wait, just to make sure, you're saying 0.7 x 0.3 = 0.21 and 0.21 x 2(since it's 100% each) = 0.42 soo.... 0.3 x 0.4 = 0.12 x 2 = 0.24 0.24 + 0.42 = 0.26...? which makes what sense and not equal 1? err....explaination?
Typo. I corrected it. Chance of A win = 0.3 Chance of B win = 0.4 Therefore, Chance of both A and B win = 0.3 x 0.4 Chance of both A and B lose = 0.7 x 0.6 Chance of A win and B lose = 0.3 x 0.6 Chance of A lose and B win = 0.7 x 0.4 You will find that it all sums up to 1.0. You can also say that the chance of at least one of them winning is 1 - (chance of them both losing) = 1 - 0.42 = 0.58.
oops, correction on MY part too, i typed the wrong thing as well XD....but anyways, thanks for clearing that up! i forgot about the different situations that could happen!
I don't really care about their rankings as long as they put up a really good fight in the coming WC.And Bananaboy...umm...I just want to say that Malaysia's not as bad as you think.Anyways, have faith in the players but don't be too overconfident.Anything can happen.
Thank hcyong for your detailed explanations. I thought my calculation was quite simple and understandable, but ....
Young Indonesian player that Indra thinks will wreck havoc on everyone. If I spelled his name wrong, it is because I am not his fan.
You know what is funnier is the way he asked whether I know about scrambled eggs. I can see the humour Cooler. Great one!
This title is confusing. KKK/TBH will become world no.1 if they win the title right ? So if FHF/CY win the title, then they'll still be world no.1. Tell me if I'm wrong. Anyways. I feel that FHF/CY chances of winning are not as good as KKK/TBH. If you compare the draws between the two, KKK/TBH have it much easier than FHF/CY. The Chinese Pair has to face 3 pairs that they've had trouble in the past with. This includes Jonas Rasmussen/Lars Parske, Candra Wiyaya/Tony Gunawan and Choong Tan Fook/Lee Wan Wah. All very good pairs that have defeated the Chinese Pair in the past. In particular, the Denmark Pairing and the Malaysian pair have a good chance. I say this because FHF/CY has succumb to petty tactics from their opponents before. This includes fist pumping, non-courteous servicing, and provoking. In particular, Jonas and Lee Wan Wah. I'd be very impressed if they actually got through these pairs. Now if you look at KKK/TBH, the Malaysian pairing, they only have the Korean Pair to contend with, who haven't been doing well and the Danish Pair, who they haven't had significant problem in dealing with. So if I were to predict who's going to win the title, it's going to be KKK/TBH, simply because they don't have to shove out their skills until the Finals. Where as FHF/CY are looking at very difficult pairings. And by then, I think they would of been studied quite well.
No doubt KKK/TBh has a better draw compare v FHH/CY.. Sometimes.. U needs to facing some tough match for u to excel in the latter stages.. There were of couple of instances whereby some pair having a easy ride towards final. But then tumble in their last hurdle against much stronger opponents. Furthermore, I fink the stamina will not be a big problem for CHN pair if some of those games lasting for 3 sets..
Good analysis Angelou. For the same reasons, I am voting KKK-TBH to be world champions this year. I like to say sometimes the draws are kind but other times, they are not. CAI-FU would have to come up with their best but KKK-TBH should not let their guards down either,which I don't think they will.
I don't think cai/fu will get past ctf/lww cos it's at the QF stage and stamina factor will not count.ctf/lww may want to recall how they beat cai/fu in HKO06, ie play relax and don't burn up 80% of the fuel in the first game.Actually CTF would now be on an emotional high with ZJW's presence. Who knows he might be in such inspired form, ctf/lww may end up winning the WC title!
You can never beat hcyong for clarity. What took me one paragraph to put across he did it in 2 sentences and effortlessly clearer. Absolutely humbling. Often wonder and rather curious about his training background.
Haha... Interesting.. It reminds me of my probability lesson. Mathematically, probability of an event X can be denoted by P(X). Suppose A = Player A winning the trophy & B = Pair B winning the trophy Then P(A) = 0.3 & P(B) = 0.4 P(A') = 0.7 & P(B') = 0.6 (a) By formula on independent events, Probability of at least one winning the trophy = P(AUB) = P(A) + P(B) - P(A).P(B) = 0.3 + 0.4 - 0.3*0.4 = 0.7 - 0.12 = 0.58 # OR (b) By summing all possible outcomes, Probability of at least one winning the trophy = P(A|B) + P(A|B') + P(A'|B) = P(A).P(B) + P(A).P(B') + P(A').P(B) = 0.3*0.4 + 0.3*0.6 + 0.7*0.4 = 0.12 + 0.18 + 2.8 = 0.58 # OR (c) By Complement Rule, Probability of at least one winning the trophy = P(AUB) = 1 - P(A∩B)' = 1 - P(A').P(B') = 1 - 0.7*0.6 = 0.58 # (as explained by hcyong) Note: All three methods above should give u the same answers. Method A requires formula, if u don't remember it then that's it. Method B is good solution for events that have a few outcomes. Method C is the simplest & easiest.
