Calculate the odds?
If you have a group six people. Every year for three years there is a random drawing for two $1000 prizes. What are the odds that the exact same two people would be drawn every year?
If you have a group six people. Every year for three years there is a random drawing for two $1000 prizes. What are the odds that the exact same two people would be drawn every year?
ignoramus
Well, it depends upon what exactly you mean by "odds". This is an odd (!) expression to use outside of a betting shop on horse or dog-racing, and is not strictly defined - it may depend on the local interpretation.
Perhaps the "official" definition refers to the chance of an event NOT occurring versus the chance of it actually happening. So the odds of throwing a 3 on a standard die would be stated as "5 to 1" (or, maybe "5 to 1 against", to make it perfectly clear), since there are 5 chances of not getting a 3 against 1 chance of it being a 3.
So, in your scenario, we need to calculate the chances of getting the same two winners every year out of the 6 players.
There are 6 choices for the first winner, and 5 choices for the second, which means there are 6 x 5 = 30 possible pairs of winners. But if the two prizes are the same, and it does not matter which of the pair is drawn first, and which second, then the possible number of pairs is only 15 (because A followed by B is the same as B followed by A).
The first year it does not matter which pair is selected; you only seek to know the odds of getting the same pair in years 2 and 3.
In year 2, therefore, out of the 15 possible pairs, the odds on getting the same pair as in the first year are 14 to 1 against (fourteen wrong pairs, only one "good" pair ).
The same situation applies again in year 3. The odds of getting the same pair again are 14 to 1 against.
So the odds of the event happening in both years 2 and 3 (14 to 1) times (14 to 1) , which comes to 196 to 1 against.
(Think of it as if you had bet $1 on it happening in year 2 : you should win $14 if it comes up.
You then bet your $14 on it happening again the 3rd year. So for every dollar you win another dollar, making total winnings of 14 x 14 = $196 for your original $1 bet.) So the odds on your winning are 196 to 1 against.
But, like I said, it depends upon the way that you state the "odds", and the way that the term is interpreted.
Steve A
Assume the rules are two people are chosen each year; i.e., the person who wins one $1000 prize cannot win the other prize that same year. Two people are chosen the first year. P = 1.0
Probability they are chosen the second year: (2/6)(1/5) = 1/15
Third year: 1/15
P(3 years with same 2 people): (1)(1/15)(1/15) = 1/225