Probability question - Page 1
Forum section: electronics / general

author: Leuven
author_rank: Regular Contributor
timestamp: February 09, 2017, 04:52:28 pm
content: I'm trying to figure out the following: what are the odds for 2 people in a random group of 7, to both have an identical twin? The average identical twin incidence is about 3-4%.3% times 7 is 21% chances to have just one twin in a group. Chances to have 2 in the group (not related to each other) are roughly 1/5 x 1/5 = 1/25, or 4%. Is this correct?

author: Assafl
author_rank: Frequent Contributor
timestamp: February 09, 2017, 05:29:38 pm
content: The likelihood that two out of the 7 is 0.03*0.03.The likelihood of not having a twin brother in a group of 5 (1-0.03)^5So the likelihood of only 2 is 0.03^2*(1-0.03)^5.But there are multiple permutations possible (the above answers only one permutation. So you have to multiply the above by the number of permutations: 10*9/2!.I hope this is right.

author: tszaboo
author_rank: Super Contributor
timestamp: February 09, 2017, 05:46:39 pm
content: https://en.wikipedia.org/wiki/Binomial_distributionWhat are the chances that we both live in Leuven?

author: Leuven
author_rank: Regular Contributor
timestamp: February 09, 2017, 05:54:00 pm
content: Quote from: Assafl on February 09, 2017, 05:29:38 pmOf course, it's always 3% no matter how big or small the group is. I've written a bunch of rubbish above.So it's 0.03*0.03=0.0009 or 0.09%. Good enough for me.

author: Leuven
author_rank: Regular Contributor
timestamp: February 09, 2017, 06:03:45 pm
content: Quote from: NANDBlog on February 09, 2017, 05:46:39 pmInteresting question. Maybe I just like beer?

author: snarkysparky
author_rank: Frequent Contributor
timestamp: February 09, 2017, 07:31:06 pm
content: So for two people to have identical twins is 0.03.The question is how many opportunities are there for two people to share a characteristic in a group of 7 people.Any of these combinations is possible.1,21,31,41,51,61,72,32,42,52,62,73,43,53,63,74,54,64,75,65,76,7So there are 21 ways for this to happen in a group of 7.Octave monte carlo trialP = rand(30000,21); % generate a large group of random numbers 21 in a row.k = P < 0.03001; % find all that are less than 0.03j = sum(k,2); % sum the matches along columns.p = (j == 2); % identify the indexes of trials where the match number equals 2r = find(p); % count the number of trials where the match number equals 2[rw cl] = size(r);ratio = rw/30000 % divide the successes by the number of trialsratio = 0.10617ratio = 0.10597ratio = 0.10780lets call er 10.6 %should have been 7 instead of 21

author: orolo
author_rank: Frequent Contributor
timestamp: February 09, 2017, 07:54:02 pm
content: Quote from: NANDBlog on February 09, 2017, 05:46:39 pmThat solves the problem, except for the details.\$ p \ = \ \binom{7}{2} \cdot 0.03^2\cdot 0.97^5 \ = \ 0.01623 \ = \ 1.623\% \$

author: suicidaleggroll
author_rank: Super Contributor
timestamp: February 09, 2017, 07:59:19 pm
content: Quote from: Leuven on February 09, 2017, 05:54:00 pmThat's the probability that one SPECIFIC pair out of a group of 7 are members of twins. The question is what is the probability that there is ANY pair out of a group of 7 that are members of twins. It's much higher than 0.09%.My naive approach would be that there's 0.09% probably that any specific pair are both members of twins, that means there's a 99.91% chance they're not. There are 21 possible pairings (see snarkysparky's post), so there's a 0.9991^21=0.98127 = 98.127% chance that none of the pairs are both members of twins, which gives a 100-98.127=1.873% chance that at least one of the pairs are both members of twins.I suspect the difference between this result and orolo's is that his equation is calculating the probability that there is exactly one pair in the group that are both members of twins, while my approach is calculating the probability that there is at least one pair in the group.Then again, it's been a long time since I was in a stats class.

author: grumpydoc
author_rank: Super Contributor
timestamp: February 09, 2017, 08:11:15 pm
content: Well, for starters the probability of twins is around 4 per 1000 births so 0.4% not 4%The probability of 5 people not having a twin is 1-(1-0.004)^5 which is approx 2%

author: tszaboo
author_rank: Super Contributor
timestamp: February 09, 2017, 11:29:14 pm
content: Quote from: orolo on February 09, 2017, 07:54:02 pmYou know teach a fish to man up. Or something like that.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: February 10, 2017, 12:20:53 am
content: Probability questions require extreme clarity in the question.There are a number of possible questions here consistent with the English wording:What is the probability that two people selected out of a group of 7 people from population having x percent twins are among those that have twins.What is the probability that a group of 7 people taken from a population having x percent twins has two members that have twins.What is the probability that a group of 7 people taken from a population having x percent twins has at least two members with twins.Various answers on this thread have used approaches appropriate for each of these questions. I suspect that the second question stated here was the intended one, making the binomial distribution answer the correct one.