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Ok so here is my question. If the Tigers (team A) win 66% of the games they play on Tuesdays and the Sharks (team B) win 55% of the games they play on Tuesdays. Now when they play each other on a Tuesday what are the true chances of each team winning since they both cannot win?

Are these events considered disjoint events even though they are playing each other on a Tuesday?

Beto
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  • whats your thoughs on this? – Bak1139 Jul 09 '14 at 20:57
  • The only way I was thinking about this was by using manipulatives. I figured that the .45 of Team B losing would be of benefit to Team A somehow. I was thinking getting one jar and putting in 66 red marbles that represent the chances of A winning and adding 45 more red for the B team losing which is the same as Team A winning and then adding 34 blue marbles for Team A losing which is the same as Team B winning and 55 blue for team B winning. Then 111/200 gives 55% which was wrong. – Beto Jul 10 '14 at 02:35

1 Answers1

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There is no way to find an answer without making unreasonable assumptions. It is conceivable that in this type of game, the "better" team always wins.

So let us make unreasonable assumptions. One possible model is that each team tosses a coin that has the appropriate probability of landing "heads." We calculate the probability that A's coin lands heads, given that exactly one coin landed heads.

The probability exactly one coin landed heads is $(0.66)(0.45)+(0.34)(0.55)$. It follows by the usual defining formula for conditional probability that the probability A's coin landed heads, given exactly one did, is $\frac{(0.66)(0.45)}{(0.66)(0.45)+(0.34)(0.55)}$.

Remark: If A and B are playing each other on a Wednesday, the events "A wins" and "B wins" are disjoint. That, however, is of no significant help in computing the probability that A wins.

André Nicolas
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