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Statistics Question/Riddle
Rent
I'm always rightFuckin' deal with itRegistered User regular
I'm always rightFuckin' deal with itRegistered User regular
No this isn't a "do my homework" thread
So I was sitting on the crapper, playing pokemon, which lead me to think about statistics/game theory (no really). Specifically, the game "draw straws".
Like so:
Now, me being all clever I knew the "trick" to maximizing your win condition chances (assuming one short straw in the group, of course), and that is to draw first.
Let us for the sake of example take situation A, with 8 players in a game of "draw straws", assuming one short straw which none of the players desire to obtain. Player One has a 1/8 chance of drawing the short straw. Assuming the short straw has not been drawn, Player Two therefore has a 1/7 chance of drawing it- an increased chance of drawing the short straw. Player Three has a 1/6, Player Four a 1/5, and so on. Therefore whenever playing "draw straws", to maximize your win condition it is imperative to draw first (assuming that drawing the short straw is the lose condition, of course).
However, what if there were multiple short straws? For instance, let us modify situation A with five lose conditions- five short straws. Player One would have a 5/8 chance of drawing the short straw- clearly a disadvantage (a 62.5% chance of drawing a short straw, as a matter of fact). If Player One drew a short straw, Player Two would have a 4/7 chance of drawing the short straw- a clear improvement from Player One's chances (from 62.5% to roughly 57%- a five percent increase in win probability) . However, if Player One were to draw a not-short straw, Player Two would have a 5/7 chance of drawing it- from 62% to 71% chance of meeting a win condition.
My theory in the case of multiple short straws- multiple lose conditions- is to choose whoever's turn has a 50% or better chance of drawing the short straw and insert yourself as that position in the drawing(for instance, in modified situation A, Players One and Two are statistically likely to draw a lose condition, making Player Three with their 3/6 chance of drawing a short straw the one most likely to achieve his or her win condition). Is this right?
Finally, what if drawing the short straw was the win condition, not the lose condition (or, put another way, what if situation A had 7 short straws?)? Therefore, every single player would have a less than 50% chance of drawing the necessary straw-except for the second to last drawer (Player Seven). They would have a 50% chance, making theirs the most optimal slot.
Or does it truly matter? If you were player Eight in the above scenario, do you also have a 50% chance of drawing the necessary straw? When you think about it, Player Eight is 'choosing' whatever straw Player Seven doesn't. Is that a 50% chance, then? Or, does Player Eight have a 1/8 chance of getting the win condition? Player Eight has no influence whatsoever on the draw, simply 'picking' what no one else does- making Player Eight's 'draw' a 1/8 chance from the start. Or am I completely wrong and it's something else?
So I was sitting on the crapper, playing pokemon, which lead me to think about statistics/game theory (no really). Specifically, the game "draw straws".
wikipedia wrote:The group leader takes a number of straws or similar long cylindrical objects, and makes sure one of them is physically shorter than the rest. He then grabs all the straws in his fist such that all of them appear to be of the same length.
The leader then offers the clenched fist to the group. Each member of the group draws a straw from the fist. At the end of the offering, the person with the shortest straw is the one who must do the task.
Like so:
Now, me being all clever I knew the "trick" to maximizing your win condition chances (assuming one short straw in the group, of course), and that is to draw first.
Let us for the sake of example take situation A, with 8 players in a game of "draw straws", assuming one short straw which none of the players desire to obtain. Player One has a 1/8 chance of drawing the short straw. Assuming the short straw has not been drawn, Player Two therefore has a 1/7 chance of drawing it- an increased chance of drawing the short straw. Player Three has a 1/6, Player Four a 1/5, and so on. Therefore whenever playing "draw straws", to maximize your win condition it is imperative to draw first (assuming that drawing the short straw is the lose condition, of course).
However, what if there were multiple short straws? For instance, let us modify situation A with five lose conditions- five short straws. Player One would have a 5/8 chance of drawing the short straw- clearly a disadvantage (a 62.5% chance of drawing a short straw, as a matter of fact). If Player One drew a short straw, Player Two would have a 4/7 chance of drawing the short straw- a clear improvement from Player One's chances (from 62.5% to roughly 57%- a five percent increase in win probability) . However, if Player One were to draw a not-short straw, Player Two would have a 5/7 chance of drawing it- from 62% to 71% chance of meeting a win condition.
My theory in the case of multiple short straws- multiple lose conditions- is to choose whoever's turn has a 50% or better chance of drawing the short straw and insert yourself as that position in the drawing(for instance, in modified situation A, Players One and Two are statistically likely to draw a lose condition, making Player Three with their 3/6 chance of drawing a short straw the one most likely to achieve his or her win condition). Is this right?
Finally, what if drawing the short straw was the win condition, not the lose condition (or, put another way, what if situation A had 7 short straws?)? Therefore, every single player would have a less than 50% chance of drawing the necessary straw-except for the second to last drawer (Player Seven). They would have a 50% chance, making theirs the most optimal slot.
Or does it truly matter? If you were player Eight in the above scenario, do you also have a 50% chance of drawing the necessary straw? When you think about it, Player Eight is 'choosing' whatever straw Player Seven doesn't. Is that a 50% chance, then? Or, does Player Eight have a 1/8 chance of getting the win condition? Player Eight has no influence whatsoever on the draw, simply 'picking' what no one else does- making Player Eight's 'draw' a 1/8 chance from the start. Or am I completely wrong and it's something else?
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The person that draws last only has a 1/1 chance of drawing the short straw if you ignore the previous 7 picks. Everyone's chances in this scenario is 1/8 regardless of the order.
This.
More specifically, the first guy has a 1/8 chance of getting the straw.
The second guy has a 7/8 chance of actually taking his turn, in which he has a 1/7 chance of getting the straw. Thus having a 1/8 chance of getting the straw!
With this taken into account you'll find that order of drawing straws never effects any of the situations given, whether or not there is 1 'lose' straw, or 7
Person 1 has a 1/8 chance of getting the short straw.
Person 2 has a 1/7 chance of getting the short straw if the first person doesn't. That means they have a 1/7*7/8 chance = 7/56 = 1/8 chance, exactly the same as the first person.
Person 3 has a 1/6 chance if the first two didn't. That means 1/6*(6/7*7/8)=1/8 chance, exactly the same.
Etc.
Now, if the drawing happened sequentially, and each drawing was preceded by an ante, then betting would get steeper the more straws were drawn. But that's not the case.
This is one of the many reasons I had to take statistics twice (and accounting 3 times) before passing.
And I'm usually pretty good at math too.
Assuming you're talking about the Monty Hall problem switching actually doubles your chances to win and doesn't really have anything to do with the problem given in the OP.
that's not a trick
that's an illusion