🎯 Q&A: Probability That All Four Seasons Occur Among 7 Birthdays

❓ Question

In a game, each of 7 players is randomly assigned one of four seasons (Spring, Summer, Fall, Winter), each equally likely.

What is the probability that all four seasons occur at least once among the 7 birthdays?

A common incorrect attempt uses:

$<br /> 1 - \frac{\binom{4}{3}3^7 + \binom{4}{2}2^7 + \binom{4}{1}1^7}{4^7}<br />$

which gives the wrong result (around 0.42 instead of 0.513).

Why does this formula overcount?
How does the Inclusion–Exclusion Principle fix this?
And what deeper pattern explains the alternating signs in the formula?

✅ Answer

The task is to compute the probability that all four seasons appear among the 7 birthdays.
Because the events “Spring missing,” “Summer missing,” etc., overlap, the Inclusion–Exclusion Principle (IEP) is required.

🌟 1. Total Birth Assignments

Each of the 7 people falls independently into one of 4 seasons, giving:

$<br /> |\Omega| = 4^7 = 16384<br />$

🌟 2. Define the “Bad” Events

Let:

$A_1$ : Spring missing
$A_2$ : Summer missing
$A_3$ : Fall missing
$A_4$ : Winter missing

The desired probability (all 4 seasons appear) is the complement of the union of these events:

$<br /> P(\text{all 4 appear}) = 1 - P(A_1 \cup A_2 \cup A_3 \cup A_4)<br />$

So the goal is to compute the size of that union using Inclusion–Exclusion.

🌟 3. Why Inclusion–Exclusion Is Needed

The sets $A_1,A_2,A_3,A_4$ overlap heavily.

For example, if all 7 birthdays fall in Summer:

Spring is missing → outcome lies in $A_1$
Fall is missing → in $A_3$
Winter is missing → in $A_4$

So one single outcome belongs to three sets at once.
If we simply added the sizes of the four sets, this outcome would be counted three times.

Inclusion–Exclusion systematically fixes this by alternately subtracting and adding overlaps.

🌟 4. Inclusion–Exclusion for Four Sets

The union size is:

$<br /> |A_1 \cup A_2 \cup A_3 \cup A_4|<br /> = \binom{4}{1}3^7</p> <ul> <li>\binom{4}{2}2^7</li> </ul> <ul> <li>\binom{4}{3}1^7</li> </ul> <ul> <li>\binom{4}{4}0^7<br />$

Interpretation:

Missing exactly one season

Choose the missing season → $\binom41$ choices.
Remaining 3 seasons available → $3^7$ assignments.

Missing exactly two seasons

Choose 2 missing seasons → $\binom42$ choices.
Remaining 2 seasons → $2^7$ assignments.
Subtract because these were counted twice earlier.

Missing exactly three seasons

Choose 3 missing seasons → $\binom43$ choices.
Only 1 season remains → $1^7$ .
Add back because they were oversubtracted.

Missing all 4 seasons

Impossible → $0$ .

🌟 5. Compute the “Bad” Outcomes

$<br /> 4(3^7) - 6(2^7) + 4(1^7)<br />$

Substituting:

$<br /> 4(2187) - 6(128) + 4 = 8748 - 768 + 4 = 7984<br />$

🌟 6. Compute the “Good” Outcomes

$<br /> 8400 = 16384 - 7984<br />$

🌟 7. Final Probability

$<br /> P(\text{all 4 seasons occur})<br /> = \frac{8400}{16384}<br /> = \frac{525}{1024}<br /> \approx 0.5127<br />$

Thus:

There is about a 51.27% chance that all four seasons appear at least once among 7 birthdays.

🌟 8. Why the Original Formula Overcounted

The incorrect expression:

$<br /> \binom41 3^7 + \binom42 2^7 + \binom43 1^7<br />$

counts many outcomes multiple times.

Example: all 7 birthdays in Summer.

That one outcome is counted:

in 3 “missing one season” counts
in 3 “missing two seasons” counts
in 1 “missing three seasons” count

Total = 7 counts, but it is only one outcome.

Inclusion–Exclusion corrects these overlaps.

🌟 9. The “Reverse Containment” Insight

(An important educational point)

Let:

$U_1$ = outcomes using 3 seasons
$U_2$ = outcomes using 2 seasons
$U_3$ = outcomes using 1 season

Look at what each IE term covers (as sets, ignoring multiplicity):

$T_1 = \binom41 3^7$ covers $U_1, U_2, U_3$
$T_2 = \binom42 2^7$ covers $U_2, U_3$
$T_3 = \binom43 1^7$ covers only $U_3$

Thus the sets satisfy:

$<br /> U_3 \subseteq U_2 \subseteq U_1<br />$

This explains the alternating signs:

3-season outcomes → counted once
2-season outcomes → counted twice then subtracted once
1-season outcomes → counted 3 times, subtracted 3 times, added once

Every bad outcome ends up counted exactly once, which is precisely what Inclusion–Exclusion ensures.

🎯 Final Takeaway

Using Inclusion–Exclusion:

$<br /> P(\text{all four seasons appear}) = \frac{525}{1024} \approx 0.5127<br />$

This method correctly handles overlapping missing-season outcomes, and the reverse-containment structure shows why the signs alternate.