Care for one last (for now) probability intuition problem?
This is more relevant than the hatcheck problem. Two teams are scheduled to play two games against each other. For simplicity, assume each of their probabilities of winning the second game are the same as they are for the first game, irrespective of the results of the first game. One of these is true. Which one?
1. The most likely outcome will be that the two teams split, each winning one game.
2. It is always more likely that one of the teams wins both games than having a split.
3. It depends. If one team is a whole lot better then it will probably win both games. Otherwise, the most likely outcome is a split.
The algebra for this is fairly simple, but check in with your intuition before doing any algebra. What it says will likely surprise you.
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2 out of three times it will be won by the same team.
I feel like my intuition (though I have spent a lot of my life thinking about probabllity in games) was right first time. Though I had a bit of confusion because of the wording and what I think is an unstated assumption.
If the odds are even,i.e., 50/50, 2 out of 3 times the same team will win both games.
If the odds favor one team, that team will more likely win both games.
#2 is your best choice answer.
To clarify, if both teams have a 50% chance then the chances of a sweep by either team is (.50)**2 + (.50)**2 = .50, so in this case the chance of a swap is the same as the chance of a sweep. The 3 possible assertions should be taken to refer to the other cases.
Right, that’s what I figured. So that means the way 2 is worded with “always”, it’s not true in the case of perfectly matched teams, since that would give a 50% chance of a split. So really the answer should be part-way between 2 and 3: The most likely result is that the better team will sweep, and it’s more likely the more uneven the teams are. But with exactly even odds in both games, there would be an even chance the games would be a split or a sweep.
Excpet that there are three out comes:
Team A wins both
Team B wins both
They each win one game.
So even if the odds are 50/50. two out of three scenarios one of the teams wins both games.
^ There are four cases: A wins both games, B wins both games, A wins the first and loses the second, and B wins the first and loses the second.
I would go with #2. The first game is immaterial as one of the two teams is going to win it. In the second game the team with one win will play the team with no wins. The team that has already won a game has an advantage and so it is unlikely there will be a split.
Unless it’s possible they could be evenly matched, in which case the chance of them winning the second time would still be 50%, so the “always” in 2 is incorrect.
The probability of a sweep is always greater than or equal to the probability of a split. If the probabilities for teams A and B are p and q, we want to show that p**2 +q**2 >= 2pq, which is the same as saying p**2 + q**2 – 2pq > 0, which is the sane as saying (p-q)**2 >= 0, which indeed is always true, with equality occurring only when p=q = .5.
Consider the case where p = .6. The chances of A sweeping are only (.6)**2 = .36. The chance of a sweep by B is (.4)**2 = .16. The combined probability is .36 + .16 = .52, which is indeed greater than .5, though not by much.
Well . . . “with equality occurring only when p=q = .5.”
In that case, the unnecessary algebra indicates the chance of each sweeping is (.5)^2 = .25. The combined probability is .25 + .25 = .5, which is NOT greater than .5.
Therefore the wording “always . . . more” for option 2 is inaccurate, as I mentioned in my answers above.
And another story problem bites the dust!
I caught the error and made a correction, to which you responded. Sorry for the original misstatement.
My original intuition was option 1, figuring that there would be more heterogeneous cases than homogeneous ones.
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