Match Point. 2. Find the probability that Holly wins the game by getting 4 points before Joe does?

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Match Point om and olly decide to play 1 game of tennis. he first player who gets to 4 points wins the game. olly has a 60% probability to win any given point against om.

1 Match Point om and olly decide to play 1 game of tennis. he first player who gets to 4 points wins the game. olly has a 60% probability to win any given point against om. Note: he actual rules of a tennis match are different. he winning player must win 4 points but must also win by two points. With this rule, 4 points to 3 points is not a win and they must play on until one player gets 2 point more than the other. In the game Sandy and Joe play the person wins when they get to 4 points making 4 to 3 a win. 1. he probability that olly wins the game by getting 4 points before om does is A) 60%! B) More that 60%! C) Less than 60%!!! 2. Find the probability that olly wins the game by getting 4 points before Joe does? 3. If the probability that olly wins any given point increase to 63% how much does the probability that olly wins the game improve? om olly 2018 Joseph Eitel! Page 1! amagicclassroom.com

2 Solutions Question 1. Answer: C. Greater than 60% olly has an advantage over om to win every point in the game. As the game progresses it becomes less and less likely that Blake will be able to win. You can say that the probability of Blake getting 4 "upsets" p(4) is less likely than a single upset P() =(40%). Question 2. Answer. olly has about a 71% chance of winning 4 0 Can happen 1 way.! p Can happen 4 different ways! p 4 1 ( ) = 1i.6 4 i.4 0 =.1296 ( ) = 4 i.6 4 i.4 1 = Can happen 10 different ways! p( 4 2) = 10 i.6 4 i.4 2! ( ) = 20 i.6 4 i.4 3! ( )! ( )!.7102 ( )! 71% 4 3 Can happen 20 different ways! p 4 3 Question 3. Answer: olly has about a 76% chance of winning. An increase of 5%. ( ) = 1i.63 4 i.37 0!.1575 ( ) = 4 i.63 4 i.37 1!.2331 ( ) = 10 i.63 4 i.37 2!.2157 ( ) = 20 i.6 4 i.4 3!.1596 ( )! ( )!.7659 ( )! 76% An increas3 of 5% 4 0 Can happen 1 way.! p Can happen 4 different ways! p Can happen 10 different ways! p Can happen 20 different ways! p ways 4 can happen if the sequence starts with a 4-0! 4-1!!! 4-2!!!!!!! 4-3!!!!!!!!!!!!! here are 15 ways 4 s can happen if the sequence starts with a !!! 4-3!!!!!!!!!! 2018 Joseph Eitel! Page 2! amagicclassroom.com

3 here are 20 ways 4 s can happen if the sequence starts with a 20 ways 4 can happen if the sequence starts with a 4-0! 4-1!!! 4-2!!!!!!! 4-3!!!!!!!!!!!!! 2018 Joseph Eitel! Page 3! amagicclassroom.com

4 here are 15 ways 4 s can happen if the sequence starts with a 15 ways 4 can happen if the sequence starts with a !!! 4-3!!!!!!!!!! 2018 Joseph Eitel! Page 4! amagicclassroom.com

5 Is there an easier way to count the number of outcomes? It is a lot of work to design the branching diagram and then count the number of outcome for olly and om. One thing we can see from this diagram is that we need to count the number of ways that and can occur is that order matters. follows a different branch and produces a different outcome so the order of the outcomes matter. hat means we are counting the number of Permutations. he permutation formula most students are familar with is shown below. Permutation Formula for n different items he number of Permutations of r items selected from n different items without replacement where each different ordering of the same items is counted as a different permutation is found by n P r = n! (n r)! his formula does not apply here as we do not have n different items. When we count the number of ways olly can win 4 points and om wins 2 we have 6 points and we want to occur 4 times and to occur 2 times. here is a permutation formula that computes the number of permutations of n items where some of the items are repeated. Permutation Formula for n items where some of the items are repeated he total number of Permutations of n items using all n items without replacement where one item occurs n 1 times and a second item occurs n 2 times. = n! n 1!i n 2! owever this formula has a problem. It computes the all the permutations of and for a given 3! number of and. If we have 2 and 1 it counts and and 2!i1! = 3 If we want to count the number of ways olly can win 2 point so we cannot include permutations that end in a. We also need do not want to include as olly will have already won the game after the first 2 points so cannot happen. he branching for 3 games is shown on the left below. It shows all the outcomes of 3 points. If we only look at the number of ways we can get to 1 and 1 then we can use the formula to compute the number of these. IF we are at 1 and 1 then each of these branchings will need 1 more to end that brance. 3 ways for 2 1 = 3! 2! = 3 2 ways for 1 1 = 2!i1! 1!i1! = 2 P(2,1) = P(2,1)i P() We can use this technique to find the number of outcomes for the branch just before the winning branch and then add the last branch with an onto any branches where an is needed. his branch adds 1 more to the outcome so the probabilities for the branches can be adjusted by the probability of getting the last Joseph Eitel! Page 5! amagicclassroom.com

6 What do we do now? Lets start with the case where we want to find the probability of olly winning 4 to 3. P(4,3) For olly to win 4 to 3 the score must have been 3 to 3 the step before, then she must win the next point. If we find the probability of P(3,3) and then multiply by P() we will have found P(4,3). At each point in the branch where there are 3 and 3 the permutation can end in either or. We do not care. We only need to find the number of ways that 3 and 3 can happen If we use the formula for finding the number of ways the 3 and 3 happen we can find the probability of P(3,3) and then multiply by P(). We use the same idea for finding the probability of P(4,2) by looking at the branches that have 3 and 1. At this point the branch will end in a or and the next branch MUS end in an Likewise we find the probability of P(4,1) by looking at the branches that have 3 and 1. At this point the branch will end in a or and the next branch MUS end in an. And finally we find the probability of P(4,0) by looking at the branches that have 3 and 0. At this point the branch will end in a or and the next branch MUS end in an Look at the branches that ends in! Look at the branches that ends in 3 and 3! 3 and 2 P(4,3) = P(3,3)i P( P(3,3)i P() = 6! 3!i 3! i.63 i.4 3 P(3,3)i P() = 6! 3!i 3! i.64 i.4 3 P(4,3) = 20 i.6 4 i.4 3 P(4,3) = P(4,2) = P(4,2) i P( 5! P(3,2)i P() = 3!i1! i.63 i.4 2 P(3,2)i P() = 5! 3!i 2! i.64 i.4 2 P(4,2) = 10 i.6 4 i.4 2 P(4,2) = Look at the branches that ends in! Look at the branches that ends in 3 and 1! 3 and 0 P(4,1) = P(3,1) i P( 4! P(3,1)i P() = 3!i1! i.63 i.4 1 P(3,1)i P() = 4! 3!i1! i.64 i.4 1 P(4,1) = 4 i.6 4 i.4 1 P(4,1) = P(4,0) = P(3,0) i P( 3! P(3,0)i P() = 3!i 0! i.63 i.4 0 P(3,0)i P() = 3! 3!i 0! i.64 i.4 0 P(4,0) = 1i.6 4 i.4 0 P(4,0) =.1296 p( olly wins)! !.7102! 71% 2018 Joseph Eitel! Page 6! amagicclassroom.com

7 he actual rules require that you win by 2 points. he winning player must win 4 points but must also win by two points. With this rule, 4 points to 3 points is not a win and they must play on until one player gets 2 point more than the other. You win if the score is 4 0, 4 1 or 4 2. If the score is 3 3 then you cannot win on the next point as you have scored 4 points but you do not lead by 2 points so the game continues until one of you has a 2 point lead. he 3 3 score is call deuce. 6! he number of ways that 3 3 can happen 20 different ways 6 c 3 or 3!i 3! At this point the game could end at 5 3 if a player wins the next two points. In other cases the score could be 6 6 and a winner is still not determined. Each time the score has both platers with an equal number of points after the first deuce the score is also called deuce so deuce can happen many times in a game. Under these rules, the probability of winning after you reach deuce is p 2 1 2p(q) ennis Facts he most points in a single game occurred in a match played in 1975 between Anthony Fawcett and Keith Glass. It took 80 points to complete 1 single game game and the played to 37 deuces. he Isner Mahut match at the 2010 Wimbledon Championships holds the record for the longest tennis match both in time and games played. he match took 1 hours and 5 minutes. John Isner def. Nicolas Mahut, 6 4, 3 6, 6 7, 7 6, (3 days). he longest rally in tournament play to score 1 point was one of 643 times over the net between Vicky Nelson and Jean epner at Richmond, VA in October Joseph Eitel! Page 7! amagicclassroom.com

GCSE MATHEMATICS Intermediate Tier, topic sheet. PROBABILITY

GCSE MATHEMATICS Intermediate Tier, topic sheet. PROBABILITY. In a game, a player throws two fair dice, one coloured red the other blue. The score for the throw is the larger of the two numbers showing.