Need help calculating odds of a dependent event

Scenario: There are ten numbers in a hat, you choose one at random and do not replace it. Suppose 6 numbers are considered "Bad" and 4 are considered "Good." Is the odds of getting a Good number the same for each of the ten positions (picks)? I have been discussing this scenario with my colleagues but we come up with different ways of looking at it. Someone suggested calculating Expected Value. Please help me with the math behind this.

Quote: betonace

Scenario: There are ten numbers in a hat, you choose one at random and do not replace it. Suppose 6 numbers are considered "Bad" and 4 are considered "Good." Is the odds of getting a Good number the same for each of the ten positions (picks)? I have been discussing this scenario with my colleagues but we come up with different ways of looking at it. Someone suggested calculating Expected Value. Please help me with the math behind this.

The odds of getting a "good" number are increased with each previous "bad" number picked, and decreased with each previous "good" number picked. So the odds change in one direction or the other with each of the ten positions. I think you can see this most easily when you get to the 10th position; by then your odds are either 100% or 0% of getting a "good" number, because all the others are known by then, and it's either a "good" number or "bad" number left for that position. (You started with a 40% chance of selecting a "good" number in the first position).

But if 10 people are all drawing 1 number from the hat (this seems to be implied in the question?), then each position has the same chance of getting any number. Doesn't change before you start drawing.

It might get more interesting if you are allowed to choose when to draw as you watch other people draw.

Quote: dwheatley

But if 10 people are all drawing 1 number from the hat (this seems to be implied in the question?), then each position has the same chance of getting any number. Doesn't change before you start drawing.

It might get more interesting if you are allowed to choose when to draw as you watch other people draw.

Yeah, I took it that you would know what the results were prior to each draw (or else why even mention that the numbers would not be returned to the pool?). You're talking about a simultaneous scenario. I agree it would be static odds in that case; 40% of the people drawing would get a "good" number.

You dont know the results of the previous picks. What I'm saying is what position would you pick? We agree the first position has a 40% chance of picking a good number, and the 10th position could have either a 0% or 100% chance of being good. so is there some sort of mathematical value that encompasses the many different cases for each position?

If you don't know the previous picks, then every pick has a 4/10 chance of being a good number. This is the same premise for how basic strategy for blackjack is calculated.

It does not matter what position you pick. Your probability of getting a good number will always be 40%.

One way to show this mathematically…first realize there are C(10,4) = 210 total combinations. And there are C(9,3) = 84 combinations with a good number in a certain position. So the probability that a good number will be in a certain position is 84/210 or 0.4.

In fact, even if you could choose your position with KNOWING the results of previous picks—any strategy will still give you a 40% chance of a good number. It does not matter.