Introduction to the problem

A set $C$ contains $c\in \mathbb{N}$ elements of three different kinds: There are $\alpha\in\mathbb{N}$ elements of kind $A$, $\beta\in\mathbb{N}$ elements of kind $B$, and $\gamma\in\mathbb{N}$ elements of kind $G$, and $|C|=c=\alpha+\beta+\gamma$. We suppose $\alpha,\beta,\gamma>0$ and we perform $n>0$ independent trials (i.e. with replacement) of one element at a time from $C$.

We define the event $\mathscr{L}_n^{\alpha_A}$ as "to get, in $n$ independent trials, at least one of the $\alpha$ elements of kind $A$", so that $P(\mathscr{L}_n^{\alpha_A})=1-\left(\frac{c-\alpha}{c}\right)^n$. Similarly, we define the events $\mathscr{L}_n^{\beta_B}$ and $\mathscr{L}_n^{\gamma_G}$, such that $P(\mathscr{L}_n^{\beta_B})=1-\left(\frac{c-\beta}{c}\right)^n$ and $P(\mathscr{L}_n^{\gamma_G})=1-\left(\frac{c-\gamma}{c}\right)^n$.

We finally define the event-intersection $\mathscr{I}_{n}^{C}=\mathscr{L}_{n}^{\alpha_A}\cap \mathscr{L}_{n}^{\beta_B}\cap \mathscr{L}_{n}^{\gamma_G}$.

By means of the principle of inclusion-exclusion, and observing that $P(\mathscr{L}_{n}^{\alpha_A}\cup \mathscr{L}_{n}^{\beta_B}\cup \mathscr{L}_{n}^{\gamma_G})=1$, we easily find that

$$P(\mathscr{I}_{n}^{C})=1-P(\mathscr{L}_{n}^{\alpha_A})-P(\mathscr{L}_{n}^{\beta_B})-P( \mathscr{L}_{n}^{\gamma_G})+P(\mathscr{L}_{n}^{\alpha_A}\cap\mathscr{L}_{n}^{\alpha_B})+P(\mathscr{L}_{n}^{\alpha_A}\cap\mathscr{L}_{n}^{\alpha_G})+P(\mathscr{L}_{n}^{\alpha_B}\cap\mathscr{L}_{n}^{\alpha_G}).$$

We can also explicit this expression in terms of the variables $\alpha,\beta,\gamma$, obtaining (with the help of Bayes' theorem),

$$ P(\mathscr{I}_{n}^{C})=1-\left(\frac{\alpha+\gamma}{c}\right)^n-\left(\frac{\beta+\gamma}{c}\right)^n-\left(\frac{\alpha+\beta}{c}\right)^n+\left(\frac{\alpha}{c}\right)^n+\left(\frac{\beta}{c}\right)^n+\left(\frac{\gamma}{c}\right)^n.$$

PROPERTY K: The probability of the event $\mathscr{L}_{n}^{\alpha_A}\cap \mathscr{L}_{n}^{\beta_B}\cap \mathscr{L}_{n}^{\gamma_G}$ is invariant by switching the number of elements of any two given kinds, i.e. performing the substitutions $\alpha\leftrightarrow \beta$, $\alpha\leftrightarrow \gamma$, and $\beta\leftrightarrow \gamma$.

The proof of this property (which is ultimately based on the symmetry of the binary operator $\cap$ and on the fact that $\mathscr{I}_{n}^{C}$ involves all the kinds of elements present in $C$) is trivial.

Let's make an example to understand how it applies: If we switch the $\alpha$ elements of kind $A$ with the $\gamma$ elements of kind $G$, we obtain a new set $C^*$, which contains $\gamma\in\mathbb{N}$ elements of kind $A$, $\beta\in\mathbb{N}$ elements of kind $B$, and $\alpha\in\mathbb{N}$ elements of kind $G$, and $|C^*|=c=\alpha+\beta+\gamma$. However, the probability of the event $\mathscr{I}_{n}^{C^*}=\mathscr{L}_{n}^{\gamma_A}\cap \mathscr{L}_{n}^{\beta_B}\cap \mathscr{L}_{n}^{\alpha_G}$ is still

$$ P(\mathscr{I}_{n}^{C^*})=P(\mathscr{L}_{n}^{\gamma_A}\cap \mathscr{L}_{n}^{\beta_B}\cap \mathscr{L}_{n}^{\alpha_G})=P(\mathscr{L}_{n}^{\alpha_A}\cap \mathscr{L}_{n}^{\beta_B}\cap \mathscr{L}_{n}^{\gamma_G})=P(\mathscr{I}_{n}^{C}). $$

In fact, the chance to get at least one element of each kind, in $n$ independent trials, does not depend on which elements are of a certain kind, but only on how many elements of the three different kinds are present in $C$, independently on the labels we use to distinguish the $c$ elements in the three different kinds.

A peculiar Urn

We study the Property K in case of an urn $C^\dagger$ such that, in $n$ independent trials, the probability to get at least one element of kind $A$ is equal to the probability not to get any element of kind $B$. In this urn, it must hold the constraint

$$ P(\mathscr{L}_n^{A})=P(\overline{\mathscr{L}_n^{B}}). $$

Clearly, this request will result in a particular choice of $\alpha,\beta,\gamma$, and $n$. In fact, if we explicit the probabilities of the events $\mathscr{L}_n^{A}$ and $\overline{\mathscr{L}_n^{B}}$, we find that the above constraint corresponds to the relation

$$ 1-\left(\frac{c-\alpha}{c}\right)^n=\left(\frac{c-\beta}{c}\right)^n, $$ which can be written $$ 1-\left(\frac{\beta+\gamma}{c}\right)^n=\left(\frac{\alpha+\gamma}{c}\right)^n $$ and $$ 1-\left(\frac{\alpha+\gamma}{c}\right)^n-\left(\frac{\beta+\gamma}{c}\right)^n=0. $$ If we substitute the constraint in the general expression of $P(\mathscr{I}_{n}^{C})$, we find

$$ P(\mathscr{I}_{n}^{C^\dagger})=-P( \mathscr{L}_{n}^{\gamma_G})+P(\mathscr{L}_{n}^{\alpha_A}\cap\mathscr{L}_{n}^{\alpha_B})+P(\mathscr{L}_{n}^{\alpha_A}\cap\mathscr{L}_{n}^{\alpha_G})+P(\mathscr{L}_{n}^{\alpha_B}\cap\mathscr{L}_{n}^{\alpha_G}), $$

and, in terms of the variables $\alpha,\beta,\gamma$,

$$ P(\mathscr{I}_{n}^{C^\dagger})=-\left(\frac{\alpha+\beta}{c}\right)^n+\left(\frac{\alpha}{c}\right)^n+\left(\frac{\beta}{c}\right)^n+\left(\frac{\gamma}{c}\right)^n. $$

My question is:

Does the Property K hold in the urn $C^\dagger$?

From one side I would say "no", because if we perform the substitution $\alpha\leftrightarrow\gamma$ in the expression of $P(\mathscr{I}_{n}^{C^\dagger})$ we do not obtain the same probability (unless $\alpha=\gamma$), although the meaning of the event $\mathscr{I}_{n}^{C^\dagger}$ is still the intersection of the three events $\mathscr{L}_{n}^{\alpha_A}$, $\mathscr{L}_{n}^{\beta_B}$, $\mathscr{L}_{n}^{\gamma_G}$. A similar argument applies if we perform the substitution $\beta\leftrightarrow\gamma$ (which would imply that the Property K holds only if $\beta=\gamma$). Since both these substitutions should not modify the probability of the event-intersection, it would follow that it must be $\alpha=\beta=\gamma$, a condition that is inconsistent with the constraint, unless $\alpha=\beta=\gamma=0$.

On the other hand, if we first perform the substitution $\alpha\leftrightarrow\gamma$ in the general expression of $P(\mathscr{I}_{n}^{C})$ and then we impose the constraint, we still have that $P(\mathscr{I}_{n}^{C^\dagger})$ is invariant by performing any switch of the variables, i.e. the Property K seems to hold.

This post is related to these other posts A weird problem of probability!, A problem of conditional probability, An urn with three kinds of balls... and a weird constraint!

  • $\begingroup$ The validity of the Property K cannot depend on the fact that one first decides which elements are of a certain kind and then he applies a constraint to their numbers, or vice versa. Therefore, in order to keep the Property K also in the peculiar urn $C^\dagger$, in relation with the $n$ trials which satisfy the constraint, it must be that the probability of the intersection is $0$, and $n\leq 2$: In this case, the probability of the event-intersection is the same for any $\alpha,\beta,\gamma$ and it is unaffected by the switching of any pair of numbers of elements. But is this rigorous? $\endgroup$
    – user559615
    Jun 24, 2018 at 3:31
  • $\begingroup$ Does it make any sense to switch the numbers of kinds of elements before having decided which ones are of a certain kind? Is there a way to make the previous question more rigorous? $\endgroup$
    – user559615
    Jun 24, 2018 at 3:41

1 Answer 1


Since Property K holds in the case of any general urn $C$ (as you've demonstrated), it must hold in the case of $C^\dagger$, since $C^\dagger$ is just a case of $C$ satisfying one extra constraint. So it remains to explain why it seems like Property K does not hold.

This can be explained as follows: upon performing a substitution, the new urn ${C^\dagger}^*$ no longer satisfies the given constraint! Thus, the formula $$ P(\mathscr{I}_{n}^{C^\dagger})=-\left(\frac{\alpha+\beta}{c}\right)^n+\left(\frac{\alpha}{c}\right)^n+\left(\frac{\beta}{c}\right)^n+\left(\frac{\gamma}{c}\right)^n. $$ which was derived based on the constraint does not remain valid upon switching $\alpha$ and $\gamma$, i.e.,

$$ P(\mathscr{I}_{n}^{{C^\dagger}^*})\neq-\left(\frac{\gamma+\beta}{c}\right)^n+\left(\frac{\gamma}{c}\right)^n+\left(\frac{\beta}{c}\right)^n+\left(\frac{\alpha}{c}\right)^n. $$

  • $\begingroup$ Thanks!!! But then, what is the value of $P(\mathscr{I}_{n}^{{C^\dagger}^*})$? Or, better, the value is for sure $P(\mathscr{I}_{n}^{{C^\dagger}^*})=P(\mathscr{I}_{n}^{{C^\dagger}})$, but what is its expression? $\endgroup$
    – user559615
    Jun 28, 2018 at 14:25
  • $\begingroup$ My impression is that the constraint is not compatible with the event $\mathscr{I}_{n}^{{C^\dagger}}$, but I cannot realize if this is true and, in case, why. $\endgroup$
    – user559615
    Jun 28, 2018 at 14:29
  • $\begingroup$ @andrea.prunotto You can certainly still use the general expression you derived above. You may also notice that the new urn ${C^\dagger}^*$ satisfies a new constraint that the probability to get at least one $G$ equals the probability of no $B$, and derive an expression using that new constraint. $\endgroup$
    – BallBoy
    Jun 28, 2018 at 14:32
  • $\begingroup$ True. Let's say: The constraint is clearly not invariant by switching $\alpha$ and/or $\beta$ with $\gamma$, whereas the event $\mathscr{I}_{n}^{C}$ is. This is the point I cannot get. Can you help me to clarify this? $\endgroup$
    – user559615
    Jun 28, 2018 at 14:35
  • $\begingroup$ You may be interested also in this post math.stackexchange.com/q/2833367/559615, which is strongly related to this problem. But... unfortunately, no more +50 : ) $\endgroup$
    – user559615
    Jun 28, 2018 at 14:42

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