Beta wrote:well, we now have proof that all four answers can be backed up with an explanation (even if it is highly improbable). (PyTom is in fact partly responsible for answer A being shown possible.)
for Answer A to be even remotely correct, we'd have to assume there's an error with the question's wording. if the question was, in reality, to find the probability of a team winning THREE games in a set of three games, then this is the only answer possible, since the probability of one team winning ALL THREE GAMES is 1 in four, as PyTom (and a few others) has inadvertantly shown.
PyTom (others have shown this as well...) wrote:...Let's consider a sport which has a best-of-three championship series, but which, for whatever reason, demands that they play all three games, even if the outcome is preordained. We'd get the following possible outcomes:
HHH *
HHT *
HTH
HTT
THH
THT
TTH *
TTT *
The lines I've starred are the ones in which the games are won in two flips. All of the lines have the same probability, so the chance of someone winning in 2 games is 4/8 == 1/2.
note how 2 of the eight outcomes (which I have marked in green) is the same team winning all three times. (Gamma brought this to my attention.)
answer B, as I have shown and argued for a while about, assumes that the series ends after one team wins two games (contradictory to the explanation of answer A above). this results in only two of six possible ways that it could end after the second game with the third game not played at all.
for answer C, follow a similar logic but discard the probability calculation on the third game altogether. (this is how Zeta and most of his math class arrived at that solution...)
answer D has the weirdest (but still followable) logic behind it, so I'll just quote the person who explained it...
herenvardo wrote:forget the coin flip and go back to the sport event. Assuming most neutral conditions, as playing in a neutral field and weather not affecting results, then it's most probable that the best team wins each game. So, after team A wins the first game, there is an important chance that A is better than B, so its chances to win the second game are higher than 50/50. With such an argumentation, the answer should be somewhat higher than 1/2, and the only option that fulfills that is 2/3. I have had many teachers that, provided an argumentation like the one above, would accept as good that option (but it will be taken as wrong if you mark it without any explanation).
this leaves me to wonder if the people who wrote that question still have their jobs.
Well, you're right: all four answers can be argued for; but that doesn't make them right. In a pure math context, the only one that would be right is 1/2, and there have been a lot of different explanations, most of them unbreakable, about that one. About your arguing for B, I've found where does it fall, and I'll try to clearly explain it. You make a case-breakdown, like that:
HH
HTH
HTT
THH
THT
TT
So, you had 6 cases and assumed that each one would have a 1/6 chance to happen. The error is that the cases you list have not the same chance to occur, since the ones of 3 matches have half the chances of occuring than the ones of 2 matches.
Anyhow, if you're still unhappy with that, let me know and I'll give you a math demonstration of 1/4 = 1/3 = 1/2 = 2/3 (of course, it'll be cheaty, but it may be fun).
Now, no joking, I insist you on what I suggested on my previous post: once you have the first 2 results, they have to be enough to decide if a 3rd match is required. So, make the case breakdown for after the second match. You should see it clear then.
beta wrote:...and it seems like nobody remembered that there would be a third game or couldn't figure out how it would affect the answer.
Ok, you're right on the fact that there can be a third game, but due to the nature of the problem, we got the answer (there is a 3rd match or not)
before the 3rd match, if it's finally done. This means that, since you by force know the answer before that 3rd match, the outcome of this one cannot affect that answer.
And this is my limit... I've teached maths to many people since 4 years ago or so (starting at secondary school, helping classmates, and afterwards helping other people for a nice price), but I don't feel able to explain you what's your mistake in many more ways. Try to think about that, and you should understand it.
Edit: PD: Wow! I just found another arguing for D! If you care the chance of the third game being cancelled, the chances of the event finishing after 2 matches go higher. ie: a game could be cancelled due to weather, violence on the playfield, too many harmed, sanctioned or ill players on both teams, and so on... the only limit is imagination XD Anyhow, both this arguing and the previous one 'bout this option are non-math, so they shouldn't be taken in account... I've put them just for fun
Edit: apologies for the "bump". I was just reviewing old posts, and I think I clicked the button by accident.
forget the coin flip and go back to the sport event. Assuming most neutral conditions, as playing in a neutral field and weather not affecting results, then it's most probable that the best team wins each game. So, after team A wins the first game, there is an important chance that A is better than B, so its chances to win the second game are higher than 50/50. With such an argumentation, the answer should be somewhat higher than 1/2, and the only option that fulfills that is 2/3. I have had many teachers that, provided an argumentation like the one above, would accept as good that option (but it will be taken as wrong if you mark it without any explanation).