Originally posted by PBE6Run the simulation. Or read the entire thread. Or go away and learn something about conditional probability.
LemonJello is right about this one. I recommend running the game yourself and checking the results. I think you'll be surprised.
I can't face going through this again. You guys have more patience than I have. 🙂
Originally posted by PBE6Well, here's my approach. First, solve the general case.
I think I made a mistake, because I got a much different answer. I'll take a look and post what I find.
T = Positive test, D = has the disease
Accuracy a = P(T|D)
Specificity s = 1 - P(T|!D)
Rate of infection in population = r
P(D|T) = 1/[1 + [(1 - s)/a](1/r - 1)]
That gave me the 1/50506 answer earlier.
Now, run the test twice. I'm assuming that if we do this then only two positive tests constitutes a final positive.
In this case:
Accuracy = a^2
Specificity = 1 - (1 - s)^2
This is just the equivalent of squaring the (1 - s)/a term in the final answer. In fact, run the test n times and you just have to raise that term to the power n.
I am amazed the three card problem hasn't died yet either..
I am honestly pondering a fool-proof way to explain the darn thing, but nothing seems to work..
Let's try a 3x2 matrix, shall we?
One axis is the chosen card, the other axis is the color of the visible side. This will only be slightly more complicated than the 2x2 matrices we've seen for other problems. Each value indicates, on average, how many times out of 6 each card will be picked with the specific visible side up.
Visible
Gold Silver
G/G - 2 0
G/S - 1 1
S/S - 0 2
Please note here the following:
1) Each card is likely to come up twice or 1-in-3 times.
2) Each color is likely to come up thrice or 1-in-2 times.
This is the basic setup without additional information on the results.
Now, we are told a silver side was showing, so we take out the Gold column and reassess odds.
Silver
G/G 0
G/S 1
S/S 2
And out of the 3 times (out of 6) that silver shows up, it shows up on the Silver/Silver card twice, and on the Gold/Silver only once. This means that the opposite side is gold only 1 out of 3 times.
For those arguing that this is different because it is a one-time event, we assume one-time events are fair unless the set-up dictates otherwise.
For the "Two Test" problem, the answer assumes that the test has zero consistency (in other words, that a subsequent retesting is independent of the first test).
This is a doubtful premise to me, but if it is assumed to be the case, then altering the matrix for results to include two positives, two negatives, or a mix, and then finding the appropriate ratio, two positives would raise the odds of having the disease to a whopping 1-in-2,551.76 ( or 0.0392% ).
If the test has some consistency (that is to say it tends to give consistent results for the same person to some degree), then the chances of being infected don't raise as much.
EDIT: Exact odds calculated to be 33 / 84,208
Originally posted by broblutoIt's a "can't see the forest for the trees" type of thing, beautiful problem but badly worded.
Forget about sides. Your confusing the point.
Fact 1: He has one card in his hand
Fact 2: That card is 1 of 2 possible cards.
Outcome 1: It is the right card
Outcome 2: It is the wrong card.
50/50 chance.
Originally posted by LemonJelloLemonJello for the umpteenth time that "randomness" is just not there. It's a contrived problem and a very contrived solution, i'm surprised that you are defending this. No additional information, no randomness, no bayes theorem, surely you must understand by now.
No. Think about it this way, brobluto. It is supposed that you have chosen one side randomly. Now, given the additional information that this side happens to be silver, do you not see how it is twice as likely that this side belongs to the silver/silver bar than the silver/gold bar?
Originally posted by mtthwFail. If anyone needs to learn anything about conditional probability it is yourself.
Run the simulation. Or read the entire thread. Or go away and learn something about conditional probability.
I can't face going through this again. You guys have more patience than I have. 🙂
Originally posted by broblutoCorrect.
No. It's not. If you have only two bars in the hat, silver/silver and silver/gold, then the chances that you bring out a silver facing side is 3 in 4 (75😵 chance. The chance that you bring out the silver/silver bar is 1 in 2 (50😵 and the chance that it's the silver/gold is 1 in 2 (50😵.
Originally posted by kbaumenreword the problem, this is getting rather silly.
Exactly. So there are three cases when you bring out silver side. And only in one case the other side can be golden.
EDIT: Look at it this way. You can't see the whole card, you see only a side, so you have to consider the number of sides, not cards. And there are three silver sides, where to only one of them, the reverse is golden.
Originally posted by geepamoogleFail.
I am amazed the three card problem hasn't died yet either..
I am honestly pondering a fool-proof way to explain the darn thing, but nothing seems to work..
Let's try a 3x2 matrix, shall we?
One axis is the chosen card, the other axis is the color of the visible side. This will only be slightly more complicated than the 2x2 matrices we've seen f ...[text shortened]... is a one-time event, we assume one-time events are fair unless the set-up dictates otherwise.
It's not a three card problem, the gold card is "automatically" excluded by definition.
eldragonfly, the fact that the simulation is run once is irrelevant.
Unless there is some sort of dynamic context where the odds somehow shift, then the short term or single event odds and ratios will match the long term odds and ratios, and ergo we should expect the odds for a single event to match the ratios of results if the same scenario is repeated any significant number of times (ignoring those events where the precondition does not occur).
Let's take this one step at a time then.
Let us therefore determine where our analysis diverges, shall we?
Remove from the problem the result of which side is facing up.
We have three cards in a hat (gold, silver, and gold/silver). One is selected randomly, and placed on the table without glancing at the sides.
What are the odds for the following events?
1) Gold side up, gold side on reverse.
2) Gold side up, silver side on reverse.
3) Silver side up, gold side on reverse.
4) Silver side up, silver side on reverse.
(We'll get to the additional information in a moment.)
Or is it your theory that this simpler starter has no bearing whatsoever on the original problem?
Originally posted by eldragonfly😴
LemonJello for the umpteenth time that "randomness" is just not there. It's a contrived problem and a very contrived solution, i'm surprised that you are defending this. No additional information, no randomness, no bayes theorem, surely you must understand by now.
Well, I see that you still don't get it. In the future you may want to have at least some semblance of understanding before you declaim in the forums. I mean, conditional probability clearly isn't your forte. Plus, you're just awesomely terrible at absorbing the intended content of relatively straightforward problem statements (PBE6 made it clear enough in the problem, along with his first additional clarification post, that the "randomness" is there).
Cheers!
Originally posted by geepamooglei said it was an interesting problem, if poorly worded. This is going nowhere, i gave the correct solution long ago. Let's move on...
eldragonfly, the fact that the simulation is run once is irrelevant.
Unless there is some sort of dynamic context where the odds somehow shift, then the short term or single event odds and ratios will match the long term odds and ratios, and ergo we should expect the odds for a single event to match the ratios of results if the same scenario is repeat ...[text shortened]... is it your theory that this simpler starter has no bearing whatsoever on the original problem?