algabraic probability,,,no numbers

oxrigby · #1 $Old$ November 16th, 2008, 06:18 PM

Hi, I'm a little confused by the concept of this Question. It is as follows:
you are told that a particular team has won w games, drawn d games and lost l games during the course of a season. Calculate the probability that (a) the team won both the first and the last games of the season, (b) the team lost both the first and the last games of the season.
for (a) I got the answer $\frac{w^2}{(w+d+l)^2}$ as this would be the multiplication of two independent events P(win)[LaTeX Error: Syntax error]
where $P(win)=\frac{w}{w+d+l}$
$P(draw)=\frac{d}{w+d+l}$
$P(lose)=\frac{l}{w+d+l}$
First of all to notice at this point is the wording of the question,'calculate the probability', clearly I didn't get a numerical answer but an algebraic one. Assuming my answer is correct then this point is merely pettiness on my part however if my answer is incorrect, presuming there is a numerical solution then I would much appreciate any further insight anybody has!!!,,,
Secondly I have further constructive criticism of my answer. On the basis that since the first game not only has the number of total games (w+d+l)reduced by one altering in theory the denominator in my equation to w+d+l-1 but also the numerator(which I am not sure about) to 1,,,,,,as all the other w's would have been 'cancelled out' in the terms in the middle of the equation. In other words Im not sure as to whether the answer is not just the p(win)p(win) or p(win)p(draw)p(lose),,,,,,in some kind of order,,p(win) in which case considering the middle terms would involve some kind of algebraic permutation or combination as well as a confusing denominator and numerator for the w terms. The draw terms and lose terms contained in the middle though I am pretty sure the numerator for both draw and lose could be simplified into binomials of some sort.
As you can see I remain very confused as to how to solve this basic question and some insight and clarification of my thinking would be much appreciated. I just hope my first answer was correct. As for now I will continue to try obtain a solution to (b) thanks very much

mr fantastic · #2 $Old$ November 16th, 2008, 11:24 PM

Quote:

Originally Posted by oxrigby $View Post$

Hi, I'm a little confused by the concept of this Question. It is as follows:
you are told that a particular team has won w games, drawn d games and lost l games during the course of a season. Calculate the probability that (a) the team won both the first and the last games of the season, (b) the team lost both the first and the last games of the season.
for (a) I got the answer $\frac{w^2}{(w+d+l)^2}$ as this would be the multiplication of two independent events P(win)[LaTeX Error: Syntax error]
where $P(win)=\frac{w}{w+d+l}$
$P(draw)=\frac{d}{w+d+l}$
$P(lose)=\frac{l}{w+d+l}$
First of all to notice at this point is the wording of the question,'calculate the probability', clearly I didn't get a numerical answer but an algebraic one. Assuming my answer is correct then this point is merely pettiness on my part however if my answer is incorrect, presuming there is a numerical solution then I would much appreciate any further insight anybody has!!!,,,
Secondly I have further constructive criticism of my answer. On the basis that since the first game not only has the number of total games (w+d+l)reduced by one altering in theory the denominator in my equation to w+d+l-1 but also the numerator(which I am not sure about) to 1,,,,,,as all the other w's would have been 'cancelled out' in the terms in the middle of the equation. In other words Im not sure as to whether the answer is not just the p(win)p(win) or p(win)p(draw)p(lose),,,,,,in some kind of order,,p(win) in which case considering the middle terms would involve some kind of algebraic permutation or combination as well as a confusing denominator and numerator for the w terms. The draw terms and lose terms contained in the middle though I am pretty sure the numerator for both draw and lose could be simplified into binomials of some sort.
As you can see I remain very confused as to how to solve this basic question and some insight and clarification of my thinking would be much appreciated. I just hope my first answer was correct. As for now I will continue to try obtain a solution to (b) thanks very much

(a) I think there are two things you need to consider:

How many ways can you arrange w Wins, d Draws and l Losses without restriction.

How many ways can you arrange w Wins, d Draws and l Losses such that there is a Win at the start and a Win at the end.

Then divide the two expressions.

(b) Similar.

oxrigby · #3 $Old$ November 18th, 2008, 12:05 AM

Oh I see thanks! hears is what I did-for the restrictions I did $\frac{(w-2+d+L)!}{d!(w-2)!L!}$ and for the first condition without restriction just (w+d+L)! Then the probability will ultimately be $\frac{(w-2+d+L)!}{(w-2)!(w+d+L)!d!L!}$ Could someone confirm if this is correct thanks. Ill do (b)

oxrigby · #4 $Old$ November 18th, 2008, 05:23 PM

is this correct?

oxrigby · #5 $Old$ November 24th, 2008, 08:30 PM

apologies for the lateness of this reply but Im still unsure if my anser is correct! I figured the number of combinations with the restriction at either end would be $\frac{w-2+d+L)!}{(w-2)!d!L!}$ the denominator came from the fact that the wins looses draws are indistinguishable. Then I came up with the unrestricted case being $\frac{(w+d+L)!}{w!+d!+L!}$ again the denominator being because the winds etc are indistinguishable. So to obtain the probability of having a win at the end and win at the start i divided the restricted on by the unrestricted one giving me $\frac{(w-2+d+L)!w!}{(w-2)!(W+d+L)!}$ NB i cancelled top and bottom the d!L! My only worry is that for some reason this didnt properly take into consideration the w at the start and the end(ie the order, as i didnt properly consider any specific permutaion)....I can't honestly get my head around it though! I would really appreciate any advice on the answer for this question thanks!..