If 5 people have different birthdays, what is the probability that none of them are born on adjacent months?

Question

Problem: If 5 people have different birthdays, what is the probability that none of them are born on adjacent months?

In this question consider December and January as adjacent.

I know that the probability will be no. of ways not on adjacent months over total, being 12P5.

But at this point I have no idea how to calculate all the scenarios in which the 5 aren’t born on adjacent months.

Can anyone help?

Answer 1

I assume each birthday month and each selection of five birthday months is equally likely.

If we arrange five squares into a circle to represent the five birthday months, and we place an x between all the squares to represent at least a single month with no birthday, we are left with two xs. The number of ways we can place those remaining two xs is:

$$5+4+3+2+1 = 15$$

Or, $5$ ways to place both xs between two squares plus $\binom{5}{2} = 10$ ways to place two xs into $5$ spaces. $$5+10 = 15$$

Now, if we take it that one of the square months is always January, it is pretty obvious that January occurs 15 times and because of symmetry, so does every other month. That means we have a total of $15\cdot 12 = 180$ month slots to fit into a series of $5$ month sequences. Hence, the number of ways you can arrange $5$ months such that no two months are adjacent is: $$\frac{180}{5}= 36$$

Hence the probability is:

$$p = \frac{36}{\binom{12}{5}} = \frac{1}{22}$$

Answer 2

Let $\Omega$ be the set of all tuples with different (see comment) entries $$ t = (t_1,t_2,t_3,t_4,t_5) $$ of integers modulo twelve, $t_1,t_2,t_3, t_4,t_5\in\{0,1,2,\dots,10,11\}$ (taken modulo $12$).

Let $A\subset \Omega$ be the subset of all tuples, such that $t_j\ne t_k\pm 1$ for any different indices $j,k$.

$\Omega$ has $12\cdot 11\cdot 10\cdot 9\cdot 8$ elements.

We count the elements in $A$. We let $t_1$ have an arbitrary value, fixed now. Because of the symmetry / of the action of $\Bbb Z/12$ on the set of values of the components of a $t\in A$, we may and do assume $t_1=0$. Then we have to choose the other components of $t$, $t_2,t_3,t_4,t_5$ in the interval $2,3,\dots,10$. We rearrange them, acting with the group of permutations with $4!$ elements, so that $t_2<t_3<t_4<t_5$. Consider the different (since $t\in A$) numbers $$ (*)\qquad\qquad 1\le t_2-1<t_3-2<t_4-3<t_5-4\le 6\ . $$ There are $\binom 64$ possible choices. So $A$ has $12\cdot 4!\binom 64=6!/(6-4)!$ elements.

If we consider that each month is equally probable as a birth month of the five people, then the probability is $$ \frac {12\cdot 6!/(6-4)!} {12\cdot 11!/(11-4)!} = \frac {6\cdot 5\cdot 4\cdot 3} {11\cdot 10\cdot 9\cdot 8} =\frac 1{22} $$ (Idea again: By symmetry we can fix the first component. The whole space has now (with this restriction) |Arrangements$(11,4)$| elements, the space of favorable events via $(*)$ similarly |Arrangements$(6,4)$| elements, the probability is the quotient.)

Computer check, here sage:

sage: ZZ(len(Arrangements([1..6], 4))) / ZZ(len(Arrangements([1..11], 4)))
1/22
sage: # explicitly
sage: Omega = Arrangements( Zmod(12), 5 )
sage: J = range(5)
sage: def f(a):
....:     return [ a[j]-a[k] for j in J for k in J ]
....: 
sage: A = [ a for a in Omega if 1 not in f(a) ]
sage: len(A), len(Omega), ZZ(len(A)) / ZZ(len(Omega))
(4320, 95040, 1/22)

So $A$ has 4320 elements, $\Omega$ has $95040$, and the probability is $$4320/95040=1/22\approx 0.045454545454545\dots\ .$$

---

If we are now more exact, consider each month with the weight corresponding to its days (and February has 28 days), then we have to program. (The symmetry is broken.) Each tuple $t=(t_1,t_2,t_3,t_4,t_5)$ gets the "mass" $$ m(t) = m(t_1)\; m(t_2)\; m(t_3)\; m(t_4)\; m(t_5) \ ,$$ the total mass is $M=\sum_tm(t)$, and thus we have the weight $w(t)=m(t)/M$.

We can compute now the (slightly more accurate) probability $937273620/20604544751\approx 0.0454886837504387\dots$ :

sage: days = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
sage: sum(days)
365
sage: def m(t):
....:     return prod( [days[tk] for tk in t] )
....: 
sage: m( Omega[0] )
25024440
sage: M = sum( [m(t) for t in Omega] )
sage: M
2472545370120
sage: prob = sum( [m(t) for t in A] ) / ZZ(M)
sage: prob
937273620/20604544751
sage: prob.n()
0.0454886837504387

(Use above the "right number" instead of days[1], $28$, maybe the inaccurate $28\frac 14$ first, to get the better probability when the problem does not depend on century, we only need as constraint the existence of the five persons and of the Sun working as these days.)