Author: fca
Date: 16:06:59 08/18/98
Go up one level in this thread
On August 18, 1998 at 08:40:19, Bruce Moreland wrote:
>On August 18, 1998 at 03:34:05, fca wrote:
>>Say the mathematical point expectation for Ferret per game against this
>>opponent is 0.51.
>>So, how much advantage does the punter have from the ability to "walk away" as
>>soon as he is "ahead" (whatever "ahead" means - let us forget grading as then
>>the question is too complex for me)? Is this an advantage at all? Please
>>ignore all psychological considerations, this is a "straight" ( ;-) ) >>math-chess question.
>If I really was scoring 51% against the pool in individual games, but the
>members of the pool want to win matches, not games, and I will accept any
>request for a game, and never make my own requests, then I will lose most of the
>matches. Assuming no draws (bad assumption, but this just changes the numbers,
>but not what they mean)
No need for this assumption, as draws could be ignored in analysing the sequence
since they *cannot* affect the outcome of a series match whose duration is
solely determined by one, single-minded ("I must win") combatant. Of course
then we need to state 51% is the expected game-score _ignoring draws_. I'll
interpret*everything* that follows with this simplification (please also do)...
I guess this is what you meant also.
>, and that everyone will stick around for exactly three games
>, or until they are ahead
if earlier
>, or until they can't win
also if earlier than 3
>, I get the following
>breakdown, W being a win for them and L being a loss for them:
>49% W (win)
>26% LL (loss)
>13% LWL (loss)
>12% LWW (win)
>So in this case I lose about 61%
yup, 61.2451% exactly.
>of the matches even though I win 51% of the
>games. It can only get worse if they are willing to play more games.
Absolutely correct.
(a) Does the 61%-type number tend to 100% as progressively longer (finally
indefinite length) sequences are permitted? If not, to what number
(clearly a function of the 51%-type number) does it asymptote?
>So I think that if it is your goal to win a match, it is a major advantage to be
>the one who decides when the match ends.
Also of course correct. But now here is the build-up question... it is
"random-walk" stuff as you probably knew.
(b) What is the *minimum* game-winning % ignoring draws (i.e. the number that
earlier was deemed to be 51%) (hereafter abbreviated to GWPID) that gave Ferret
at least a 50% expected match result against such a strategy in such a match of
*indefinite* duration?
If the answer to (a) was "yes", some might suggest the answer here will need to
be GWPID=100%.
A *very* relevant question IMO, as the GWPID, together with data on draw
frequency for that opponent (I postpone the 'pool' concept, please, which Bruce
introduced - let me understand the single opponent problem first), translates
into a direct ELO difference through a well-known integral (or look-up table).
To show goodwill, I compute the answer x to (b) in the 3-games-at-most case ;-)
(1-x) + x*(1-x)^2 <= 0.5 ( where 0 <= x <= 1 )
which on solving the cubic and disregarding the two false roots gives x=
59.6968283237+%
So, if Ferret's GWPID is >= 59.6968283237%, Bruce is "OK" to promise the
opponent 3 games (ignoring draws, here as ever) with the opponent having the
option to truncate earlier if he chooses. By "OK" is meant, Bruce has at least
50% chance of winning the match. For skeptics >>
40.3031716763% W
09.6968283237% LWW
--------------
50.0000000000%
All I ask here is the extension of this to the general case i.e.
W + LWW + LWLWW + LLWWW + LWLWLWW + LWLLWWW + LLWLWWW + LLWWLWW + LLLWWWW
1 3 ----5---- -------------------7-------------------
etc.
Note it is the number of outcomes for each match-length that is the key, as each
will always have the same basic composition. i.e. If 2k-1 is a match length
(they are laways odd of course) then there will be k "W"s in it and k-1 "L"s in
it. So if the GWPID=x, each constituent of the 2k-1 long match will have a
probability of x^(k-1) * (1-x)^k. So, how many different flavours of (2k-1)
length match are there?
Length of match Number of ways match could be that long
1 1
3 1
5 2
7 5
' '
2k-1 ???
Get me "???" as f(k), (the "never ahead until" problem) and I'll sum "our"
series... I tried to see a quickie fit, but I am tired, so it might actually be
easy (I suspect it is not).
>Which is one reason I don't want to
>play a non-specific number of games against Morovic, I might add.
Good reason. I wondered whether his name would be mentioned.. ;-)
btw This sub-thread is not reserved for Bruce or myself AFAIK. Dann, Dan,
someone else?
Kind regards
fca
PS: This is unrelated mathematically, but it "sounds" analogous. Draws are
ignored for *all* purposes (i.e. as if they never occurred). Say I score <50%
against Ferret (number is not critical here - say 49%), but have the right to an
match of upto N games which only I can truncate at any earlier point. We bet
(always evens) 1 chekel on the first game, 2 on the 2nd, 4 on the 3rd, 8 on the
4th and so on (the "doubler's strategy"). Assume I have an endless supply of
chekels. Since I have the right to end at any time, I clearly have an advantage
because as soon as I win I can quit and I will be ahead (as 4 > (1+2), 8 >
(1+2+4) etc.). So, how much should I pay you for the privilege of such a match
(veniality assumed on both our parts)?
My expectation in chekels if I stop as soon as I win is less than many might
imagine:
0.49*1 + 0.51*0.49*(-1+2) + 0.51*0.51*0.49*(-1-2+4) + .... to N terms
= 0.49 * (1 - 0.51^N) / (1 - 0.51) = 1 - 0.51^N
So, if we set N=3, and if I paid you more than 0.86735 chekels for the privilege
of such a match, I'd have been diddled. If we set N=infinity this rises.... to
just 1 chekel.
Of course I have better strategies... like waiting till I am J > 1 game ahead,
for example... now the fun maths starts... And the value of GWPID starts
playing a big role. Please feel free to take a punt here too...
Till next time.
This page took 0 seconds to execute
Last modified: Thu, 15 Apr 21 08:11:13 -0700
Current Computer Chess Club Forums at Talkchess. This site by Sean Mintz.