Tournament Question
Patrick D. Rockwell
tournament. Each one
is ranked according to his/her skill. Player 1 can beat all others.
Player 2 can beat all
except player 1. Player 3 can beat anyone except players 1 and 2, etc.
In the first round, the players form 9 groups of 3 players each. When
it's over, the 9 winners
of each group form 3 groups of 3 players and compete again.
What is the probability that player 3 will make it out of the first
round as one of the 9
winners. What is the probability that he will make it out of the
second round as one of
the 3 finalists?
Thanks in advance.
Richard Heathfield
Patrick D. Rockwell said:
The question can't be answered without making some assumptions. I'll assume
that all the groups are chosen at random at every stage.
The probability that Player 3 will make it out of the first round is equal
to the number of ways in which he can survive, divided by the number of
outcomes. For this purpose we can actually ignore the other eight groups,
since their outcomes have no bearing on his survival or otherwise.
Whatever group he's in, there are only two people who can beat him, and 24
who can't. There are two slots to fill in his group (other than his own).
He survives if the first slot is filled by any of the 24 people, and the
second by any other of the 24: 24 * 23. There are 26 * 25 possible ways to
fill the slots, so his survival probability is (24 * 23) / (26 * 25) which
is 552/650 which is 0.84923+. It's surprisingly low, isn't it? I mean,
okay, it's third-highest - but it's still lower than I thought it would
be.
Conditional probabilities can get murky (can't they, Monty?), so the
following may be flawed. To make my working more obvious to the
probability-minded, I haven't bothered cancelling common factors.
Let us look at ways in which #2 and #3 can both survive Round 1. We have
already filled the first group (in one of 24 * 23 ways). We place #2
firmly in the second group, so we've now filled four places, leaving 23
players, but #2 can't face #1, so #3 *and* #2 survive with probability
552/650 * (22*21)/(23*22) = 552/650 * 462/506 = 255024/328900 = 0.77538+.
We deduce that player #3 survives without #2 with probability 552/650 *
44/506 = 24288/328900 = 0.07384+.
Assuming #2 did NOT survive, player #3 *will* survive to the final unless
he meets #1 in the semis. He has seven suitable opponents out of eight,
and we need to draw two of them, so that's (7*6)/(8*7) = 42/56. We need to
multiply this by the probability that #3 survived but #2 did not, which
gives us 42/56 * 44/506 = 1848/28336 = 0.06521+
Assuming that #2 DID survive, player #3 must avoid not only #1 but also #2.
So he has only six suitable opponents out of eight. We need two, so that's
(6 * 5) / (8 * 7) = 30/56. We need to multiply this by the probability
that both #2 and #3 survived, which gives us 30/56 * 255024/328900 =
7650720/18418400 = 0.41538+.
#3 survives to the final if either of these mutually exclusive events
occurs, so we add:
1848/28336 + 7650720/18418400 = (34037203200 + 216790801920) / 521903782400
which is 250828005120 / 521903782400 or 1437 / 2990 or 0.480602+.
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
ken
1) For P3 to win in round 1,
P1 and P2 must be in other groups.
p(P3 avoids P1 and P2)
p(P3 has 2 weaker in group of 3)
p=(24/26) x (23/25)
p= 0.849230769231
2) For P3 to win in round 2,
P3 must avoid P1 and P2,
or simply avoid P1 if P2 eliminated.
p = p(P2 into round 2) * p(P3 avoids P1 and P2)
+ p(P2 eliminated) * p(P3 avoids P1)
-p(P2 into round 2) calculated as
-p(P2 avoids P1) in the 8 non-P3 groups,
-same for p(P2 eliminated)
p= ((22/23) x (21/22)) * ((6/8) x (5/7))
+((1/23) + (1/22)) * ((7/8) x (6/7))
p= 0.555830039526
I see my answer (part 2) differs from Richard Heathfield so I'll try
and find his obvious mistake :-)
ken
But first my own obvious mistake :-(
1) For P3 to win in round 1,
P1 and P2 must be in other groups.
p(P3 avoids P1 and P2)
p(P3 has 2 weaker in group of 3)
p=(24/26) x (23/25)
p= 0.849230769231
2) For P3 to win in round 2,
P3 must avoid P1 and P2,
or simply avoid P1 if P2 eliminated.
p = p(P2 into round 2) * p(P3 avoids P1 and P2)
+ p(P2 eliminated) * p(P3 avoids P1)
-p(P2 into round 2) calculated as
-p(P2 avoids P1) in the 8 non-P3 groups,
-same for p(P2 eliminated)
XXX p= ((22/23) x (21/22)) * ((6/8) x (5/7))
XXX +((1/23) + (1/22)) * ((7/8) x (6/7))
XXX p= 0.555830039526
should be
p= ((22/23) x (21/22)) * ((6/8) x (5/7))
+(1-(22/23)x(21/22)) * ((7/8) x (6/7))
p= 0.554347826087
Still different.
BTW What sort of price could I get on P1 winning this tournament?
Mike Terry
news:cc392876-3237-43a8...@y18g2000pre.googlegroups.com...
Yes this looks good. (I haven't checked the calculation)
>
>
> 2) For P3 to win in round 2,
> P3 must avoid P1 and P2,
> or simply avoid P1 if P2 eliminated.
>
Or more simply - for P3 to survive round 2,
P1 and P2 must not be in his "group of 9".
p(P3 avoids P1 and P2)
p(all 8 opponents in his group of 9 are weaker)
p=(24/26)(23/25)(22/24)(21/23)(20/22)(19/21)(18/20)(17/19)
i.e. this is really just like the first question but with a larger group...
Regards,
Mike.
ken
<news.dead.person.sto...@darjeeling.plus.com> wrote:
>
> Or more simply - for P3 to survive round 2,
> P1 and P2 must not be in his "group of 9".
>
> p(P3 avoids P1 and P2)
> p(all 8 opponents in his group of 9 are weaker)
>
> p=(24/26)(23/25)(22/24)(21/23)(20/22)(19/21)(18/20)(17/19)
>
> i.e. this is really just like the first question but with a larger group...
>
> Regards,
> Mike.
Lovely.
Nice and simple.
p(24/26)...(17/19)= 0.470769230769
which is different from my obviously wrong answer.