I'm not really good at maths, but here's my guess. Take note that we win if the ball we put out is green.

__Theory 1__:

After extracting 9 balls we are confronted with the two possible situations:

1) The 10th extracted ball is green, so it is certain that we had 0% chances of extracting a non-green ball, and so 100% chance of winning.

2) The 10th extracted ball is not green, so now we know that jar contained 9 green balls and a non-green one. By replaying the game we have 1:10 odds of selecting the one non-green ball out of the 10 possible balls.

You said that you win if the last ball is green. So the worst chance of selecting a non-green ball is 1:10, meaning we have 90% chances of winning.

__Theory 2__: To simplify the problem, we must pick from a jar which contains either a green ball or a non-green ball. The odds here would be 1:1. But I think this reasoning is wrong because seeing 9 green balls taken from the jar, we would have to think that the frequency of green balls is way bigger than that of the non-green balls.

__Theory 3__: I belive "the person offering the wager has no information about what's in the set" is crucial to knowing the right answer. Because this is mentioned, we could abstractly assume that the ball color is generated as we pick, just as cards are generated in poker

, and not modify the result of the problem.

If the first ball we pick is green, and the next one is not green, we can assume that there is a 1:1 green ball-non green ball generation ratio so we would have 50% chances of the next ball picked being green.

If the first ball we pick is green,the 2nd ball is green, and the 3rd ball is not green we can assume that the generation ratio is 2:3. This is correct "on the long run" since I'm guessing somehow that we would have a chance of (0.33)^9 ( I hope this is correct) chance to hit 9 green balls with these odds.

If the first 9 balls are green and we don't know anything about the last ball, we must come back to the two cases: 50% chance that we won 100% and 50% chance we have a 9:10 chance of winning. From here I would conclude we have 95% chance of winning.

T3 seems like a deeper explanation of T1, so the odds of 95%( or 19:20) being the correct answer are 2:1

D.

I notice the problem is asking for "worst" odds so I'm wondering if my answer is that they want.

PS: The problem does not include the option of having 0 chance of winning.

As silent21 said, we care only if the ball is green or not. That's two options, even if the non-green could mean 16million colors.