Probability Question?

[Sethual]

We have been discussing this question at work and as with all things nobody can agree.

3 of us are drawing straws and there are 3 straws, one shorter than the other two and you are trying to draw the short straw.
Here's the question:
Does the first person have a statistical disadvantage against the other two?
or does the fact that he "may" draw the short straw the first time negate the last two's advantage?

Here's my thinking: the first persons odd's are 33.3%, the second guy has just increased his odds to 50% along with the third guy. The fact that the first guy could draw the short straw on his first pick reduces the last two guys total percentages down to 66%, then they have 50% of the total 66% which would leave them 33% each making everyone equal.
But I can't help but think drawing to 50% as opposed to 33% would give me some kind of advantage in the long run.

Any help or new ideas on this would be greatly appreciated.
I realize that this might be the most obvious answer to anyone that knows anything about this subject but i just can't get my mind to settle on one answer.

Thanks, Sethual

[JKetzer]

The first person has a 33% chance of drawing the straw. If he does, the 2nd has a 0% chance of drawing it; therefore, the 2nd person has a 0% chance 33% of the time.

The other 66% of the time, he has a 50% chance of drawing it.

Therefore:

(1/3)*0 + (2/3)*(1/2) = 1/3.

Everyone has an equal 1/3 chance of drawing it because the first guy will still guess correctly sometimes.

[niin]

Another way to look at it is to, instead of going one by one and drawing a straw, have each person just indicate which one they wish to pick, but not actually draw it yet. Then all people pull the straw out at the same time.

[jeffnc]

[bgaineshunter@yahoo.com]

The first person has a 1/10 chance
2nd has 9/10 * 1/9 = 1/10
3rd has 9/10 * 8/9 *1/8 = 1/10
4th has 9/10 * 8/9 * 7/8 * 1/7 = 1/10

[dev_cat]

Firstly, I agree with all the previous responses and think the first person doesn't have an advantage. But it did remind me of the Monty Hall paradox, which maybe what your workmates are thinking of (?)

****

A thoroughly honest game-show host has placed a car behind one of three doors. There is a goat behind each of the other doors. You have no prior knowledge that allows you to distinguish among the doors. "First you point toward a door," he says. "Then I'll open one of the other doors to reveal a goat. After I've shown you the goat, you make your final choice whether to stick with your initial choice of doors, or to switch to the remaining door. You win whatever is behind the door."

You begin by pointing to door number 1. The host shows you that door number 3 has a goat.

Do the player's odds of getting the car increase by switching to door number 2?

[Wenona]

If the host knows where the car is and always opens a door with a goat before asking you if you want to change, you should always change as the remaining door you haven't picked has a higher probabilty of being the car than the one you picked originally.

Why, because you have a 33.33% chance of picking the right one originally. Given the assumptions stated, if you didn't pick the car originally, that is 66.66% of the time, the car will be the other door left after the host knowingly reveals a goat.

[cybrarian]

Some earlier discussion on this

[dev_cat]

[the_hawk]