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May 25[edit]

Pairing letters and numbers[edit]

A machine is given eight letters, from A to H, and eight numbers, from 1 to 8. It has to form eight pairs, each containing a letter and a number, on a random basis, with the following limitations:

A may not be paired with 2 or 3.
1 may not be paired with B, C, or D.

What is the probability that A be paired with 1? Does this value change if there are extra limitations regarding other letters and numbers (e.g., G may not be paired with 7, etc)? --Theurgist (talk) 15:50, 25 May 2012 (UTC)[reply]

It all depends upon how you "deal" the pairings. Running 100,000 deals of 1..8 paired with a random letter from the list of letters that remain unpaired, I divided the categories into "invalid" (55,453), "Non-A1" (35,791) and "A1" (8756). That suggests a probability of 8756/35,791 = 0.244642.

If you "deal" in the order of tightest constraint first then you 'Pair 1 with AEFGH Giving "A1" 0.2'.

If you "deal" A first, you get 'Pair A with 145678' Giving "A1" 0.16667'.

So more information is required to be able to answer this any further. -- SGBailey (talk) 17:04, 25 May 2012 (UTC)[reply]

Sorry, I couldn't quite understand: specifically what information is required? I made up this problem, though it's inspired by real-life situations. --Theurgist (talk) 18:45, 25 May 2012 (UTC)[reply]

SGBailey: There's no need to do random experiments. There are 8! = 40320 pairings in total, so you can enumerate all of these (easier than doing 100000 random experiments anyway). You find that exactly 19440 of them satisfy OP's criterions (A is not paired with 2 or 3, and 1 is not paired with B or C or D), and of these 5040 have A paired with 1, so the probability is 5040/19440 which is approx 0.259. – b_jonas 18:51, 25 May 2012 (UTC)[reply]

and exactly 7/27. —Tamfang (talk) 21:05, 25 May 2012 (UTC)[reply]

Theurgist, the extra information that is needed is excatly what SGBailey asks. What is the procedure that the machine uses to pair letters with numbers? This matters because the enumeration that b_jonas describes assume that each legal pairing has the same probability. And this might not be the case. Take a pairing that first assigns A to a legal number, then assign B to a legal number and so on until it is done or until it reaches a point where it fails. This machine will assign A to 1 in 1 of 6 cases. Whereas a machine that first selects a pairing totaly at random, and then rejects if it violates your criteria will assign A to 1 in 7/27 cases. Extra limitations can affect these probabilities, as a trivial example if 7 can not be paired with B,C,D,E,F,G or H then 1 is paired with a in 0 cases.(Because a will always be paired with 7) 176.11.10.213 (talk) 20:00, 26 May 2012 (UTC)[reply]

Yes, phrasing the question in terms of probabilities creates an ambiguity. A better way of phrasing the question would be "what proportion of valid pairings include the pair (A,1) ?", in which the answer is unambiguously 7/27, as Tamfang says. Gandalf61 (talk) 20:14, 26 May 2012 (UTC)[reply]

Left handed redheads[edit]

If a given population contains one percent red-haired and ten percent left-handed people, what proportion would be left-handed redheads?
Not homework, I'm trying to explain in simple terms to a friend how a baby could have two entirely unrelated congenital problems at the same time. Roger (talk) 19:04, 25 May 2012 (UTC)[reply]

0.1%. $1/10$ * $1/100$ --Gilderien Chat|List of good deeds 19:12, 25 May 2012 (UTC)[reply]

Thanks! I'm glad it's a simple easy to understand formula/calculation. Roger (talk) 19:26, 25 May 2012 (UTC)[reply]

Yeah, with any two random unrelated factors, the probability of having both is the individual probabilities multiplied together.--Gilderien Chat|List of good deeds 20:05, 25 May 2012 (UTC)[reply]

Whoa. Aren't we confusing two unrelated things: proportions and probabilities? The question was about proportions. It is not safe to say that (1% * 10% =) 0.1% is how many left-handed redheads there'd be, without assuming a uniform distribution. It's possible that all the redheads were left-handed, and the remaining 9% of left-handers had some other hair colour. In that scenario, the answer is 1%. -- ♬ Jack of Oz ♬ ^{[your turn]} 21:28, 25 May 2012 (UTC)[reply]

Unless someone goes out and finds a population in which 1/10 are lefties and 1/100 are redheads, and counts the left-handed redheads, all we can ask for is an expected proportion, which is the same as a probability. —Tamfang (talk) 21:47, 25 May 2012 (UTC)[reply]

That caveat needs to be made explicit. Gilderien stated 0.1% as the definite answer to the question, which was about proportions. The word "probability" did not appear in his first answer at all, and only in his second post as an aside. He's suggesting that if the population is 5,000, then under the conditions outlined by the OP, the number of left-handed redheads is exactly 5. It's certainly not as clear-cut as that. That's the expected figure assuming a uniform distribution, but any given population may not happen to be exactly uniform (real life populations rarely if ever are), so you can't bet your house on it being exactly 5. -- ♬ Jack of Oz ♬ ^{[your turn]} 22:18, 25 May 2012 (UTC)[reply]

If I were you, I'd step gingerly when describing those two traits as if they were sinister congenital problems. StuRat (talk) 02:46, 26 May 2012 (UTC) [reply]

Indeed - I specifically used innocuous traits in the "thought experiment" so that my friend's thinking would not be clouded by the "poor suffering baby" aspect of the situation. It could just as well have been "trait A and "trait B". BTW sorry I forgot to specify the assumption that the population is large enough to ignore the distinction between proportion and probability. Roger (talk) 11:56, 26 May 2012 (UTC)[reply]

Actually I don't think that the issue is that distinction. It's the fact that we have not established that redheadedness and lefthandedness is statistically independent. And this means that we can't automatically tell either the probability or the proportion of redheaded lefties. 176.11.10.213 (talk) 20:09, 26 May 2012 (UTC)[reply]

The most relevant article is probably Independence (probability theory). As others have pointed out, this applies to probabilities, not the actual numbers of people with given characteristics in a population (the answer to your question as stated is 'somewhere between 0% and 1%'). Oddly enough, there does actually appear to be a correlation between handedness and hair colour - this paper ("Associations of handedness with hair color and learning disabilities") says "Among blonds, the frequency of non right-handedness... was 44% compared to 24% of non-blonds". 130.88.73.65 (talk) 14:01, 29 May 2012 (UTC)[reply]

One out of twenty possible correlations passes a 5% significance test, even if the correlation is not genuine. The accuracies of the figures 44% and 24% are not known. Bo Jacoby (talk) 09:13, 31 May 2012 (UTC).[reply]