Film, Media & TV1 min ago
BOY/GIRL AGAIN!
13 Answers
While pondering on the discussion in the Odds On question below, I visit old friends who I know have one set of twins, but I don't know their genders. I guess they have a boy and a girl, which is one out of three possibilities (2 girls, 2 boys, girl/boy). When I knock on the door it is answered by a little girl, so now the other child could be another girl (in which case my guess is wrong) or a boy (in which case it is correct). So suddenly the chances of my guess being correct seem to have changed from 1 in 3 to 1 in 2. But precisely the same logic would apply if a boy had come to the door - so the probability of boy/girl seems to be 1 in 3 if an adult comes to the door; but 1 in 2 if it is either one of the twins. Any thoughts on this?
Answers
1 to 13 of 13
Best Answer
No best answer has yet been selected by Sylday. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.
That's right.
The question changes once you know the sex of one of the children. If a girl answers the door, you then know that the boy/boy possiblity is excluded.
For the purposes of the question, we are of course assuming that among pairs of twins, identical and non-identical twins are equally likely.
i'm not sure as its very confusing but the odds do apparently change when one of the "answers" is revealed to you- the problem works in 2 stages. It sounds similar to something known as the "Monty Hall problem" which i recently read about in "The curious incident of the dog in the night time" by Mark Haddon (which is excellent btw!). it invoves goats & cars & is explained far more clearly than i can muster here
http://www.worldhistory.com/wiki/M/Monty-Hall-problem.htm
(like i said its not exactly the same theory but if nothing else it will probably interest you!)
Thanks andytreble99
I'd not seen that particular page
A simpler illustration(perhaps)
deal the four aces from a pack of cards face down.
The probability of turning over a red ace is (2 in 4) = (1 in 2)
Shuffle and deal again
Turn over the first card.
If it's red, the probability of now turing over the other red card is (1 in 3)
If it's black, the probability of turning over a red card is (2 in 3)
We are notoriously bad at dealing with probabilities:
if a test is accurate to 1 in a million, what does that mean to someone who tests positive?
if a drug can be detected in sea water is it relevant?
I've just climbed onto a hobbyhorse.
Jabba I think you're probably right, and the fallacy in my proposition above is that the probability was ever 1 in 3; it should have been 1 in 2 to begin with. But I can't quite get my head round the difference between girl/boy and boy/girl! Anyone explain that? As for the Monty Hall problem - like lots of others, I was initially sceptical, but after using an interactive website now accept that switching choices is the right decision.
Sylday - you were right with your initial 1 in 3 probability. If,like Jabba, you make the "order" of the twins relevant then there are in fact six possibilities - two of which are a boy & a girl - therefore still 1 in 3.
As a different slant on the problem, imagine there are three sets of twins behind three doors, one set of boys, one of girls and one boy & girl. you must find the boy/girl. After you choose, one of the other doors is opened to reveal a same sex set of twins - should you change your mind about the door you have chosen? [Copyright Monty Hall]
Funnily enough, although it's called the Monty Hall Problem (and it's a classic puzzle in recreational mathematics, one of the first things anybody reading up on probability is likely to learn about), it actually never had anything to do with Monty Hall at all - he was simply the best-known US TV quiz show host when the puzzle was devised, so he was used as the host of a fictional gameshow just to bring the puzzle alive.
I'm not sure whether I am agreeing or disagreeing here, but the chances of the twins being a boy and girl is 1 in 2. Not because of all the fictitious possibilities being bandied around, but because the options are; same sex or one of each! (This assumes that there is an equal probability of a boy or girl, which I think everyone else has assumed as well.) This can be explained as; given one child, it is 50/50 whether the second is the same sex, or different. Therefore the chance of a mixed set is 1 in 2. Behind that, the chance of 2 boys is 1 in 4, and 2 girls is 1 in 4, because it is 1 in 2 for each child = 1 in 4.
No, Dsg is incorrect. In a family of four children, the probability that there is one child of each sex is 1/2. You can try an empirical experiment to convince yourself of this, if your probability skills are not good enough to work it out.
Take two coins, and let heads represent a boy and tails a girl. Toss the two coins repeatedly and keep a record of the number of times each combination occurs. If you repeat this experiment a number of times, you will see that the proportions settle down in the ratios of the probabilities. That is, roughly 1/2 of the "families" will have one child of each sex, 1/4 will have two boys and 1/4 will have two girls.
You were wrong in the first place when you said that the probability is 1 in 3. Just because there are three possibilities (2 boys, 2 girls, 1 of each) does not mean that those three possibilities are equally likely. It would be just as logical to say that there are 2 possibilities (same, different) or that there are 4 possibilities (boy-boy, boy-girl,girl-boy, girl-girl) or that there are 2 possibilities (2 boys, not-2-boys). To work out the probability of each thing happening, you have to work out the individual events which make up the combined event. The possibility of "one of each" is actually two possibilities combined (boy-girl,girl-boy). The possibility of "not-2-boys" is actually 3 possibilities combined. In this case, there are actually 4 possibilities with a 1-in-4 chance of each.
1 to 13 of 13
|
|
|
|
|
|
 |
|
|