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The probability of winning all games is inherently conditional on the probability of reaching the quarter finals. To equate this to common examples of probabili
by seabee 12y ago
The probability of winning all games is inherently conditional on the probability of reaching the quarter finals. To equate this to common examples of probability, winning at the knockout stage is like flipping a biased coin and getting 4 heads in a row (= four matches won).
As such, there is no way the probability of winning can exceed the probability of reaching the quarter finals.
- StavrosK 12y agoThe coin won't get any advantage from playing in its home field, for example. I specifically said that I'm not arguing that this is the case here, but one can conceive of an example where it would.
- seabee 12y agoI cannot conceive of an example. In fact I argued that it is impossible because of the fundamental rules of conditional probability.
- sp332 12y agoProbability of losing is not transitive. If A is likely to lose to B, and B is likely to lose to C, it does not follow that A is likely to lose to C.
- pedrosorio 12y agoUnless we're using ELO to determine these probabilities.
- sp332 12y agoAw, you're right. I didn't read closely enough to understand their model.
- Someone 12y agoThis is a different situation. We have P(Belgium reaches quarters) >= P(Belgium wins overall) P(USA reaches quarters) >= P(USA wins overall) P(USA reaches quarters) >= P(Belgium reaches quarter) That does not imply P(USA wins overall) >= P(Belgium wins overall)