# The Dog Dilemma

1. Jun 15, 2009

### bucher

This is a probability problem that I thought up one day after being inspired by the Boy or Girl Dilemma I read in the news (a while back).

Say a breeder has five dogs. A customer shows up one day and wants to know about the dogs. The customer is interested in a female dog and so asks the breeder if there is at least one female dog in the litter of five. The breeder says, "Yes".

After this, the customer leaves the store and goes to another one. There are five dogs at that store as well. The customer asks the breeder if the oldest dog in the litter of five is female. The breeder says, "Yes".

What is the probability of the first breeder having all females?
What is the probability of the second breeder having all females?

Since this is a probability question, I'll tell you the answers and you can determine why.

The probability of the first breeder having all females is 1/31

The probability of the second breeder having all females is 1/16

2. Jun 15, 2009

### Tibarn

It's just an application of Bayes' Theorem:

$$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$

First breeder: Let A denote "all females" and B denote "at least one is female". Clearly, P(B|A) = 1 and P(A) = 1/32. Now, the probability that at least one is female is one minus the probability that all are males, so it comes out to 1 - 1/32 = 31/32.

By Bayes' Theorem, P(A|B) = 1/31

Second Breeder: Similar thing, except that this time we let B denote "the first one is female", so in this case P(B) = 0.5. Again using Bayes' Theorem, we find that P(A|B) = 1/16

3. Jun 16, 2009

### HallsofIvy

Staff Emeritus
Writing "M" for male and "F" for female, there are 25= 32 ways to write 5 letters, each being either an "M" or an "F". Of those 32, exactly one is all "M"s and exactly one is all "F"s.

Knowing that there is at least one female drops "MMMMM" leaving 31 possibilities. "FFFFF" is one of those: probability 1/31.

If we write the dogs in order of age, then knowing that the oldest dog is female means that we are looking at lists of 5 letters, from "M" or "F", with the first letter being "F". Obviously in all possible lists of 5 "M"s or "F"s, exactly half, 16 start with "F" and half with "M" so knowing that "the oldest dog is a female" throws out half the possible lists, leaving 16. one of those is "FFFFF" so the probability of "all females" is now 1/16.

There is no "dilemma" here. "The oldest dog is female" gives you more information than "at least one of the dogs is female".

4. Jun 16, 2009

### fleem

First breeder:

Since its given that there is at least one female, the question, "what is the probability that all five are female when one is female" means the same thing as the question, "What is the probability that the remaining four are female". The probability that the remaining four are female is 1/16, and thus the probability that all five are female (given that one is female) is 1/16.

Second breeder:

The temptation here is to assume there might have been intelligent pre-selection of dogs in the group, by the store owner. However the OP makes it pretty clear, I think, that there was no such pre-selection, and that these are ALL the dogs from a given litter. So the sex of a randomly selected dog does NOT change the probability of the sex of another dog in the litter. Each is a separate coin toss. So the fact that the oldest is female works the same as "there is at least one female". So like with the first breeder case, the probability of the remaining four all being female is 1/16.

5. Jun 16, 2009

Fleem, the answer to the "first breeder problem" is indeed 1/31, with precisely the explanation HallsofIvy gave. You can also do this:

\begin{align*} \Pr(X = 5 | X \ge 1) & = \frac{\Pr((X =5) \cap (X \ge 1))}{\Pr(X \ge 1)} \\ & = \frac{\Pr(X = 5)}{\Pr(X \ge 1)} \\ & = \frac{\dfrac{1}{32}}{\dfrac{31}{32}} = \frac 1 {31} \end{align*}

6. Jun 16, 2009

### fleem

Thanks, statdad, I stand corrected. I should have taken a closer look at tibarn's and HallsofIvy's posts.

7. Jun 16, 2009