mathematical reason why China always win Uber Cup by 4-1. here is a mathematical explanation why China always win by 4-1 instead of a dominating 5-0 in the Uber final events. as we all know that winning an individual event isn't a 100% matter. there are many factors contributing to the win and loss of a match. if we are allowed, let's peg a probability to China winning a match. say that's P(win). a Uber cup team tie has 5 events. as established above, we have a simple Binomial distribution of the probably P(n/N), ie. the probability of winning n trials in N Bernoulli trials. the graph of the probabilities are as plotted in the graph. this is when P(win) = 0.84. and that's where P(4/5) ~= P(5/5). in laymen terms, the probability of winning 5 out of 5 events is the same as the probability of winning 4 out of 5 events are the same. if we increase P(win) to > 0.84, then P(5/5) is higher, if we decrease P(win) to < 0.84, then P(4/5) is higher. and to me, a probability of P(0.84) is quite high in general. that's requires a very high level of consistency as well as external factors like illness and injury. and more importantly, their opponents are no push-overs either, as we can see, Korea (the previous finalists) are quite strong as well. thus that explains why China always lose one match out of 5 possible. and why LiYongBo always emphasize consistency. for the ppl who do not understand the above mathematical explanation, it is ok, it is not any mathematical breakthrough but just some small insight. basically it means that to win 5/5 is quite hard.

Cool Why didn't I think of that before? But we need the combinations of 3-0 and 3-1. Never get 5-0 since not all matches will be played once 3 wins are reached. Sheesh, the things badminton fanatics do..............

Binomial distributions...statistics...haha good stuff Kwun. Maybe I can include some of this in my presentation in my statistics class. Well...only if you give me permission to

Binomial is what contributes to the bell curve of averages. It concerns with the calculation of probabilities of something happening within the distribution curve, given the number of samples and some other controlled elements. ie. The distribution of player skill levels in badminton out of 1000 samples may yield 200 beginner levels 350 intermediate 300 advanced 100 professionals 50 world class thus, the bell curve. I don't quite remember how to explain the P(N/r) calculation thingy. Darn it, it's been eons since I did statistics and I don't regret it one bit - I never liked statistics and probabilities! I'm the calculus/integration type of person.

Binomial Badminton In mathematics, the binomial distribution is a discrete probability distribution which describes the number of successes in a sequence of n independent yes/no experiments, each of which yielding success with probability p. Such a success/failure experiment is also called a Bernoulli experiment. A typical example is the following: 5% of the population are Badminton-freak. You pick 500 people randomly. How likely is it that you get 30 or more freaks? The number of freaks you pick is a random variable X which follows a binomial distribution with n = 500 and p = .05. There's a standard way to compute the chances but it's difficult to write down in formulas here.

Final Thomas Cup Hey Kwun, estimate some chances for the Thomas Cup Final match and calculate the score for us - Mark.

The number you are looking for so that P(5-0) = P(4-1) is 5/6 or 0.833333 recurring. Why? There are 5 ways of winning 4-1 ( you can lose the 1st,2nd.3rd,4th or 5th match). So when the P(win) = 5 x P(Lose) we get what we want. And this is when P(win) = 5/6 and P(Lose) = 1/6 (I had gone into more detail, but my PC crashed and I can't do it all again because TC final has started now)

does the mathematical model look different if they play to 3 wins instead of playing all 5 matches regardless of results?

There are only 20 possible outcomes instead of 32 (2**5) assuming each pair has 84% chance of winning a match Win 3-0 59.3% Win 3-1 28.4% Win 3-2 9.1% Lose 2-3 1.7% Lose 1-3 1.0% Lose 0-3 0.4% 96.8% chance of winning the tie (so the 84% looks too high) reduce the 84% to 66.666% ( 2/3 ) and you get Win 3-0 29.6% Win 3-1 29.6% Win 3-2 19.8% Lose 2-3 9.9% Lose 1-3 7.4% Lose 0-3 3.7% 79.0% chance of winning the tie for a 3-1 win to be more likely than a 3-0 win, each pairs chance of winning a match has to be below 66.666%

Neil what about the fact that the % chance of winning would be higher if you were leading 2-0 instead of at 1-1, i.e. the effect of the match score on the player performance, Can we have some figures on that please. Only kidding Perhaps Kwun and yourself can do a PHD in "Statistical Probablity of an Uber Cup tie being won by China" I calculate the probablity of England winning a Thomas Cup tie as follows England win: MS1 1% MS2 2% MS3 3% MD1 50% MD2 15% Please tell me if we can win a Thomas Cup in my lifetime

the USA was strong in badminton the 50's (or was it 40's). if even that is possible, and only a mere 50-60 years ago, there has got to be some small hope for England.