Nancy figures this one out quite quickly, as she's very mathematical.

The odds are 75% (3/4) that the other puppy is male. 

Puppy Puzzle Solution 1 (From Brian Lowe)

Puppies are born in order A B C

A and B are male (you said)

C can be male or female

 

Therefore you have: 

 

    M M M   or   M M F

If the first option happens, you can be looking at:

 

    AB  or  AC  or BC

and ditto if the second option happens.   So.... there are 6 possible solutions

Because the two you looked at are M M you can eliminate AC and BC where C was female.

Therefore you have 4 possibilities left:

AB , AC, BC from the 3rd puppy is a male option
and AB from the 3rd puppy is a female option

Therefore 75% chance male and 25% chance female

 

Puppy Puzzle Solution 2 (from Andreas Huster)

Similar to Brian's solution, but in table form.  There are the equally likely cases:

3rd pup is actually | Nancy sees | Other is | Case is consistent with
                    |            |          | Nancy's observation

       male         |  1 and 2   |   male   |        yes

       male         |  1 and 3   |   male   |        yes

       male         |  2 and 3   |   male   |        yes

       female       |  1 and 2   |  female  |        yes

       female       |  1 and 3   |   male   |        no

       female       |  2 and 3   |   male   |        no

So, of the four cases consistent with Nancy's observation, three imply that the other pup is male.

Puppy Puzzle Solution 3 (also from Andreas)

Solve using conditional probability (Baysian Statistics / Bays Law)

 

Let A be the event that all three puppies are male.

Let B be the event that Andy sees two male puppies.

Then we are looking for the probability of A given that B is true:  P(A|B)

 

Bay's Law:  P(A|B) = P(B|A) * P(A) / P(B) 

 

P(B|A) = 1

P(A) = 1/2

P(B) = 2/3

 

So, P(A|B) = 1 * 1/2  /  2/3 = 3/4

 

Puppy Puzzle Solution 4 (from Richard)

The following problem generalizes the puppy problem in a nice way.

Consider three different die that roll either an M or an F, with probability of M being d1, d2 and d3 respectively. We roll all three die, and look at two of them (but we don't know which two) and see two Ms. What is the probability that all three are Ms? 

There are eight cases: MMM, MMF, MFM, MFF, FMM, FMF, FFM, FFF with probabilities of each being a product of individual probabilities.

combo| prob

-----+----------------

 MMM | (d1)(d2)(d3)

 MMF | (d1)(d2)(1-d3)

 MFM | (d1)(1-d2)(d3)

 MFF | (d1)(1-d2)(1-d3)

 FMM | (1-d1)(d2)(d3)

 FMF | (1-d1)(d2)(1-d3)

 FFM | (1-d1)(1-d2)(d3)

 FFF | (1-d1)(1-d2)(1-d3)

From each combo, how many ways are there to choose two Ms? For the first case, there are three; for the second, third, and fifth cases, there is one each; for the remainder, it's impossible. So the odds seeing MM after rolling all three die and looking at two randomly selected die are 1/3 (3(d1)(d2)(d3) + (d1)(d2)(1-d3) + (d1)(1-d2)(d3) + (1-d1)(d2)(d3))

We KNOW this happened, so we use the conditional probability formula, where P(A|B) = P(A and B)/P(B), where A = "all three die are M" and B = "we saw two Ms from two randomly chosen die". [Remember, P(A|B) is probability that A is true given we know B is true.]

So P(all three M | saw two M) = 
     P(all three M and we saw two) / P(saw two M) =
(d1)(d2)(d3)/[1/3((3(d1)(d2)(d3)+(d1)(d2)(1-d3)+(d1)(1-d2)(d3)+(1-d1)(d2)(d3))]

This collapses to the generalized puppy problem with d1 = 1, d2 = 1 (since we KNOW 100% that they are Ms, they are equivalent to die fixed to always roll "M") and d3 = M (this M is the probability that the third pup born is male, given no other information). This simplifies to

3M / (3M + 1 - M) = 3M / (2M+1)

Let f(M) = 3M/(2M+1). This populates the table nicely:

  M  | f(M)

-----+-------

  0  |  0

 1/2 | 3/4

  1  |  1

Works well for extreme values M = 0 (if the last pup is male with probability 0, we get probability 0 that all three are male) and M = 1 (the last pup certainly is male, yielding probability 1 that all three are male), and agrees with your calculation f(1/2) = 3/4.  

 

Puppy Puzzle Solution 5 -- Showing that 5/6 is NOT the Solution
(also from Richard)

> Solution 2
> He saw either A+B, A+C or B+C (1/3 chance of each). In 2 of those cases, we
> know that the unknown pup is male. In the third case, he saw the two that we
> already knew were male, so the unknown one is still completely unknown and
> 50:50 either sex. That gives us 1/3 + 1/3 + (1/2 * 1/3) = 5/6 chance that
> it's male.

Let's start by showing that 5/6 is incorrect.  Let M = probability that the last pup is male out of the womb, with no additional information. Implicit in the problem is M=1/2, but it's instructive to see what happens if we vary this.

Let g(M) = probability that all pups are male using the reasoning of the Solution 2 in your e-mail. Applying this (flawed) reasoning, we see that

g(M) = 2/3 + (1/3)M

Validate this for yourself, following his reasoning, and seeing that it coincides with his answer for M=1/2. Note the following table:

  M  | g(M)

-----+-------

  0  | 2/3

 1/2 | 5/6

  1  |  1

In other words, according to this reasoning, if the probability that the third pup is born male is 0, there's still a 2/3 chance that all three pups are male! Shhh-yeah! The other user-submitted solution yields the same results. The error is the assumption that each pup -- the first, the second or the third -- are equally likely to have not yet been sexed -- which doesn't take into account conditional probability.

 

Easy, right?

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