In a game of drawing straws, why are all turns equally good? For example, there are 3 straws in a pile - 2 long and 1 short. The person who draws the shortest straw loses. When a straw is drawn, it is removed from the pile. Drawing first, second, or last all apparently result in the same probability for drawing the shortest straw.
For me, this is counter-intuitive. I originally thought that drawing the straw later would be more beneficial because the previous drawers would have to not draw the short straw before you have a chance to. Is there a reason why drawing on all turns results in the same probability for drawing the short straw?
 A: It has to do with the difference between conditional and unconditional probability. A conditional probability is one where you are only looking at situations with a certain amount of known information, whereas the unconditional probability looks at the overall probability. For example, if you're the second person to draw, then there are two conditional probabilities:
(1) The probability that you draw the short straw, given that the first person didn't draw it.
(2) The probability that you draw the short straw, given that the first person did draw it.
The probability of (2) is clearly zero, since if the first person drew the short straw then you have no chance of doing so. By comparison, the probability of (1) is 1/2, since after taking one long straw out there are two straws left and one of them is the short one.
But then what's the probability, overall, of you drawing the short straw? In other words, what's the unconditional probability? There's a formula for that, which boils down to taking each of the conditional probabilities, multiplying them by the probability of that condition happening, and adding them together. In symbols, it looks like this:
$P(A) = \sum_B P(A|B)P(B)$ where $P(A|B)$ is the probability of event A occuring, given that event B occured, which is then multiplied by $P(B)$, the probability that event B occured, summed over all of the possible event Bs*. In our case:
Probability you draw the short straw = (Probability you draw it given the first person didn't)×(Probability the first person didn't draw it) + (Probability you draw it given the first person drew it)×(Probability the first person drew it = 1/2 × 2/3 + 0 × 1/3 = 1/3
Because the first person had a 1/3 chance to draw the short straw, and a 2/3 chance to draw one of the long ones. Basically, the order of drawing doesn't matter because those two events are weighted in such a way that everything cancels out.
You can see what happens for the third person to draw as well - since you know the probability that the first person, or the second person, draws the short straw, and you should be able to work out the conditional probabilities of the third person drawing the short draw given that either the first person drew it, or the second person drew it, or neither of them drew it.
Another way to look at it, consider the case where everyone draws a straw simultaneously, but reveals one at a time. Clearly whether you won or lost was determined as soon as you took the straw, so it can't be influenced by the other people's choices, but functionally it works exactly the same as if you take turns drawing.
*One small clarification - those event Bs must be exhaustive, meaning that all possible results must be covered by at least one event, but also mutually exclusive, meaning that there's no overlap between the two. If you roll a six-sided die, then the events of rolling 1, 2, 3, 4, or 5 are mutually exclusive, but not exhaustive (because rolling 6 isn't covered), but the events of rolling an even number and rolling a number greater than 4 are not mutually exclusive, because rolling a 6 counts for both.
A: There are three ways to arrange 2-long and 1-short straws.  Thus short straw is equally likely to be in any order of draw.
If you go first, there is a one third probability of drawing short.
If you go second, there is also a one third probability of drawing short.   Now it's true that if the first person drew short the second cannot, and if the first person did not the second person has a 1/2 probability of drawing short.   But the total probability adds up:$$\tfrac 1 3 \cdot\tfrac 0 2 + \tfrac 2 3\cdot\tfrac 1 2 = \tfrac 1 3$$
Likewise if you go last, there is a one third probability of drawing short.   Alternatively, that is a one third probability that nobody will draw short before you draw.
A: Consider any number $n$ of straws (for convenience numbered $1$ to $n$) which are to be drawn one at a time by $n$ people.  Since (if there is no cheating) the process of straw-drawing is not affected by the numbers on the straws, straw number $1$ has the same probability of being drawn in any position (first, last or whatever) as straw number $2$ (or any other number from $1$ to $n$), namely $1/n$, and thus there is 
no advantage or disadvantage in drawing first, last or in any other position.
A: Imagine the straws to be placed (hidden) randomly in a line.
Since short straws have no preference for position you are equally likely to draw the short straw, whichever one you draw (or in whichever order you are made to draw)
