Twins puzzle

How many were female?

 

Not sure - I was doing a head count.

 

You should have counted the legs and divided by 4!

 

Yes, I know that. I should of course have said 'Foetus 1' and 'Foetus 2'.

 

I'd say it's still 50/50, sheep are most likely to have dizygotic which means non identical. At least one is male, the other could be either male or female, which would mean 50:50. Also the test has not eliminated that they are both female as it could be a false positive.

 

Take a representative flock of 12 sheep all having twins with each foetus being equally likely to be male or female. Although it's a small sample we are assuming that it's representative so the proportion of each possible combination is equal. We have MM MF FM FF MM MF FM FF MM MF FM FF.
Now ignore the 3 FFs and leave
MM MF FM MM MF FM MM MF FM
and 6 out of the 9 have an F

 

I resoundingly agree with the reasoning that the chances of the other twin being female are 2 in 3. However, as others have intimated, this requires the assumption that the twins are dizygotic (= fraternal). If the rate of monozygotic (= identical) twinning in sheep were substantial, the 2 in 3 chance would have to be adjusted accordingly.

As it happens, quite a bit is known about sheep twinning. Twin lambs mean that a farmer gets two for the price of one, so to speak, and if both are female they potentially grow into ewes that will twin again. Hence the impetus for twin research in sheep. It appears there's yet to be a definitive study on MZ vs DZ sheep twinning but from preliminary work it's stated on good authority in this post (after the cute pictures) that the MZ rate is less than 1%.

Thus the 2 in 3 solution effectively stands.

 

The answer that was given with the Monty Hall problem and now with his sheep twin problem are both wrong according to all the maths taught at school. My son had this type of problem last year in grade 8 and I remember them from a very long time ago! Once one option is known, it no longer plays a role in the odds of future choices because it is a known result and no longer a chance.

So once you state what one lamb is sex wise, it becomes eliminated from the odds and we are left with simply whether the second is male or female, in other words fifty : fifty or one in two.
With the door problem, identifying one failed choice certainly does not affect the remaining two choices, which are then one in two, not two in three, since one door has been eliminated. If you would like, I can explained why the logic given in 2013 to explain the answer you gave was flawed.

I will say that recently a local radio station (in Brisbane Australia) gave a very similar quiz and quoted a youtube mathematician as proof of the alternative answer, at the same time stating that every maths teacher at a local school they contacted told them they were wrong. So maybe mathematics , a universal constant, has changed in some way in recent times! Not likely.

This type of question was common in high school exams 40 years ago and still is today. Most people got it wrong then and most still do today, the difference today is that those that are wrong can maintain they are correct on the internet and then they become magically correct through popular belief ;-)

 

Perhaps a hands-on simulation will help. Go to https://betterexplained.com/articles/understanding-the-monty-hall-problem/

 

I found your posting very interesting and I have long thought that knowing one option simply rules it out of the equation; thus 50/50 (in the cases discussed and leaving out the identical versus fraternal question) remains (to me) the only viable answer.

Being somewhat aged, I have come across the same or similar many times (my favourite being the 2 Door puzzle. One leading to Doom the other Freedom; Each door guarded by a sentry. One always lies the other tells the truth and you know not which. You are allowed one question to the sentry of choice and from the answer can choose the door to Freedom. But that puzzle - capable of at least two answers - does not challenge mathematical principles, nor rely on somewhat doubtful statistical reasoning whether or not supported by YouTube cognoscenti!

Edit: after posting I chanced on the link posted by MeganN and remain unmoved having seen the same reasonings many times before. Of course I could be wrong..it has happened before

 

Well at least I got that bit of my initial answer correct! (#2)

 

I believe you are confusing two different concepts. It is true that if a fair (repeat fair) coin is tossed 10 times and always comes up heads then this does not affect the chances of the next toss, which is 50:50. However the two problems discussed here are different. They are both cunningly constructed to leave a choice of two and one's first reaction is that it must be 50:50 but I always advise when thinking about chance to play around with some actual numbers. If you do this in both problems the correct answer becomes clear. I strongly recommend "Reckoning with risk" by Gerd Gigerenzer who explains it much better than me.

 

I first saw this puzzle in New Scientist when it was originally published a few weeks ago but I have not been part of this discussion up to now for lack of a working computer, as explained in a new discussion elsewhere on this forum.

Initially, I reasoned along similar lines to Helen7 in post #1 but then thought that there was possibly more to be considered and decided that Pauline in post #2 was more correct.

After the genetic test result is known, I think that everyone believes that the possibility FF becomes invalid and needs to be discarded. However, perhaps one of the MF and FM possibilities should be discarded at the same time, leaving the odds as evens (sorry, I should reword that to be 50% or 1/2).

FF is not discarded just because there is no male in the combination but rather because neither gender of the lambs is male. A subtle point but possibly better understood by considering the gender of each lamb individually. The first letter indicates the gender of lamb #1 and the second letter indicates the gender of lamb #2. If the first lamb is the one that was shown to be male then FM should also be discarded. If the second lamb is the one that is male then MF needs to be discarded along with FF. In either case, that leaves just two possibilities with equal probabilities. Hence I believe that the answer should be 1/2 and not 2/3.

 

It does say bloodstream... now this is where I am overthinking things. But a blood test could show y dna from a previous pregnancy, plasma on the other hand would be from the pregnancy - just a thought...

 

I fear this puzzle is generating some woolly thinking.

 

I don't see how you can discard either FM or MF as you don't know which twin gave rise to the fragments of Y-DNA in the mother's blood.

 

All the sheep carrying twins can be divided into 4 groups, MM MF FM and FF. If the chance of gender is equal and ignoring the biological diversions then each group should be of equal size (subject only to random variation). The sheep in question has been shown not to be in group FF, so must therefore be in one of the other 3 groups. 2 out of these 3 contain a F.
Gigerenzer says always work with natural frequencies rather than probabilities.

 

But that is exactly the problem with your analysis. Why have you eliminated only FF? One of the MF/FM possibilities should also be eliminated, depending on which of the two lambs is assumed to be male. We don't know which of the 2 lambs is male so have to consider each possibility in turn. If the first lamb is male then there are only 2 possible outcomes, MM and MF. If the second lamb is male then there are also only two possible outcomes, MM and FM. Hence the probability of one lamb being female is 2/4, ie 1/2.

Any Monte Carlo simulation that you conduct will only reinforce your incorrect assumption and lead you to believe that the probability is 2/3. Using my extended analysis, a Monte Carlo simulation will indicate a probability of 1/2. It all depends on the conditions defined for the simulation.

That is why one has to consider all valid possibilities and ignore any impossible situations when calculating the probabilities.