Unfortunately, this is wrong. Two mistakes:
- You are accounting for short games incorrectly
- Ties matter
Take the simple case of a max of two games in a match. Let's say that
you win if you can score 1.0 and Bill wins if he scores 1.5. Just as
in this match, you don't play the second game if it isn't necessary.
So what are the outcomes? (W- bill wins, T - tied, L - bill loses)
WW B(ill wins)
WT B
WL S(am wins)
TW B
TT S
TL S
L S -- short game
Say that T occurs 50% of the time, W occurs 25% and L occurs 25%.
What are the odds?
WW occurs .25*.25 = 6.25% (B)
WT occurs .25*.5 = 12.5% (B)
WL occurs .25*.25 = 6.25%
TW occurs .5*.25 = 12.5% (B)
TT occurs .5*.5 = 25%
TL occurs .5*.25 = 12.5%
L occurs .25*1.0 = 25% -- short game
Total = 100%
The B-favorable events add up to: 6.25+12.5+12.5 = 31.25%
However, let's say T only occurs 10%, W then occurs 45% and L occurs
45%
What are the odds?
WW occurs .45*.45 = 20.25% (B)
WT occurs .45*.1 = 4.5% (B)
WL occurs .45*.45 = 20.25%
TW occurs .1*.45 = 4.5% (B)
TT occurs .1*.1 = 1%
TL occurs .1*.45 = 4.5%
L occurs .45*1 = 45% -- short game
Total = 100%
This time, the B-favorable events only add up to: 20.25+4.5+4.5 =
29.25%
Ties matter.
Sam Sloan wrote:
Boy is he stupid! Self-proclaimed computer expert George John, whose
only proof of his computer expertise is his own statements about
himself, who claims that he has worked for 25 years in the computer
field yet his name is not associated with any recognized computer
program or device, came up with the following duzie, which makes it
clear that George John has never taken even the most basic college
match or statistics course. Did he ever graduate from high school?
On 24 Jun 2005 14:07:33 -0700, "George John"
wrote:
I'm not a probability/statistics expert (far from it!), so what I'm
going to suggested is only a half-educated (at best -grin-) guess.
Assume Sloan's performance follows the Normal Distribution with a mean
of 1931 and standard deviation of 400. To break even against Bill
Brock he will need a performance of 2042 or better. The probability of
his doing so is roughly 0.390698 (using:
http://davidmlane.com/hyperstat/z_table.html)
So, based on this analysis (which may be totally flawed) I give Sloan
about a 39% chance.
Best regards,
George John
This is not the way to calculate the odds.
As any college freshman will tell you, first you decide the
probability of a win, a loss or a draw in an individual game.
Unfortunately, the USCF rating formula does not tell you that, because
it does not tell you the probability of a draw.
In the traditional formula, if two players are playing a match and one
is rated 200 points higher than the other, them the higherr rated
player should win by 7.5-2.5 in a ten game match or by 24-8 in a 32
game match..
If the players are seperated by 100 points, the higher player should
win by 20-12 in a 32 game match.
However, this does not tell you how many games will be draws.
If we can estimate the % of draws, then we can calculate exactly the
probability of victory.
For example, suppose that we decide that 25% of the games will be
draws. That will be 8 games in a 32 game match.
Of the remaining 24 games, Sloan should win 8 and Brock 16, so we
reach the final score of 20-12.
So,m according to this, Brock wins 50%, Sloan wins 25% and 25% are
draws.
Now, we take each possible outcome. W stands for a Sloan win, L stands
for a Brock win and D stand for a draw.
Now, we just list all the possible outcomes and the probability of
each.
WW means Sloan won the first two games and there fore the match. This
will happen 6.25% of the time or 25% times 25%.
Here are all of the possible outcomes:
WW
WDW
WDD
WDLW
WDLD
WDLL
WLW
WLDW
WLDD
WLDL
WLL
LWW
LWDW
LWDD
LWDL
LWL
LDWW
LDWD
LDWL
LDDW
LDDD
LDDL
LDL
LLWW
LLWD
LLWL
LLD
LLL
So, there are 28 possible outcomes. Of these Sloan wins 15 and Brock
wins 13.
Now, calculate the probability of each outcome. .For example, the
probability of LLWD (in which case Brock wins) is .5 x .5 x .25 x .25
Now, add them all up and you get Brock's chamces of winning the four
game match.
Sam Sloan