are other factors that can impact the outcome of a tournament.

Can you use the same formula to calculate the probability of winning for a team in a team table-tennis tournament?

Yes, you can use the same formula to calculate the probability of winning for a team in a team table-tennis tournament. The only difference is that you need to consider the number of players in the team and their individual probabilities of winning in order to calculate the team's overall probability of winning.

How do you account for ties or draws in the probability calculation?

If there is a possibility of a tie or draw in the tournament, you need to include these outcomes in your calculation of the total number of possible outcomes. For example, if there are two players who can potentially tie in the final match, the total number of outcomes would be the number of outcomes for a win by the first player plus the number of outcomes for a win by the second player plus the number of outcomes for a tie.

Is it possible to accurately predict the probability of winning a table-tennis tournament?

While probability calculations can give us an estimate of the likelihood of winning a tournament, it is not possible to accurately predict the outcome of any sporting event. There are many variables and factors that can influence the outcome and it's important to remember that probability is not a guarantee of future results. However, using probability calculations can help us make informed decisions and assess the potential risks and rewards of participating in a table-tennis tournament.

Similar threads

Can a Tournament Graph Define a Total Order Without Three-Player Cycles?
Replies
1
Views
1K
Optimizing Stakes in a Match: Probability Strategies for Player A and B
Replies
11
Views
2K
Probability of a team wining a best of 3 game
Replies
9
Views
14K
Probability in knock-out torunament
Replies
1
Views
2K
B Validating Darts Tournament Skill Levels: A Non-Intrusive Approach
Replies
4
Views
1K
MHB Kate's Chances of Winning a 5 Match Tennis Tournament
Replies
5
Views
2K
MHB Probability of winning a point on serve in tennis
Replies
4
Views
3K
MHB Combinatorics problem. Discrete Mathematics II
Replies
2
Views
2K
Conditional Probabilities in Tennis
Replies
7
Views
4K
Calculating Probability for a Game of Picking White Balls: Three Players
Replies
9
Views
1K

Hot Threads

Recent Insights

Back
Top