Viewing a response to: @rycharde/brainsteem-mathematics-challenges-quarter-finals-quirk
The way I see it, we can work it out like this: Suppose an English team (Man U) is drawn for spot "A"... then the possibility of them facing another English team (Arsenal) is (1/7). Arsenal's only chance to be paired with Man U is naturally in spot B. Suppose next, an Spanish team (Sevilla) is drawn for spot "C". The probability that they won't be paired with an Spanish team (Madrid) is 3/5. For the remaining Spanish team the proability that they dont' get put against a Spanish teams is 2/3. Multiplying these numbers gives: 6/105 = 2/35 or about 5.7%, not an impossibility, but a more-likely-than-not scenario! Note: Alternatively, we could calculate the probability by similarly placing Sevilla in slot C, then calculating that Madrid is placed in a favorable position: 4/5. Similarly, the remaining team has a 1/2 chance of being placed in a favorable position. This gives 2/35, the same result as above.
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rycharde | 0 | 37,922,424,762 | 49% |
Haha! I guess it depends which nation's press one reads. Yes, there was also the astonishment that the 3 Spanish teams avoided each other *and* the 2 Premier League teams were paired.
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sammy111100 | 0 | 579,394,992 | 100% |
It seems a bit odd, but I suppose if you run the numbers it's not impossible. By the way, I've posted what I hope is an interesting question on my own blog, and if you're interested, I want to invite you to take a look.
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Your calculations seem to be right. I got the same answer Freely place the first british team: propbability 1 free spaces 3x2+1 British teams together: probability 1/7 free spaces 3x2 Freely place Spanish team: probability 1 free spaces 2x2+1 Second Spanish team not with the first one: probability 4/5 free spaces 1x2+1+1 Third Spanish team with neither one of the Spanish: probability 2/4=1/2 So a total of 1x1/7x1x4/5x1/2=4/70=2/35 which indeed is approximately 5,7% I just saw it now since I was in school before. Well done!
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