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Local football association has $10$ teams. Team A has $40\%$ chance to score a win in the game against better ranking opponent, and $75\%$ against worse ranking opponent. If team A is currently $4$-th on the ranking list, define probability that A will win in the next game.

So I have done the following here $0.4 \cdot 0.3$ ($3$ out of $10$ are better ranking opponents)+ $0.75 \cdot 0.6$ ($6$ are worse ranking opponents) = $\frac{1}{12} + \frac{3}{24} = \frac{5}{24}$.

I am not sure whether or not this is correct, but I plead you to elaborate this claim of mine.

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    $\begingroup$ Team A cannot play with himself so the probability to play with a better team is $ \frac{3}{9}$, and with a worse team $1-\frac{3}{9}$, of course if there are not other teams with the same ranking as team A. Otherwise you are correct $\endgroup$
    – clark
    Jul 2 '12 at 19:54
  • $\begingroup$ Ah, you are right! I forgot that fact! Thank you @clark! $\endgroup$
    – Takarakaka
    Jul 2 '12 at 20:00
  • $\begingroup$ There's an unstated assumption in your solution that the probability to be the next team played against is equal for all other teams. Since this probability depends on against whom they already played, and their ranking isn't independent from that (the worse the teams they've already played against, the higher their expected ranking), this also enters the probability. $\endgroup$
    – celtschk
    Jul 2 '12 at 20:59
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As celtschk remarked, the question hinges on rather unrealistic assumption that the team's next opponent is chosen randomly, with each of $9$ possible opponents equally likely ($9$ rather than $10$, as clark points out). Under the above assumption, we calculate the probability of playing a better or a worse team: $$P(\text{better opponent})=\frac39,\quad P(\text{worse opponent})=\frac69$$ Using the conditional probabilities $$P(\text{win } | \text{ better opponent})=0.4,\quad P(\text{win } | \text{ worse opponent})=0.75$$ we arrive at $$P(\text{win})=\frac39\cdot 0.4+\frac69\cdot 0.75$$

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