It depends on whether the odds are in favour or against.
Generally, the odds refer to a wins versus b losses. That is a wins out of a total number of a+b outcomes and so a probability of winning = a/(a+b).
If that's the case, then the odds of the visiting team of winning is 1 to 3.
The probability is 4/15
21%
Let P(A) = 1/10; P(A) = probability of selecting one people on a basketball team P(B) = 1/35; P(B) = probability of selecting one people on a football team P(C) = 1/10 = probability of selecting one people who plays in both teams P(D) = probability of selecting from either team. P(D) = P(A) + P(B) - P(C) P(D) = 1/10 + 1/35 - 1/10 P(D) = 1/35 or 0.0286
A fair chance of winning is more than 50% chance of winning. Therefore probability = 0.5 We need to find a fair chance of winning atleast one match. 1-(5/6)^n > 0.5 hence, n=4 QED (quite easily done, :-p)
If that's the case, then the odds of the visiting team of winning is 1 to 3.
The probability is 4/15
Statistically, there is a 1 in 32 chance any NFL team can win the championship. The comparisons of team strength during the season results in "odds" that are updated by bettors in locations such as Las Vegas.
50 %
The odds that the US soccer team would win the 2014 World Cup is about 23 percent.
The odds were alot
no The probability of that happening is 10% because the Sonic Team Members are a bunch of prudes.
1/2
1/4
1/3
Such probabilities are impossible to calculate exactly or even approximately. However bookmakers will quote you odds if you wish to bet. They will base their odds on how each team is presently playing and their current records.
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