4 ms·
This conjecture seems to be easily seen as true by intuition. It's surprising that it took so long to find a proof.
by huahaiy 10y ago
This conjecture seems to be easily seen as true by intuition. It's surprising that it took so long to find a proof.
- vecter 10y agoCan you explain to me how this conjecture is intuitively true?
- andreareina 10y agoLet's say A has a 0.7 probability, and B has 0.6, and are independent. The compound probabilities are gotten by multiplying the individual ones: Neither: 0.3 * 0.4 = .12 A, not B: 0.7 * 0.4 = 0.28 B, not A: 0.3 * 0.6 = 0.18 Both: 0.7 * 0.6 = 0.42 Now let's say that they're not independent; in fact, B absolutely requires A. You'd get something like: Neither: 0.3 (this is reduced to the chance of not A) B, not A: 0 (B can't happen without A) Both: 0.6 (only other probability with B, has to make up the 0.6) A, not B: 0.1 (simply what's left to sum to 1.0) Concrete examples of independent events: (fair) coin tosses -- the chance to come up heads is always the same, no matter the prior result. Related events: being dealt a face card in blackjack, and winning the hand -- whatever your normal chance of winning a hand, the odds of winning that particular hand just went up.
- deleted 10y ago[deleted]