04 Apr 08
Originally posted by PBE6I am taking it that each urn must have at least one ball in it?
Might as well make this a new problem!
Suppose you are given 50 black balls and 50 white balls. Your task it to distribute the balls between the 2 urns in any combination you wish, provided you use all 100 balls, with the goal of maximizing the probability that a white ball will be chosen when a ball is selected at random from a randomly selected urn.
What is the optimal solution?
If that is the only constraint, then I can make the odds of selecting the color ball I want to be almost 74.75%
04 Apr 08
1 edit
Did we get an exact answer on the sports problem, because I got 7/13 as odds it was played in New York?
The calculation was for odds that the game was in New York given that the Mets won, which seemed to me the most direct way of looking at it, but the other answer seem more complicated.
EDIT: And in looking back at the problem, I see you have to given equal weight to the home and the away games, which means each of the 9 individual home games has a bit more weight then the 10 away games.
In accounting for this, I arrived at an adjusted answer of 35/62 odds the game was in New York given the Mets won, a percentage of 56.45%
The proper odds ratio for a bet against this is 35:27.
04 Apr 08
Originally posted by mtthwcan you crank through this first set of calculations for me, i'm a bit confused.
It was decided on the toss of a coin initially. But we've been given some additional information (that the Mets won), and that allows us to make a further inference, which is where conditional probability comes in.
It might be more obvious if you consider an extreme case. Let's say that instead of the win ratios given, the Mets always win in NY, an ...[text shortened]... win percentages are different in the two cities we can refine the initial probability of 0.5.
04 Apr 08
Originally posted by mtthwThe extreme case described here :
Sorry, which calculations precisely? I'm happy to try and help, but I want to make sure I'm trying to explain the right things.
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It was decided on the toss of a coin initially. But we've been given some additional information (that the Mets won), and that allows us to make a further inference, which is where conditional probability comes in.
It might be more obvious if you consider an extreme case. Let's say that instead of the win ratios given, the Mets always win in NY, and never win in LA. In that case when we hear that they win, we know the match must have taken place in NY.
[The formula I gave still gives the correct answer in this case - the answer resolves to 1 in the case where P(M|NY) = 1 and P(M|LA) = 0].
The case given isn't that extreme, but as long as the win percentages are different in the two cities we can refine the initial probability of 0.5.
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04 Apr 08
2 edits
Originally posted by eldragonflyI can help with the concept, although I'll drag in the Monty Hall problem (the bazillionth time for some of these guys).
both, i am trying to understand conditional probability.
You're playing Let's Make a Deal with Monty Hall. There are three doors on stage, and behind one there is a prize (the prize door is randomly selected before the show, 1/3 chance it could be behind any door). For simplicity's sake, there's nothing behind the two non-winning doors. You are asked to pick which door you think the prize is behind, so you pick a door at random (1/3 chance of selecting any door) but it doesn't get opened yet. Let's call this Door A.
1) What is the chance that the prize is behind Door A? (normal probability question)
Now, with you having selected Door A, Monty says "reveal what's behind one of the other doors!" and the crew opens either Door B or Door C, depending on the following:
- if you picked Door A, and the prize is behind Door B, the crew will open Door C
- if you picked Door A, and the prize is behind Door C, the crew will open Door C
- if you picked Door A, and the prize is behind Door A, the crew will select Door B or Door C at random (1/2 chance for either door) and open it
In all cases, the door opens to reveal nothing (i.e. the prize is not behind that door). Monty then asks you "do you want to stick with your choice of Door A, or switch to the other door that hasn't been revealed yet?"
2) Should you stay with Door A or switch? (conditional probability question)
Once you've answered both parts, you should have a good handle on what conditional probability means and how to use it.
04 Apr 08
1 edit
Originally posted by PBE6i will have a look at that, but here's one i clearly do not understand.
I can help with the concept, although I'll drag in the Monty Hall problem (the bazillionth time for some of these guys).
You're playing Let's Make a Deal with Monty Hall. There are three doors on stage, and behind one there is a prize (the prize door is randomly selected before the show, 1/3 chance it could be behind any door). For simplicity's sake, there should have a good handle on what conditional probability means and how to use it.
What is the probability that you will draw a king given that you have drawn a face card. "Statistics the Easy Way" gives the answer as p=1/3, using the conditional probability formula P(A|B).
4/51 seems a much more reasonable number, or 12/52 * 4/51.
04 Apr 08
Originally posted by eldragonflyOK. I'll do my best.
The extreme case described here :
The first part of the calculation doesn't assume anything about the probabilities. My previous argument showed that:
P(NY | M) (i.e. the probability it was in NY given that we know the Mets won - which is what the question is looking for)
= P(M | NY)*P(NY)/[P(M | NY)*P(NY) + P(M | LA)*P(LA)]
This is true no matter what the probabilities are. It only assumes that NY and LA are the only possible venues.
Because P(NY) = P(LA) = 0.5 (because it was a coin toss to decide the venue), this simpifies to
P(NY | M) = P(M | NY)/[P(M | NY) + P(M | LA)] *
OK. Now let's consider the extreme case. The Mets always win in NY, but always lose in LA.
So, if we know the Mets won, then it must have been in NY. We don't need to know anything else. It doesn't matter that the initial choice was made by a coin toss: we know that if it was in LA the Mets would have lost, so therefore the match was in NY. It's the only possibility.
Now, putting that argument in terms of conditional probabilities:
Mets always win in NY: P(M | NY) = 1
Mets always lose in LA: P(M | LA) = 0
Substitute these into equation * above:
P(NY | M) = 1/(1 + 0) = 1
That is, it must have taken place in NY.
04 Apr 08
1 edit
Originally posted by eldragonflyI am assuming the following.
i will have a look at that, but here's one i clearly do not understand.
What is the probability that you will draw a king given that you have drawn a face card. "Statistics the Easy Way" gives the answer as p=1/3, using the conditional probability formula P(A|B).
4/51 seems a much more reasonable number, or 12/52 * 11/51.
1) The card is a fair deck of 52 cards.
2) Any card in the deck is equally likely to be chosen.
We are told the card drawn is a face card, which means of the 52 cards it could be, we can eliminate all but 12 of them from having been chosen.
Any of these 12 cards are equally likely, because they any of the 12 were just as likely as any of the others before we were told the card selected was a face card, and nothing has yet reduced the chances of any of the 12 remaining cards it could be.
Of these 12 cards, any of which is just as likely to have been picked as any of the others, 4 of them are Kings, making the chances it is a king (given that it is a face card) 4-in-12, or 1-in-3.
The Bayes math is as follows.
P(A&B) = 4/52 or 1/13
P(B) = 12/52 or 3/13
P(A|B) = P(A&B) / P(B) = (1/13) / (3/13) = 1/3.
04 Apr 08
1 edit
Originally posted by eldragonflyLet K denote that a king is drawn; Let FC denote that a face card is drawn.
i will have a look at that, but here's one i clearly do not understand.
What is the probability that you will draw a king given that you have drawn a face card. "Statistics the Easy Way" gives the answer as p=1/3, using the conditional probability formula P(A|B).
4/51 seems a much more reasonable number, or 12/52 * 4/51.
Bayes tells us that
P(K|FC) = [P(FC|K)*P(K)]/P(FC) = [(1)*(4/52)]/(12/52) = 1/3.
04 Apr 08
Originally posted by geepamoogleyeah i understand that, that is the solution that is given but the odds of drawing a king next is 4/51. You cannot eliminate the rest of the deck minus the remaing face cards, as you propose. 😕
I am assuming the following.
1) The card is a fair deck of 52 cards.
2) Any card in the deck is equally likely to be chosen.
We are told the card drawn is a face card, which means of the 52 cards it could be, we can eliminate all but 12 of them from having been chosen.
Any of these 12 cards are equally likely, because they any of the 12 were ju ...[text shortened]...
P(A&B) = 4/52 or 1/13
P(B) = 12/52 or 3/13
P(A|B) = P(A&B) / P(B) = (1/13) / (3/13) = 1/3.