6 football teams each of different level
of play compete in a knockout fashion: the 1st pair chosen at random and in the following rounds the winner is facing another randomly chosen team.
Given that a certain team won three games so-far what is
the probability of winning the 4th game?
The question calls for Bayesian reasoning and that requires prior assumptions. We're told that all the teams have different levels of play, but not how different. If the differences are only slight, the requested probability gets closer to the 50% of pure chance, like that of a coin that has come up heads three times in a row. The more different the skill levels are, the more the three wins can be taken as evidence of skill among the five other teams. I don't know if "the 4th game" means that the three games this team won were the first three of the tournament, but I assume we can make this assumption.
Let's first assume, at one extreme, the skill levels are so far apart that the more skilled team will certainly win.
There are 360 possible orderings of the teams such that the same team wins all of the first three games:
order by strength:
654321 **
654312 **
654231 **
654213 **
654132 **
654123 **
653421 **
653412 **
653241 **
653214 **
653142 **
653124 **
652431 **
652413 **
652341 **
652314 **
652143 **
652134 **
651432 **
651423 **
651342 **
651324 **
651243 **
651234 **
645321 **
645312 **
645231 **
645213 **
645132 **
645123 **
643521 **
643512 **
643251 **
643215 **
643152 **
643125 **
642531 **
642513 **
642351 **
642315 **
642153 **
642135 **
641532 **
641523 **
641352 **
641325 **
641253 **
641235 **
635421 **
635412 **
635241 **
635214 **
635142 **
635124 **
634521 **
634512 **
634251 **
634215 **
634152 **
634125 **
632541 **
632514 **
632451 **
632415 **
632154 **
632145 **
631542 **
631524 **
631452 **
631425 **
631254 **
631245 **
625431 **
625413 **
625341 **
625314 **
625143 **
625134 **
624531 **
624513 **
624351 **
624315 **
624153 **
624135 **
623541 **
623514 **
623451 **
623415 **
623154 **
623145 **
621543 **
621534 **
621453 **
621435 **
621354 **
621345 **
615432 **
615423 **
615342 **
615324 **
615243 **
615234 **
614532 **
614523 **
614352 **
614325 **
614253 **
614235 **
613542 **
613524 **
613452 **
613425 **
613254 **
613245 **
612543 **
612534 **
612453 **
612435 **
612354 **
612345 **
564321 **
564312 **
564231 **
564213 **
564132 **
564123 **
563421 **
563412 **
563241 **
563214 **
563142 **
563124 **
562431 **
562413 **
562341 **
562314 **
562143 **
562134 **
561432 **
561423 **
561342 **
561324 **
561243 **
561234 **
543261
543216 **
543162
543126 **
542361
542316 **
542163
542136 **
541362
541326 **
541263
541236 **
534261
534216 **
534162
534126 **
532461
532416 **
532164
532146 **
531462
531426 **
531264
531246 **
524361
524316 **
524163
524136 **
523461
523416 **
523164
523146 **
521463
521436 **
521364
521346 **
514362
514326 **
514263
514236 **
513462
513426 **
513264
513246 **
512463
512436 **
512364
512346 **
465321 **
465312 **
465231 **
465213 **
465132 **
465123 **
463521 **
463512 **
463251 **
463215 **
463152 **
463125 **
462531 **
462513 **
462351 **
462315 **
462153 **
462135 **
461532 **
461523 **
461352 **
461325 **
461253 **
461235 **
453261
453216 **
453162
453126 **
452361
452316 **
452163
452136 **
451362
451326 **
451263
451236 **
432165
432156
431265
431256
423165
423156
421365
421356
413265
413256
412365
412356
365421 **
365412 **
365241 **
365214 **
365142 **
365124 **
364521 **
364512 **
364251 **
364215 **
364152 **
364125 **
362541 **
362514 **
362451 **
362415 **
362154 **
362145 **
361542 **
361524 **
361452 **
361425 **
361254 **
361245 **
354261
354216 **
354162
354126 **
352461
352416 **
352164
352146 **
351462
351426 **
351264
351246 **
342165
342156
341265
341256
265431 **
265413 **
265341 **
265314 **
265143 **
265134 **
264531 **
264513 **
264351 **
264315 **
264153 **
264135 **
263541 **
263514 **
263451 **
263415 **
263154 **
263145 **
261543 **
261534 **
261453 **
261435 **
261354 **
261345 **
254361
254316 **
254163
254136 **
253461
253416 **
253164
253146 **
251463
251436 **
251364
251346 **
243165
243156
241365
241356
165432 **
165423 **
165342 **
165324 **
165243 **
165234 **
164532 **
164523 **
164352 **
164325 **
164253 **
164235 **
163542 **
163524 **
163452 **
163425 **
163254 **
163245 **
162543 **
162534 **
162453 **
162435 **
162354 **
162345 **
154362
154326 **
154263
154236 **
153462
153426 **
153264
153246 **
152463
152436 **
152364
152346 **
143265
143256
142365
142356
ct =
360
ct2 =
288
>>
Out of these 360, in 288 cases the next team is also weaker than the winner of the first game (marked with **). So if strength of team is absolute, the probability of winning the fourth game is 288/360 = 4/5 = 0.8.
As the differences among the teams' abilities approaches zero, this probability approaches 1/2 = 0.5.
A simulation was done varying the probability of the stronger team winning from 90% to 60% in steps of 10%. In each case the sets of 6 games were repeated enough times to get 1 million cases where the winner of the first game also won the second and third games. A hit is when such a team won the fourth game also:
for strength=.9:-.1:.6
hitCt=0;
for trial=1:1000000
good=false;
while good==false
s=randperm(6);
good=true;
firstWinner=max(s(1),s(2));
r=rand;
if r>strength
firstWinner=min(s(1),s(2));
end
for j=3:4
r=rand;
if s(j)>firstWinner && r<strength || ...
s(j)<firstWinner && r>strength
good=false;
end
end
end
r=rand;
if s(5)<firstWinner && r<strength || ...
s(5)>firstWinner && r>strength
hitCt=hitCt+1;
end
end
fprintf('%3.1f %7d %7d %9.7f\n',strength,hitCt,trial,hitCt/trial);
end
Two runs of the program produced:
strength instances probability
of hits of base of winning 4th
advantage condition after first 3
0.9 720403 1000000 0.7204030
0.8 642063 1000000 0.6420630
0.7 571690 1000000 0.5716900
0.6 519505 1000000 0.5195050
0.9 719726 1000000 0.7197260
0.8 642168 1000000 0.6421680
0.7 571774 1000000 0.5717740
0.6 520188 1000000 0.5201880
So if the better team has a 90% probability of winning a game, the probability of winning the 4th after winnin all of 1-3 as about 72%.
With an 80% probability the better team wins a given game, this drops to 64%. Then 70% leads to 57% and if 60% is the probability for an individual game, 52% is the probability in question.
But, of course if the probability the better team will win increases as the difference in ability increases, this makes for a more complicated simulation:
hitCt=0;
for trial=1:1000000
good=false;
while good==false
s=randperm(6);
good=true;
firstWinner=max(s(1),s(2));
r=rand;
strength=.5+abs(s(1)-s(2))/11;
if r>strength
firstWinner=min(s(1),s(2));
end
for j=3:4
r=rand;
strength=.5+abs(firstWinner-s(j))/11;
if s(j)>firstWinner && r<strength || ...
s(j)<firstWinner && r>strength
good=false;
end
end
end
r=rand;
if s(5)<firstWinner && r<strength || ...
s(5)>firstWinner && r>strength
hitCt=hitCt+1;
end
end
fprintf(' %7d %7d %9.7f\n' ,hitCt,trial,hitCt/trial);
637739 1000000 0.6377390
637835 1000000 0.6378350
It seems if the probability the stronger team wind increases proportionally with the difference in their relative strengths, the probability in question is 64%.
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Posted by Charlie
on 2023-03-17 10:21:02 |
computer solution and discussion
