A puzzle from The Guardian:
A 17th century farmer observes that one of his sheep is pregnant. As all famers know, lambs arrive as non-identical twins, each with a 50-50 chance of being male or female. The local vet has an Elizabethan ultrasound machine and finds out the genders of the lambs: “Is it true that at least one of them will be male?” asks the farmer. “Yes, it is true” replies the vet.
“In that case,” the farmer says, “the other one will most likely be female”. Is the farmer correct?
Solution Yes!
There is a 2/3 chance that one of the lambs will be female. If we know that at least one lamb is male, then the possible pairings of the first and second lamb are male-male, male-female and female-male, and each of these pairings is equally likely. There will be a female in two of the three scenarios, hence the 2/3 probability.
Did you solve it? Art thou smarter than Shakespeare? | Mathematics | The Guardian
I can see the confusion here. When the farmer says "the other one will most likely be female", we may model that situation as the following:
As one of the lambs are male, take this one, call it lamb 1. What is the probability that the "other" lamb, call it lamb 2, is female?
Also consider this variation:
We have 100 lambs. The doctor now says, "Lambs 1 and 3-100 are male". What you don't know is that the doctor is assigning the numbers after the fact -- if there were 50 lambs he would have said "Lambs 1 and 3-51 are male", strategically leaving out 2 and filling in the rest as male. What is the probability that "lamb 2" is female?
In either case, it really depends on what we are referring to when we discuss "lamb 2", i.e. the farmer's "other lamb". Suppose the doctor knows the genders of both lambs. Saying "the other one" implies the existence of the "first one". Suppose that the doctor assigns "first one" to the first male lamb he has scanned -- he scans them in order. Say that the lambs also are born with unique genetic codes derived deterministically from the mother, regardless of whether they are male or female: call them lamb A and lamb B. As it happens, lamb B was the first to be scanned and was male, hence this is now referred to as "the first one". Therefore, lamb A is now the "other one". What is the probability that the "other one", lamb A, is female? Of course, the probability that lamb A is female is 50%.
So it really depends on what the farmer means by "the other one". What happens when there is no genetic code invariant across the different possibilities, with which we could interpret an association with the farmer's "other one"? If the farmer's emphasis is on the "other" part of "other one", then perhaps 2/3 is correct. If the farmer's emphasis is on the "one" part of "other one"... then, it gets difficult. Identification across multiple realities requires something to identify by... if that doesn't exist, isn't the identification purely local to this particular reality? And what is the farmer's reality? It depends -- it could be M-M, M-F, or F-M, which then reduces down to the computation* below, assuming that the farmer chooses on a "first one" completely random 50-50 chance.
If we were to interpret this way: that "the other will be female" is "most likely" correct, i.e. correct in 2/3 of the circumstances, then his statement is true.
Or we could argue that the lamb is either a male or female, and isn't "most likely" anything, so he's correct in 2/3 cases and wrong in 1/3 but no one would take this interpretation...
We should think about what went through the farmer's head when he said that. How can he choose which one to make the "other one"?
Maybe like this:
1. There are two lambs.
2. As we now know one of these two lambs is male, select one male lamb from the set... randomly.
3. Label this select lamb "first". Label the other, "other".
4. The probability that "other lamb" is female is 2/3.
I don't really like this because I know for sure the farmer wasn't really thinking about all this, and I'm sure the farmer didn't consciously make a random choice. Though what did happen is that he did "select out" a male. But... how??
Take this example:
There's 16 candy bars. As it happens, by weight I can tell there's at least 15 Hershey bars in there. You say, "Ah, the other one's most likely a Dark Chocolate Hershey Bar, isn't it?" As it so happens all 16 are Hershey Bars, and there are 2 Dark Chocolate Hershey Bars. So... what?
What if it's like this: with 50-50 chance it's either Hershey or a Butterfinger. And if it is a Hershey, it could either be a Dark Chocolate Hershey bar or a Milk Chocolate one, with 50-50 chance again.
What if I told you there's at least 2 Hershey bars in there? What do we mean by "the other ones"? Especially if there were actually 4? By "selecting out" 13, aren't we selecting out 2 Hershey bars as well?
How can we make using "the other one(s)" totally nonsensical? Suppose Hershey bars come with a code. Your assorted bag of candy, the probability of the ratio of Hershey bars to Butterfingers is actually determined by the included codes written on the Hershey bars. Say that for bags containing only a Hershey bar with the code "A", there's only Butterfingers for the rest of them, and for bags containing Hershey bar "B", there's Snickers. Only thing is: there's always a code "A" Hershey bar.
Given that there is at least 1 Hershey bar in this bag, how do we interpret the following statement: "the other candy bars are probably Butterfingers"? What if we "choose" Hershey bar A as the "chosen" bar, and the rest as the "other ones"? What if we "choose" Hershey bar B, then we would be wrong..?
...Is there any scenario where we could "get them" for using "other ones" without thought? Where the truth of an "other ones" statement depends on the choice of said "other ones", where the context is a quantitative problem?
*Computation: in the M-M case, the farmer might have chosen either of the two males as a "first one", hence we have M-M-1 and M-M-2 each with probability 1/6. In the M-F case, we have M-F-1 with probability 1/3, and in the F-M case, we have F-M-2 with 1/3. The "other one" is F with probability 1/3 + 1/3 = 2/3, and M with probability 1/3.
