$4$ dogs have been born in the same week. What is the probability that they were born on different days?

$4$ dogs have been born at a dog kennel in the same week. What is the probability that they were born on different days?

I did: $$\frac{^7C_4}{7^4}$$

But my book says the solution is: $\frac{120}{7^3}$

What did I do wrong?

EDIT: I copied the problem exactly as it is in my book. If it is missing information, poorly thought or doesn't make any sense, that's not my fault. Typos, mistakes and low quality abound in these schoolbooks.

• This is a specific case of the Birthday Problem. – Myridium Oct 19 '16 at 2:20
• Dogs are mostly multiple births; if a mere 4 dogs are born, they are probably all the same litter, and the odds against them being born on different days is very low. Looks like the book is very wrong. – Yakk Oct 19 '16 at 13:12
• Dog births at a dog kennel are not independent events, the question should clarify that, or you should note that as an assumption in your answer, since it appears you are supposed to make that assumption. – Brad Thomas Oct 19 '16 at 13:12
• What is really horrible about this question is not that it's missing information per se, but that it implicitly teaches students to be extremely hand-wavy when dealing with probabilities. It teaches them that you can ask about "probability" without specifying anything about a probability distribution, and that everything is uniform and IID. This leads to statements like "there is a 50:50 chance climate change is caused by humans". And, for everyone who says "assume uniform IID when nothing else is specified", this may work here, but check Bertrand's paradox. Rant over. – Sasho Nikolov Oct 19 '16 at 15:59
• Everyone worrying about multiple births is ignoring the KISS Principle. This is a math problem, not a veterinary longitudinal study requiring a lot of data points of litter size and birth timing. You must use only math and you do not have time for such research. Because of that, you may safely assume the births are independent without having to be told that. – Spencer Oct 20 '16 at 2:35

The first dog can be born on any day.

The second dog has probability 6/7 of being born on a different day.

The third dog has probability 5/7 of being born on a different day.

The fourth dog has probability 4/7 of being born on a different day.

$$\frac67\cdot\frac57\cdot\frac47 = \frac{120}{7^3}$$

• This answer fails to note the assumption of independent births, which in reality is a rather significant assumption to be making – Brad Thomas Oct 19 '16 at 13:18
• One of the most important things to check in relation to probability questions is independence, because yes, many real life and exam questions put that assumption to the test. Common sense is to check for that, not ignore it. When questions are framed in a real life (not abstract) scenario, common sense is to question how the assumptions we have to make actually apply to the real life scenario. – Brad Thomas Oct 19 '16 at 15:32
• I am dumbfounded this question has gotten so much attention. What happened? I've answered questions far more complicated than this, with far better answers than this, and don't often come up with any upvotes. – Kaynex Oct 19 '16 at 17:20
• @Kaynex HNQ. Have an upvote and enjoy the ride! – SQB Oct 19 '16 at 17:28
• It's a matter of answering a "Goldilocks question:" not too difficult that most can't understand it, not too easy that most would consider trivial. Just difficult enough that the answer is beyond their grasp, and just a little nudge gets them over the threshold. – Khashir Oct 20 '16 at 0:34

There is not enough information to solve this problem. If the dogs were from the same litter the probability is very low that they were born on different days. Normally you should be given some assumption like each day of the week is equally likely (reasonable) and independence (unreasonable - does the problem writer have any idea dogs are usually born into litters?).

In a Bayesian sense learning the four dogs were born in the same week will require updating the prior that their births are not correlated. Given the kennel is not that large.

In short, what a terrible word problem.

• Why do you assume that they are from the same litter? It seems pretty obvious to me how this question is meant, and your answer does little to help the OP. – xLeitix Oct 19 '16 at 6:07
• Also, we're not informed if one of the days had a leap second added (or less likely subtracted). If one of the days is longer or shorter than the others than that will obviously affect the probabilities, even if only by 1/ 86400. – Mike Scott Oct 19 '16 at 6:17
• @xLeitix It is only very obvious if you have already seen lots of other problems of the same kind. On the other hand, if you have experience with animals, it is very natural to assume they are born together, because dogs typically have 3-8 puppies at a time, not only one. – Federico Poloni Oct 19 '16 at 8:31
• @MikeScott I'd not care much about a leap second, but would take into account a possible $\pm1$ hour for a day light saving switch Sunday. – CiaPan Oct 19 '16 at 9:50
• The problem can be solved, but only if we make the (thoroughly unrealistic) assumption that the dog births are independent events. In short, I agree it is a poor question. However I think people who criticize this answer for NOT automatically making the "independent event" assumption would do well to recognize that highlighting that assumption is more rigorous. Because in the real world, we can't just answer questions like this while making unrealistic assumptions and get anywhere close to predictive power. – Brad Thomas Oct 19 '16 at 13:21

The denominator, $7^4$, counts the ways to select a day for each dog. So you must do the same in the numerator: count ways to select a day for each dog (although, distinct days).

You counted ways to select 4 distinct days for the dogs to be born.

However, there are $4!$ ways to assign the dogs to each of these days.

$$\dfrac{{^7\mathsf C_4}\cdot 4!}{7^4} = \dfrac{7!/3!}{7^4} = \dfrac {120}{7^3}$$

• Right, so my answer would be correct it order didn't matter, right? – SilenceOnTheWire Oct 19 '16 at 0:13
• Yes. You could also use $^7P_4$ to reflect the importance of the order.. – Kaynex Oct 19 '16 at 0:15
• No; @SilenceOnTheWire The denominator, $7^4$, counts the ways to select a day for each dog. So you must do the same in the numerator: count ways to select a day for each dog (although, distinct days). – Graham Kemp Oct 19 '16 at 0:17
• @SilenceOnTheWire Order doesn't matter in this question either- you're not given any ordering. In a way, if you include ordering in both the numerator and the denominator, they cancel each other out. So no, your answer wouldn't work, because order already doesn't matter. – Kevin Long Oct 19 '16 at 0:22
• ... seven distinguishable categories, and if these were all equally likely, then the probability that the items were assigned to different categories would be $\dfrac{{^7\mathsf C_4}}{{^{10}\mathsf C_4}}=\dfrac{1}{6}$. This calculation might be appropriate in a generalization of Einstein's non-interacting gas, where four indistinguishable, non-interacting bosons could occupy any of seven states of equal energy. It would not, however, be appropriate for dogs' birthdays. – Will Orrick Oct 19 '16 at 10:29

You're mistakenly giving an ordering when you use $7^4$. $7^4$ gives the number of $4$-tuples $(w,x,y,z)$ where $1\leq w,x,y,z\leq 7$ with each number representing a day. You get because there are $7$ choices for each place. But in the numerator, you have no ordering- instead, you're counting the number of subsets (and by definition, sets are unordered) of the set $\{1,2...7\}$ with cardinality $4$. As before, each number represents a day, but you're not assigning a day to each dog.

Each set of $4$ days, e.g. $\{1,2,3,4\}$, which we can think of as Sunday, Monday, Tuesday, and Wednesday, represents multiple assignments of days to dogs. The first dog could be born on any of those $4$ days, the second dog could be born on any of those $3$ remaining days, the third dog could be born on any of those $2$ remaining days, and the last dog has no choice. If we multiply the number of sets of size $4$ by $4!=24$, we get all of the permutations of days represented by those sets, since each one has $24$ permutations. This additional factor gives us the given answer.

• I like the point that this hits on, as I believe it's what the OP needs to understand to overcome this problem. But I think it's not clearly explained; if you can improve the clarity then I'll upvote. – Myridium Oct 19 '16 at 2:22
• @Myridium Thanks for the suggestion. Is this better? – Kevin Long Oct 19 '16 at 2:49
• I think it may be too complicated for the OP at this point. I like answers that are understandable on a first pass by dummies like myself. Nonetheless, I appreciate your effort and the focus of your answer, so you get my upvote :) – Myridium Oct 19 '16 at 2:55
• "You're mistakenly giving an ordering when you use $7^4$." Why "mistakenly"? The rest of your post suggests that it is $C(7,4)$ that is mistaken, not $7^4$. – Will Orrick Oct 19 '16 at 10:35
• @WillOrrick That's true, thanks. I thought about your other post, and then I realized that if the ordering cancelling out part was true, then we'd need $7^4$ to be divisible by $24$. I'd argue, however, that, by having one ordered and one not, you have a mistake, and that you can correct that by ordering the numerator, though I suppose intuitively that you should just have both unordered. Strangely enough, I thought that the number of unordered $7$-tuples of nonnegative integers whose sum is $4$ was $10 \choose 6$, but $\frac{7\choose 4}{10\choose 6}$ doesn't give me the same answer... – Kevin Long Oct 19 '16 at 14:14

Obviously, the eldest one has nothing to care about getting birth on a 'different' day, so he/she has $C(7,1)$ choices out of $7$ days of a week with the probability equal to $\frac{C(7,1)}{7}=1$.

• You should reconsider your answer. What you want to say? The question is asking something else. – Reeshabh Ranjan Oct 19 '16 at 15:52
• @Reeshabh Ranjan..I just stated the point where the OP has committed mistake. – Nitin Uniyal Oct 19 '16 at 17:01