Talk:Probability/Conditional Probability

example
This is from a game show, maybe "The Price Is Right" or "Wheel of Fortune".


 * 1) Three doors are shown. A nice price is behind one of the doors.
 * 2) A contestant is asked to choose 1 of 3 doors to open.
 * 3) Before opening the selected door, one of the unselected doors is opened, revealing some junk. (a goat maybe)
 * 4) The contestant is offered the option to change his choice. (they may, and should)
 * 5) The chosen door is opened, revealing either a goat or something nice.

Lots of people get this one wrong. Lots of wrong people feel really sure that they are correct. Wikipedia may have an article about this example.

AlbertCahalan 06:16, 3 October 2005 (UTC)

There is indeed a Wikipedia article about this problem: http://en.wikipedia.org/wiki/Monty_hall_problem

Credner 27. Nov 2005

um... "The probability of one die showing six is 11/36"...a die has 6 sides on it, not 11. Let's not be teaching BS with a book. This is a sure way to make people not want to look at Wikibooks. I will try to correct the mistake. But come on! since when does a die have 11 sides??? CatastrophicToad 18:26, 15 January 2006 (UTC)

I don't kno how familiar you are with probability theory, but the text was correct, allthough I made it a little more explicit. The experiment was throwing two dice and hence in 11 of the 36 cases (at least) one dice shows six. And BTW there are dice with eleven sides. Nijdam 23:46, 16 January 2006 (UTC)

The text is not correct. The probability of a double six given that you've seen one of the dice and it's a six is 1/6. Because the outcome of the two die are independent events, you can look at it as mathematically equivalent to throwing one die twice. The first comes up six; what's the probability of a six when you roll it again? Dice has no memory. Now consider what the probability is that you get at least one six when rolling two dice. This is a different problem than the one where one die is revealed. In this case, the probability is higher (11/36). Either die could have the six so the number of rolls that fit the problem statement is higher. Too bad the first example given in a "textbook" is wrong.

The point is the difference between one of the dice or dice number one. When saying "one of the dice" it may refer to any one of the two. If it were the case that die number one shows six, the probability of the other, number two, showing six would be 1/6. But that is not the case as might be see nfrom the example. But for utter accuracy I change "one dice" into "one of the dice".Nijdam 20:50, 10 March 2006 (UTC)

I don't think conditional probability is introduced in the best way. The two dice, and accidentally seeing that one of them is six is a good example (and you're right to change it to one of the dice). It might be clearer if you have somebody tell you rather then you accidentally seeing one of the dice.

However, the thing I really dislike is whenthe probabilities are pulled together and the conditional probability just stated as being 1/36 / 11/36

I feel it would be better to say: Now look at all possibilities for the two dice, where at least one of them shows a six. Note that there are 11 possibilities, all equally likely and so the chance of a double 6 given you know there is one.....--MarkyParky 23:16, 25 August 2006 (UTC)

Hello MP, the problem is that in this case we may refer to the 36 possibillities that are equally likely. But that is not the general case. What if P(B)=0.321 and P(AB) = 1/99? Maybe it is good to mention somehing like we "blow up" the probabillity of the event B to 100% etc.?Nijdam 15:22, 30 August 2006 (UTC)

The statement that accidentally seeing one of the die is a 6 gives us a 1/11 probability is wrong, becase it fails to take into account the odds that one of the die was a 6, but that the die we accidentally saw was not the 6. If we imagine one red die and one blue die, we can see that the the odds of the blue die being 6 (having seen the red die and knowing it's 6) are 1/6. The same holds true if we saw the blue die is 6. The only case where where the odds are 1/11 is when we know that EITHER the red die OR the blue die is a 6, which is not the case if we actually saw the die. Therefore, it should be changed to say "Suppose somebody looks under the cup, sees the result, and tells us AT LEAST ONE of the die is 6.

I'm sorry, this is just all wrong, and most of the above discussion is not even relevant. In the example as currently given: ONE die is known to be six; ONE die is not known; we wonder what the probability of "double-six" is. The probability that THE UNKNOWN DIE is a six is 1/6. If the unknown die is a six, given that the known die is also a six, we have "double-six"; therefor, the probability of double-sixes is 1/6. Simple. I will not edit, because I really am not sure what the point of this example is supposed to be, so I can't usefully fix it.


 * I have rewritten the example in a clearer way. Hope that the example can be understood more easily now. LeoChiukl (discuss • contribs) 10:46, 25 May 2022 (UTC)

Incoming link from Wikiversity
I'm not sure what to do. I am working on v:Bell's theorem and need to reference conditional probability. The best wiki on this is right here, so I made a link to one section (that I added). The alternative is for me to copy this page to Wikiversity and edit it. That eliminates the cross-wiki stuff, but also forces duplication. Which alternative is best?

Another option is to use a permalink from the article that needs the info on Wikibooks. I just added that to my article. --Guy vandegrift (discuss • contribs) 21:43, 26 September 2015 (UTC)See also: w:User:Guy vandegrift

The lede
In addition to creating the example with the venn diagrams and integers (even/prime), I changed the lede:

1. I tried to put more information into less space. The pair of equations appears on one line, but split with a space (&nbsp) so that people with cell phones will be able to see two lines.
 * $$P(A|B) = \frac{P(A \cap B)}{P(B)} \, \Leftrightarrow $$  $$P(A \cap B)  = P(A|B) P(B) =  P(B|A) P(A)$$

I see no reason to use an entire paragraph to show what is algebraically trivial.

2. Wikipedia calls it Bayes' theorem, so I noted the controversy and linked to Wikpedia. Is that OK? If not, I have no objection to removing the link to Wikipedia and reference to Bayes.

yours truly --Guy vandegrift (discuss • contribs) 22:22, 26 September 2015 (UTC)