demolay,
How far did you get? If you show your work we'll be happy to help.
This is a part of a problem. I decided to ask it here after hours of trying to solve it...
Two tennis players make a match composed of 5 sets. Player A is better than Player B. Before tha match starts, PA's probability of winning the match (winning 3 sets out of 5) is 60%. And of course PB's probability of winning the match is 40%. According to statistics of previous 250 tennis matches done in that city, same player won the first two sets by 100 times. It means that a player started to match with 2-0 on sets. Out of this 100 (2-0)s, same player won the match 90 times.
Now, lets look at the match between PA and PB. If we know that PA won first two sets by 2-0, what is the probability of his winning of the match?
demolay,
How far did you get? If you show your work we'll be happy to help.
actually i am stuck. i guess we cannot calculate PA's probability of winning a single match. so i decided to go on with conditional probability. but i could not find the probability of PA's winning first two sets.Originally Posted by quark
In general winning first two sets probability is 100/250. and winning first two sets + winning the match's probability is 90/250. these are independetn probabilities and not connected with PA's ability. It is clear that PA's probability of winning the match should be higher than %60 after he wins first two sets. but how much it should increase? i also checked bayne's theorm but i miss something important and i cannot find it.
It is given that:
P(A winning match)=0.60
P(winning first 2 sets)=100/250=0.4
P(winning match | winning first 2 sets)=0.9
We need to calculate P(A winning match | winning first 2 sets). The probability should be greater than 0.9.
I am not sure, but it seems that P(A winning match | winning first 2 sets) = 0.6*0.9=0.54
but 0.54 is smaller than 0.6. Before the match starts probability of PA's winning is 0.6. After the match starts and he wins first two sets, his probability of winning the match should increase. am i wrong? how can someones probability of winning a match decreases if he winnes the sets?Originally Posted by quark
how can i find the exact answer?
Here is a way to look at the problem.
Player A has won the first two sets of the match. There are three ways that Player A can win the match.
1. He can win the third set. That probability is 0.6.
2. He can lose the third set and win the fourth set. That probability is 0.4 * 0.6 = 0.24.
3. He can lose the third and fourth sets but win the fifth set. That probability is 0.4 * 0.4 * 0.6 = 0.096.
The probability of winning the match is the sum of the probabilities of the three events described above: 0.6 + 0.24 + 0.096 = 0.936 .
Harvey Berman
[url]http://stattrek.com/[/url]
I also considered this as a possibility, using a tree diagram to lay it all out, but I abandoned it because:
P(A wins match vs player B) is not the same as P(A wins a set vs player B).
If you assume that in any given set within a match, the probability that A beats B is 0.6, then probability of player A winning an entire match against player B is higher than 0.6, which is specified at the beginning of the posted problem....
In fact if P(A wins any given set vs B) = 0.6, then the probabilities are for the various match-winning combinations:
W-W-W = .6^3 = .216
W-W-W-L = .6^3 * .4 * 3 = .2592
W-W-W-L-L = .6^3 * .4^2 * 6 = .20736
so the overall chance of player A winning a match against B would be .6826, which is not .6
so I don't think .936 is the answer.....
I wonder how bet ratios are calculated during a match. that situation is same with this question. we know the probabilities of winning match. but we dont know the probability of winning a single set. during the match, when a player makes is 2-0 on sets, how his probability of winning the match changes? how bet ratios are calculted? It should not be a hard question (except for me)
They're made on historical probabilities, recent player performance, and also playing on bettors' psychology (what they "think" the "informal" odds of winning are).....
ok but when they add all those data, they get some statistics. in the question we know historical data. also before the match starts we know the probability of winning. any idea?
John,
i love a slippery probability problem and sometimes i get them right. so here goes:
you wrote:
let me take that as a model and go forth.Originally Posted by JohnM
since we do not know that .6 is P(A wins a set vs B), but rather we believe that it is P(A wins a match vs B) let us allow p=P(A wins a SET vs B), now following your lead, we have:
p^3 + p^3*(1-p)*3 + p^3*(1-p)^2*6 = P(A wins a match vs B) = .6
or
p^3 + p^3*(1-p)*3 + p^3*(1-p)^2*6 - .6 = 0
which has the approximate solution p = .5537 = P(A wins a SET vs B)
now moving to the situation where two sets have allready been won by A we find possible end of match scenarios:
ww w with probability of .5537 for the final win
ww Lw with probability of .4463*.5537 for the final win
ww LLw with probability of .4463^2*.5567 for the final win
thus the probability of winning having allready won two sets is
.5537 + .4463*.5567 + .4463^2*.5567 = .911
i got the same answer when i took a slightly different attack on the problem by allowing the each set to be a bernoulli trial in a binomial experiment. in either case you MUST make the assumption that the winning of individual sets are independant events, which the original poster points out is not a real-life occurance.
that's the best i got tonight, its been fun
cheers
jerry
Jerry,
My overall point was that, as stated in the original problem, P(player A wins a match vs player B) is 0.6.
When Harvey posted a potential solution to the question "what is the probability that A wins the match vs player B, given that they are already ahead 2 sets to 0."
I had attempted the same approach, using a tree diagram, but Harvey assumed that P(A wins a single set vs player B) = 0.6, which I asserted as incorrect, because, if we use that probability for the outcome of a single set, player A's overall chances of beating player B in a match compute to 0.68, which is incompatible with the information given in the original problem...
...and it appears that you've gone "backwards" and gotten a very good estimate of the P(winning a single set), then used that estimate for the P(wins the whole match)
nice...but please tell me you'd NEVER give a problem like this for homework
John
not on my homework assignments,
i once had a colleague teaching econ who for some reason gave a crazy, and poorly stated, yahtzee probability problem for homework. he gave a solution for the problem that he intended to ask, not the one he actually asked.
he never has admitted that his solution was wrong, go figure
the other thing about this that caught my eye is that you could also consider that player A and player B are specific individuals and the .6 applies to their specific matchup, whereas the 100/250 and 90/250 would apply to any random match-up. that would further complicate things i think.
cheers
jerry
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