# Counterexample against binomial assumptions

The question is from a Master-level Probability Course.

It is well known that the underlying assumption for the binomial distribution is that there are n independent Bernoulli trials. More specifically, the assumptions are:

(1) The number of trials, $n$, is fixed.

(2) There are two and only two outcomes, labelled as "success" and "failure". The probability of outcome "success" is the same across the n trials.

(3) The trials are independent. That is, the outcome of one trial doesn't affect that of the others.

My question is, are there any counterexamples which just violate one of those three assumptions? Particularly, are there cases where Assumption (2) holds but (3) doesn't, or vice versa?

• It looks more like you're after generalisations of the binomial. Sep 27 '14 at 4:00
• This sounds like routine bookwork. Is it for some subject? Sep 27 '14 at 6:52
• Consider an experiment with $n=2$ tosses of a "fair" coin in which 3) fails miserably in the sense that the outcome of the second trial is guaranteed to be the complement of the first. Thus, there are only two (equally likely) outcomes in the sample space: $\Omega = \{HT, TH\}$. Obviously, the probability of Heads on the first trial is the same as the probability of Heads on the second trial. Thus, 1) and 2) hold but 3) does not. Of course, the distribution of the number of Heads on the two trials is not a binomial distribution. Oh wait, you want a binomial distribution but 3) to not hold? Sep 27 '14 at 14:21
• Thanks Dilip. Your example makes sense to me. Of course we should not expect a binomial distribution when 3) doesn't hold. Additionally, any clues for (3) holds but (2) doesn't? Sep 27 '14 at 14:37
• (3) and (1) hold but (2) doesn't when you're flipping a coin that can land on its edge.
– whuber
Feb 26 '17 at 20:05

Breaking each one, singly:

(1) The number of trials, $n$, is fixed.

The experiment continues until $k$ successes are observed. Or until $m$ successes in a row. Or until the number of successes exceeds the number of failures by 2.

(2) There are two and only two outcomes, labelled as "success" and "failure". The probability of outcome "success" is the same across the n trials.

P(success) is drawn from a beta distribution with mean $p$. Or P(success) alternates between $p_\text{A}$ and $p_\text{B}$.

(3) The trials are independent. That is, the outcome of one trial doesn't affect that of the others.

P(Success|Success at previous trial) = $p_1$ and
P(Success|Failure at previous trial) = $p_2$

You suggested something like an urn model as a concrete example, and it's quite easy to construct several forms of urn model of this third case (if you use sampling with replacement) - or you could use dice if there's more than one die you could use.

• I can understand here. But I think my question is, how could I make up concrete examples, say drawing balls from a urn, to illustrate how (3) could be violated but (2) holds, and vice versa? Sep 27 '14 at 14:24
• I mean the example like @DilipSarwate gave. Sep 27 '14 at 15:30
• Sorry for any confusion. But this is basically a question from Intro Probability course. There is no typical subject associated with. The example could be either "tossing a coin" or "drawing balls from a urn", you know, basic example to illustrate probability question. I am not quite exactly sure what you mean by asking whether this question is for any subject. But if your subject means "course", yes, it's for a probability course. Otherwise, no, just general probability question. Does that make sense? Sep 27 '14 at 15:52
• Thanks - I'm using the word subject synonymously with your use of course. I asked because when you were asking for concrete examples, your question then seemed to have the character of routine bookwork. Please add the self-study tag to the tags on your question and read its tag wiki, then modify your question as necessary to follow its guidelines. I'll then be able to be somewhat more responsive. Sep 27 '14 at 21:48
• Thanks for the comment, @Glen_b . This was my first time to ask a question here, so that I might not have experience to follow the rules of how to ask a question the right way. Anyhow, I added the tag, and a sentence describing the origin of the question. Sep 27 '14 at 22:49

Taking your example of picking balls from an urn. Assume you have M balls in the urn, with some black and some red. You pick N balls (with N a fixed number) and count up the number B which are black. B is not binomially distributed, because the probability of getting a black ball at each pick depends on how many black balls you've already picked.

• I don't agree here. If you pick with replacement, the probability is still the same. What's your point of "counting up the number B"? Do you mean stop picking until B blacks? But you pick N balls anyway, right? I cannot see how counting the number of blacks affects the probability, given you have fixed total pick number N. Sep 28 '14 at 15:45
• I mean the everyday sense of "pick X balls from an urn". If you ask someone to take some balls out of the urn, they wouldn't put each ball back in before taking the next one out. Sep 28 '14 at 16:24
• Thanks, I think hypergeometric process is what you refer to. And that makes perfect sense now. I posted an answer based on your comment here. Sep 28 '14 at 17:10

Correct me if I am wrong here.

a) A counterexample only violates Assumption (1) would be a negative binomial process, where the number of trial is the random variable; The situation where a variable from a geometric distribution also works.

b) A counterexample only violates Assumption (3) would be a hypergeometric process. I came up with idea based on @Hong_Ooi's comment. The assumption (2) holds in a hypergeometric process setting is not that obvious at first glance though.

For example, suppose we have 10 balls, 6 blacks and 4 reds, in the urn. Suppose we draw 3 balls (fixed trials, assumption (1) holds) without replacement from the urn, and we are interested in the probability of 2 black balls out of the 3. The reason why assumption (2) holds is as follows.

Pr(1st draw is a black) = 6/10

Pr(2nd draw is a black) = Pr(BB) + Pr(RB) = 6/10 * 5/9 + 4/10 * 6/9 = 54/90 = 6/10

Pr(3rd draw is a black) = Pr(BBB) + Pr(BRB) + Pr(RBB) + Pr(RRB) = 6/10 *5/9 *4/8 + 6/10 *4/9 *5/8 + 4/10 *6/9 *5/8 + 4/10 * 3/9 *6/8 = 6/10

The same goes for the probability of "red" ball. That said, assumption (2) holds. But obviously assumption (3）doesn't. For example, Pr(2nd draw is a black, given 1st is a black) is not equal to Pr(1st draw is a black). Thus the independence assumption violates.

It is quite easy and many of the answers/comments are making this more complicated than it needs to be.

Suppose you are looking at someone taking freethrow shots. Assumption two is met because you can either miss or make the shot. But assumption three might or might not be met. If the person is shooting a bunch of shots at the same time, they will get tired, and so their success probability will decline as time goes on. But if we are taking a set of trials that occur on different days, then they are independent of one another.

Suppose we say that we will look at 10 shots, or 100 shots, or whatever. Then assumption one is met. But if we say we will look at however many shots occur in an hour, or a year, or whatever, then assumption one is violated because we don't have a fixed number of trials, so we can't use the binomial distribution to predict in advance our likelihood hitting at least a certain percentage of our shots. (If I generally have a 50% chance of making any individual shot, and I want to know how likely it is that I will hit at least 2/3 of my shots, I can't make that calculation without knowing how many shots I'll take-- if I only take 3 shots, I have a much better chance than if I'm taking 3000 shots).

1) From a healthcare perspective, consider a clinical trial. Let no. of deaths in a trial be assumed to be binomially distributed with p=prob of success/death and n=number of patients in that trial.

Above assumption will not hold if : a) Chances of death in one patient is different from another patient, happens when health conditions of patients is very varied (identical trials assumption violated) b) Disease is infectious, as death of a patient is effected by other patients' deaths. (independent trials assumption violated)

2) On a simpler note : Choosing Chinese cuisine for lunch isn't binomially distributed, because if you have it today, you won't have it tomorrow (independent trials assumption violated)

• In response to Dilip Sarwate's answer, in 2 tosses of a fair coin, the sample space contains not just HT & TH, but also TT and HH. Even if the coin is fair, probability of head is half doesn't mean that if you toss the coin 10 times, you'll surely end up with 5 heads. You'll get almost half number of heads when no. of trials is "sufficiently" large. Please see below image for details about law of large numbers : yunwah.files.wordpress.com/2009/10/… Sep 29 '17 at 17:02