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        riddles >> hard >> Maximise Uncertainty
        
(Message started by: ThudnBlunder on Jan 25th, 2011, 6:40pm)
Title: Maximise Uncertainty
Post by ThudnBlunder on Jan 25th, 2011, 6:40pm
What's this, a Hard one from Thud? :o Yes, home-brewed, too!
I'm sure it can be better explained but you should get the general idea.

In 2001 some of the rules of table-tennis were changed. One change was to increase the diameter of the ball by about 10%. This had the effect of slowing the ball down and allowing more long rallies, thus making the game more attractive for spectators. It also made it easier to follow the fast-moving ball on television. Another change was to the scoring system. Instead of playing best of 5 sets (3 for ladies) up to 21 points, they now play best of 7 or 9 sets, each up to 11 points. I might be wrong, but it seems to me that the second method tends to make the results more unpredictable and therefore the matches more interesting. In the following, the advantage of serving has not been considered.

Let each set be best of 2g+1 points, ie. first to win g+1 points, with deuce at g points each.
Let match be best of 2s+1 sets, ie. first to win s+1 sets, with no deuce at s sets each.
Let 0 < p < 1/2 = probability of weaker player winning any given  point.

Note that in terms of uncertainty, playing for example (g+1)(s+1) sets of 1 point each is equivalent to playing 1 set of (g+1)(s+1) points. And, by a discrete equivalent of Rolle's theorem, between those two extremes there must exist either a maximum or minimum amount of uncertainty.

Let F(p,g+1) = Probability of weaker player winning a set up to g+1 points, with deuce at g points each.
Let G{r,k+1} = Probability of weaker player winning k+1 sets and therefore the match,
where r = F(p,g+1) = probability of weaker player winning a set, but with no deuce at k sets each.

Then (I think)
F(p,g+1) = pg+1[1 + g+1C1q + g+2C2q2 + g+3C3q3 + ......... + {2g-1Cg-1qg-1/(p2 + q2)}]
where q = 1-p
and
G{r,k+1} =  rk+1[1 + k+1C1q + k+2C2q2 + k+3C3q3 + ......... + 2kCkqk]
where r = F(p,g+1) and here q = 1-r

Hence we seek some value of m+1 between 1 and (g+1)(s+1) which divides (g+1)(s+1) to give an odd integer
and which maximises or minimises the function G{F(p,m+1),(g+1)(s+1)/(m+1)} for given constant p and values of
g and s that determine a suitable expected match length.

Now, where is an anally retentive programming nerd when you need one?
Title: Re: Maximise Uncertainty
Post by ThudnBlunder on Jan 29th, 2011, 9:56am
For example, considering the simple case g = s = 2, we have
(g+1)(s+1) = 9
and
F(p,g+1) = F(p,3)
              = p3[1 + 3q/(p2 + q2)]
              = p3(2p2 - 5p + 4)/(2p2 - 2p + 1)
and
G{r,s+1} = G{r,3}
              = r3(1 + 3q + 6q2)
              = r3(6r2 - 15r + 10), where r = F(p,3) as calculated above.

Let p = 1/3, say.
Then F(1/3,3) = 23/135 = 0.17037......
And G{r,3} = G{23/135, 3) = 0.037675.....  

Considering the extreme scenarios, if we have no deuces then
G{r,1} = F(p,9) = (1/3)9[1 + 9C1(2/3) + 10C2(2/3)2 + 11C3(2/3)3 + 12C4(2/3)4 + 13C5(2/3)5 + 14C6(2/3)6 + 15C7(2/3)7 + 16C8(2/3)8
                     = (1/19683)[1 + 6 + 20 + 440/9 + 880/9 + 4576/27 + 64064/243 + 91530/243 + 366080/729]
                     = 1485.58/19683
                     = 0.0754...

Hmm.......so unless I have miscalculated it appears the extreme scenarios offer maximum uncertainty.  :-/


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