On 12 Oct 2002, Tanguy wrote:
> As you may be aware the Louis Vitton Cup is taking place in Hauraki
> Bay, New-Zealand. Nine challengers take part. The winner of the LV Cup
> will face the New Zealand holder of the America's Cup at the beginning
> of 2003. Each team races each other once. One point per win. Thus the
> preliminary stage totalize 72 races (9*8/2), hence 36 points to be
> shared by 9 boats.
You have not mentioned the possibility of a tie rather than a win.
Is a tie, in fact, possible? If so, the possible ties should be included
in your list of scenarios; and in order to assist you we would need to
know how a tie would be scored (1/2 point to each?).
> I would like to determine how many scenarii are likely to happen.
Presumably you actually meant "how many are possible" ? (I cannot see how
to determine likelihood without knowing, e.g., the relative probabilities
of each challenger's winning, and I cannot see how one could know that
sort of information in advance of the races themselves.)
> I have tried to answer this problem by considering 3 boats only (3
> races: A vs B, A vs C, B vs C), then 4 boats (6 races: A vs B, C and
> D, B vs C,D and finally C vs D) and so on for 5 and 6 boats (cf.
> below). Unfortunately I have not understood the logic behind my first
> findings.
>
> I will appreciate if you see a "probabilistic" way to solve the
> problem or a solution that does not involve the enumeration of all the
> solutions.
"Combinatorial", I think, rather than "probabilistic". <snip>
> Considering 3 boats (A,B,C) yields 2 scenarii to share 3 points at
> stake:
> boat identity. A B C
> points........ 2 1 0
> .............. 1 1 1
Seems to me these aren't all the possible outcomes. I see:
Boat Points for various outcomes
A 2 2 1 1 1 0 0
B 1 0 1 2 0 2 1
C 0 1 1 0 2 1 2
but I am not certain that these seven sets of points are equally likely
(if one entertains the hypothesis that all three boats are equally
speedy). Here is another way of considering scenarios, based on the
possible outcomes of each of three races: A vs B, A vs C, B vs C.
I'll use "A>B" to mean "A won against B" and "A<B" to mean "A lost against
B". (If ties are possible, one can add scenarios that have "=" in them
instead of ">" or "<".) By "score" I'll mean the points for each boat,
in alphabetical order.
Race Outcomes
A vs B A>B A>B A>B A>B A<B A<B A<B A<B
A vs C A>C A>C A<C A<C A>C A>C A<C A<C
B vs C B>C B<C B>C B<C B>C B<C B>C B<C
score 210 201 111 102 120 111 021 012
and on this enumeration we see that there are two scenarios that would
yield one point for each boat, and only one scenario for each of the
other six patterns of points. And there are eight outcomes all together,
which is correct for three trials each with two possible outcomes:
2^3 = 2 to the power of 3 = 8.
> Considering 4 boats yields 4 scenarii to share 6 points at stake:
>
> A B C D
> 3 2 1 0
> 3 1 1 1
> 2 2 2 0
> 2 2 1 1
As for three boats, the list of outcomes is incomplete. As I perceive
"scenarios", there should be 64 of them (= 2^6) for 4 boats, because there
must be 6 trials -- A:B, A:C, A:D, B:C, B:D, C:D -- and each trial has two
possible outcomes (in the absence of ties).
If by "scenario" you do not mean a particular pattern of outcomes of the
trials, but a particular type of distribution of points, so that "210"
means merely that "of three boats, one of them one twice, another one won
once, and the third lost all three races", without specifying which boat
was which, then my enumeration above is not what you want ultimately, but
rather the relative frequency of each type of distribution of points.
Then, for 3 boats:
Pattern 210 111
Frequency 6 2
Thus if each scenario is equally likely the odds for one boat winning
twice, versus all three winning once, are 3:1. But if you want to compute
gamblers' odds, you need to add to each of the 8 scenarios information
about the odds given for each race, and the problem is rather more
complicated.
For 4 boats:
3210 3201 3120 3102 3021 3012 = 6 for A winning all three of its races;
thus 24 for A,B,C, or D winning all three.
3111 = 1 x 2 for A winning all three and the other three winning one
each; see above for "x2" with three boats = 111;
thus 4 x 2 = 8 for all 4 boats.
2220: For "222" we have only 2 patterns, with D losing all 3 races:
A:B > <
A:C < > Each of 4 boats could be the loser,
A:D > > thus 4 x 2 = 8.
B:C > < So far, 24 + 8 + 8 = 40.
B:D > >
C:D > >
2211: This is the only other pattern, and the total must be 64.
By subtraction, there are 64-40 = 24 such patterns.
An exhaustive analysis, as above for 3 boats:
Race Outcome of each race; points for each boat; note each pattern.
A:B >>>>>>>> >>>>>>>> >>>>>>>> >>>>>>>> <<<<<<<< <<<<<<<< <<<<<<<< <<<<<<<<
A:C >>>>>>>> >>>>>>>> <<<<<<<< <<<<<<<< >>>>>>>> >>>>>>>> <<<<<<<< <<<<<<<<
A:D >>>>>>>> <<<<<<<< >>>>>>>> <<<<<<<< >>>>>>>> <<<<<<<< >>>>>>>> <<<<<<<<
B:C >>>><<<< >>>><<<< >>>><<<< >>>><<<< >>>><<<< >>>><<<< >>>><<<< >>>><<<<
B:D >><<>><< >><<>><< >><<>><< >><<>><< >><<>><< >><<>><< >><<>><< >><<>><<
C:D ><><><>< ><><><>< ><><><>< ><><><>< ><><><>< ><><><>< ><><><>< ><><><><
A 33333333 22222222 22222222 11111111 22222222 11111111 11111111 00000000
B 22111100 22111100 22111100 22111100 33222211 33222211 33222211 33222211
C 10102121 10102121 21213232 21213232 10102121 10102121 21213232 21213232
D 01120112 12231223 01120112 12231223 01120112 12231223 01120112 12231223
3210 ++ ++ ++ + + + + ++ ++ + + + + ++ ++ ++
3111 + + ++ + + + +
2220 + + + + ++ + +
2211 + + ++ +++ + +++ + + +++ + +++ ++ + +
Thus for 3210 there are 24 scenarios; for 3111, 8; for 2220, 8; and
for 2211, 24.
I do not have time to analyze the problem further, but this will perhaps
assist you in thinking about it. Bonne chance!
> Considering 5 boats yields 7 scenarii to share 10 points at stake:
And, as above, there are 2^10 = 1024 different outcomes possible.
The seven patterns occur with different frequencies.
> A B C D E frequency (incomplete...)
> 4 3 2 1 0 5 x 24 = 120
> 4 3 1 1 1 5 x 8 = 40
> 4 2 2 2 0 5 x 8 = 40
> 4 2 2 1 1 5 x 24 = 120
> 3 3 2 1 1
> 3 2 2 2 1
> 2 2 2 2 2
>
> Considering 6 boats yields 18 scenarii to share 15 points at stake:
For 6(5)/2 = 15 races there are 2^15 = 32768 different outcomes.
I can see why you'd prefer to deal only with the different patterns,
rather than the different individual outcomes; 2^72 is fairly
astronomical (4.7 x 10^21).
<snip, the 18 patterns ... >
-- DFB.
-----------------------------------------------------------------------
Donald F. Burrill [EMAIL PROTECTED]
56 Sebbins Pond Drive, Bedford, NH 03110 (603) 626-0816
[Old address: 184 Nashua Road, Bedford, NH 03110 (603) 471-7128]
.
.
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