Probability

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Sam96

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You are part of a football team, there are 3 possible outcomes for each match win ; lose or draw. what is the probability of winning the next 2 matches in a row?
 
Sam96 said:
You are part of a football team, there are 3 possible outcomes for each match win ; lose or draw. what is the probability of winning the next 2 matches in a row?
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As written there is absolutely no way that question has an answer.
 
pka said:
As written there is absolutely no way that question has an answer.
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Why do you say that? I think it meant to assume the likelihood of each event of winning, losing, and drawing is equally likely. Also, each game is assumed to be independent.
 
Because nothing is given about the chances, why would assume anything?
 
pka said:
Because nothing is given about the chances, why would assume anything?
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For the exact reason: "nothing is given about the chances," the equally likely assumption would be fair. Don't you think? (given if that's the exact wording of the original problem, and that's all you have to go on)
 
BigBeachBananas said:
For the exact reason: "nothing is given about the chances," the equally likely assumption would be fair. Don't you think? (given if that's the exact wording of the original problem, and that's all you have to go on)
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Do you really think it's reasonable to suppose those three outcomes are equally likely? It's somewhat like supposing that heads, tails, and edge are equally likely in tossing a coin.

My guess is that the question may be one of a series in which one possible answer is "not enough information" or "can't be determined", and that the very purpose of the exercise may be to emphasize the fact that knowledge of the probabilities (or equal likelihood) of outcomes is necessary in such problems.

In any case, I think that would be an appropriate answer.
 
pka said:
As written there is absolutely no way that question has an answer.
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BigBeachBananas said:
For the exact reason: "nothing is given about the chances," the equally likely assumption would be fair. Don't you think? (given if that's the exact wording of the original problem, and that's all you have to go on)
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pka wrote that post because the problem can't be done. If two people sit down at a blackjack table do they both have the same chance of winning?
 
Jomo said:
pka wrote that post because the problem can't be done. If two people sit down at a blackjack table do they both have the same chance of winning?
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We don't know for sure, and maybe they have equal chances of winning; perhaps one is better than the other. How do you know for sure without historical experience? But it would be a fair initial assumption to say A has an equal chance of winning as B. Why would it be unfair to say they have equal chances of winning without knowing anything about them?

Dr.Peterson said:
Do you really think it's reasonable to suppose those three outcomes are equally likely? It's somewhat like supposing that heads, tails, and edge are equally likely in tossing a coin.
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To answer your question, yes, I think it's a reasonable assumption to make as we know nothing about the football team's winning records. As we observe more games, with historical experience, we can update our initial beliefs. This idea follows Bayesian Inference, where we make a particular assumption about our model and update it with more information.

On second thought, it is possible to be true that this is not the intention of the question writer. If the question were meant to answer with prior experience, then the Pr(winning/losing/drawing) would be a missing piece. I would agree with you the answer is "can't be determined."
 
Here is the best you can do with this problem.
Suppose p(win) = w, p(lose)=l and p(tie) = t (so w+l+t =1). Of course you do not even have to define what p(lose) and p(tie) equals
If the games are independent, then p(winning two games in a row) = w^2.
If the games are not independent, the you can't compute p(winning two games in a row)
 
The crux of the issue lies in how probability is presented, at least at the introductory level.
Using the classic coin toss example: "A coin has two possible outcomes, head or tail. What's the probability of getting two heads in a row?"
When presented with such questions, are students being taught to ask the question: "Is it a fair coin or is it loaded" (as most of you suggested, and the correct way) OR
Are they being taught to ignore that question and proceed with the calculation with their common understanding a coin has two sides, therefore Pr(Head)=Pr(Tail)=1/2?
I think you would agree, like it or not; the latter is most likely the standard practice. As a result, the "common practice" is to assume the outcomes are equally likely, unless stated otherwise in such problems.
With all that said, it boils down to the precision of the educators and question's author.
 
Even if we can assume that the teams are equally matched, we cannot assume that one third of games end with a draw.
 
JayJay said:
Even if we can assume that the teams are equally matched, we cannot assume that one third of games end with a draw.
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If we assume that Team 1 and Team 2 are equally matched, then Team 1 has Pr(win)=Pr(lose)=Pr(draw)=1/3. Similarly, for Team 2. I don't see why "we cannot assume that one-third of games end with a draw."?
 
Just missed the edit window to add: Sometimes, as in a knockout competition, a drawn match is not allowed, extra time and a penalty shoot out being used to determine a winner.
 
BigBeachBananas said:
If we assume that Team 1 and Team 2 are equally matched, then Team 1 has Pr(win)=Pr(lose)=Pr(draw)=1/3. Similarly, for Team 2. I don't see why "we cannot assume that one-third of games end with a draw."?
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You could suppose that, for the sake of argument, and state it up front when you write up your answer. But to assume that it is true is not reasonable.

But your (reasonable) assumption that the teams are equally matched, in the absence of other information, absolutely does not imply that a draw would be as likely as a win. That's why I used the example of heads, tails, and standing on edge. A draw is a very different situation.

Even if we made an oversimplified model of scoring for a game, so that each team's score is determined by the roll of a die, the probability of a draw (tie) would be 1/6, compared to 5/12 each for win or lose.
 
BigBeachBananas said:
The crux of the issue lies in how probability is presented, at least at the introductory level.
Using the classic coin toss example: "A coin has two possible outcomes, head or tail. What's the probability of getting two heads in a row?"
When presented with such questions, are students being taught to ask the question: "Is it a fair coin or is it loaded" (as most of you suggested, and the correct way) OR
Are they being taught to ignore that question and proceed with the calculation with their common understanding a coin has two sides, therefore Pr(Head)=Pr(Tail)=1/2?
I think you would agree, like it or not; the latter is most likely the standard practice. As a result, the "common practice" is to assume the outcomes are equally likely, unless stated otherwise in such problems.
With all that said, it boils down to the precision of the educators and question's author.
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No, what you say is not true at all.

Suppose the way coins are made, there is a 1/3 chance of getting a tail and a 2/3 chance of getting a head. This IS the way it is (at least for this problem) and one can't argue with it.
Of course, someone can load the dice to change those probabilities listed above.

Suppose you were asked to find the probability of tossing two heads in a row with no additional information.

Would you compute (1/2)*(1/2) or (2/3)*(2/3). I suspect that you'll do the latter. Why, because you assume that the coins were not adjusted in any way.

That is the reason you concluded that the answer is (1/2)*(1/2) using the fact that p(h)=p(t)=1/2 NOT because without any additional knowledge you always assume 1/2 and 1/2.

Your example holds no water at all--sorry
 
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