Sandbox: Difference between revisions

From ChanceWiki
Jump to navigation Jump to search
No edit summary
No edit summary
Line 1: Line 1:
<center>Mean1 = <math>\frac{(4+6+8+4+5+3)}{6}=\frac{30}{6}= 5.0 </math><br><br>
To compare the amount of skill and luck in these games Sherman would have us
carry out an analysis of variance in the same way we did for our medical
example. Here the players are the groups and the games are the treatments.  The
group means are the averages of the players winnings and are in the last row of
the data. 


Mean2 = <math>\frac{(7+5+8+9+7+9)}{6}= \frac{45}{6}=7.5 </math><br></center>
The grand mean (gm) is the sum of all the winnings divided by 35
which gives us a grand mean of -.105714. The sums of squares within  groups
is the  sum of  the squares of the differences between the winnings and the
player's mean winnings:


Then a grand mean over all observations:<br>
<center><math>(-6.75 -1.934)^2 + ... + (-1.42-2.489)^2 = 758.499,.</math></center>
<center>Mean = <math>\frac{(30+45)}{(6+6)} = 6.25</math><br></center>


Variance is always a sum of square deviations divided by degree of freedom:
The sums of squares between groups is the sum of the squares of the differences
SS/df.  This is also called a mean squared deviation MS.
between the winnings and the grand mean:


ANOVA begins by expressing the deviation of each observation from the grand mean as a sum of two terms:  the difference of the observation from its group mean, plus the difference of the group mean from the grand mean.  Writing this out explicitly for the example, we have, for the placebo group:<br><br>
<math>(-6.75 - gm)^2  + ...+ (-1.42-gm)^2 = 311. 447.</math>
<center>(4 - 6.25) = (4 - 5.0) + (5.0 - 6.25)<br>
(6 - 6.25) = (6 - 5.0) + (5.0 - 6.25)<br>
...<br>
(3 - 6.25) = (3 - 5.0) + (5.0 - 6.25)</center><br>


and for the vitamin ME group:
Thus the total sums of squares is
   
<center>(7 - 6.25) = (7 - 7.5) + (7.5 - 6.25)<br>


(5 - 6.25) = (5 - 7.5) + (7.5 - 6.25)<br>
758.499 + 311.447 = 1069.95.
...<br>
(9 - 6.25) = (9 - 7.5) + (7.5 - 6.25)<br></center>


The magic (actually the Pythagorean Theorem in an appropriate dimensional space)
Sherman assumes that the variation  in the amount won within groups  is
is that the sums of squares decompose in this way.<br>
primarily due to luck and  calls this ``Random Variance'' and the variation
between groups is due primarily  to skill and calls this ``Systematic Variance".  
He then defines


<math>(4-6.25)^2 +...+(9-6.25)^2 =</math>
$${\rm Game's\ Skill\ Percentage} = \frac{\rm Systematic\
[(4-5.0)^2+...+(9 - 7.5)^2] + [(5.0 - 6.25)^2+...+(7.5 - 6.25)^2] <br>
Variance}{{\rm(Systematic\  Variance} + {\rm Random\  Variance})}\,,$$
Check: 46.25 = 27.5 + 18.75<br>
and similarly,
$${\rm Game's\ Luck\ Percentage} = \frac{\rm Random\ Variance}{{\rm(Systematic\
Variance} + {\rm Random\  Variance})}\,.$$
So, in our poker game, the Random Variance is 758.499 and the Systematic
variance is 311.477. So the Skill Percentage  is 29.1\% and  the Luck Percentage
is 70.9\%.
\par
In his second article, Sherman reports  the Skill Percentage  he obtained using
data from a number of different types of games. For example, using data for
Major League Batting, the Skill Percentage for hits was 39\% and for home runs
was 68 \%. For NBA  Basketball it was 75\% for points scored. For poker stars in
weekly tournaments it was 35\%.
\par
Sherman concludes his articles with the remarks:


In the usual abbreviations:<br>
\begin{quotation} If two persons play the same game, why don't both achieve the
 
same results? The purpose of last month's article and this article was to
<center>SST = SSE + SSG</center><br>
address this question.  This article suggests that there are two answers to this
question: Skill (or systematic variance) or Luck (or random variance).  Using
both the correlation approach described last month and the ANOVA approach
described in this article, one can estimate the amount of skill involved in any
game.  Last, and maybe most importantly, Table 4 demonstrated that the skill
estimates involved in playing poker  (or at least tournament poker) are not very
different from other sport outcomes which are widely accepted as
skillful.\end{quotation}
\vskip .2in
\noindent
Discussion questions:
\vskip .1in
\noindent
(1) Do you think that Sherman's measure of skill and luck in a game is
reasonable?  If not, why not?
\vskip .1in
\noindent
(2) There is a form of poker modeled after duplicate bridge.  Do you think that
the congressional decision should apply to this form of gambling?

Revision as of 21:37, 17 October 2007

To compare the amount of skill and luck in these games Sherman would have us carry out an analysis of variance in the same way we did for our medical example. Here the players are the groups and the games are the treatments. The group means are the averages of the players winnings and are in the last row of the data.

The grand mean (gm) is the sum of all the winnings divided by 35 which gives us a grand mean of -.105714. The sums of squares within groups is the sum of the squares of the differences between the winnings and the player's mean winnings:

<math>(-6.75 -1.934)^2 + ... + (-1.42-2.489)^2 = 758.499,.</math>

The sums of squares between groups is the sum of the squares of the differences between the winnings and the grand mean:

<math>(-6.75 - gm)^2 + ...+ (-1.42-gm)^2 = 311. 447.</math>

Thus the total sums of squares is

758.499 + 311.447 = 1069.95.

Sherman assumes that the variation in the amount won within groups is primarily due to luck and calls this ``Random Variance and the variation between groups is due primarily to skill and calls this ``Systematic Variance". He then defines

$${\rm Game's\ Skill\ Percentage} = \frac{\rm Systematic\ Variance}{{\rm(Systematic\ Variance} + {\rm Random\ Variance})}\,,$$ and similarly, $${\rm Game's\ Luck\ Percentage} = \frac{\rm Random\ Variance}{{\rm(Systematic\ Variance} + {\rm Random\ Variance})}\,.$$ So, in our poker game, the Random Variance is 758.499 and the Systematic variance is 311.477. So the Skill Percentage is 29.1\% and the Luck Percentage is 70.9\%. \par In his second article, Sherman reports the Skill Percentage he obtained using data from a number of different types of games. For example, using data for Major League Batting, the Skill Percentage for hits was 39\% and for home runs was 68 \%. For NBA Basketball it was 75\% for points scored. For poker stars in weekly tournaments it was 35\%. \par Sherman concludes his articles with the remarks:

\begin{quotation} If two persons play the same game, why don't both achieve the same results? The purpose of last month's article and this article was to address this question. This article suggests that there are two answers to this question: Skill (or systematic variance) or Luck (or random variance). Using both the correlation approach described last month and the ANOVA approach described in this article, one can estimate the amount of skill involved in any game. Last, and maybe most importantly, Table 4 demonstrated that the skill estimates involved in playing poker (or at least tournament poker) are not very different from other sport outcomes which are widely accepted as skillful.\end{quotation} \vskip .2in \noindent Discussion questions: \vskip .1in \noindent (1) Do you think that Sherman's measure of skill and luck in a game is reasonable? If not, why not? \vskip .1in \noindent (2) There is a form of poker modeled after duplicate bridge. Do you think that the congressional decision should apply to this form of gambling?