No, but they are related.
Suppose an experiment is repeated N times, and X outcomes are favourable to the event A.
Then the probability of A is X/N
The odds for A are X/(N - X)
Note that (N-X) are the number of outcomes that are favourable to "not-A".
Yes
The probability is 0.0198. The probability of that happening twice is 0.000393 Is this the odds of it happening twice with the same number?
Odds against A = Probabillity against A / Probability for A Odds against A = (1 - Probabillity for A) / Probability for A 9.8 = (1 - Probabillity for A) / Probability for A 9.8 * Probability for A = 1 - Probability for A 10.8 * Probability for A = 1 Probability for A = 1 / 10.8 Probability for A = 0.0926
For 5 to 3 odds of winning; the probability of winning is 0.375 or 37.5%
Your "odds" are your likelihood or probability of winning.
Probability is a numeric expression of the "odds" that something will happen.
Just take the probability, which is a decimal number between 0 and 1, and convert it into a fraction. For instance, a probability of 0.75 corresponds to odds of 3 in 4.
For 9 to 11 odds of winning;Probability of winning:0.55, or;Chance of winning:55%
The probability of both dice showing the same number is 1/6 and the probability of different numbers is 5/6.
Odds.
Answerodds the odds in favour are p/(1-p)the odds against are (1-p)/p
If the probability is 5/12 then that is an acceptable answer in itself however it is also the same thing as 5 divided by twelve or .4167