Conditional Probability of a Weekend Party Day
You plan to hold a party on the first sunny day of the weekend. Independently, it rains on Saturday with probability $0.70$ and on Sunday with probability $0.60$. The rule is simple: if Saturday is sunny, the party is Saturday. If Saturday is rainy but Sunday is sunny, the party is Sunday. If both days are rainy, the party doesn't happen.
You fast-forward to Monday and learn that the party did happen -- your friends had a great time. Given this information, what is the conditional probability that the party was held on Saturday?
Hints
- Think about what "first sunny day" means -- Saturday gets priority. Write out the probability of each scenario: party on Saturday, party on Sunday, no party.
- The party happens on Saturday if and only if Saturday is sunny. The party happens on Sunday if and only if Saturday is rainy AND Sunday is sunny. These are disjoint events.
- Apply Bayes' theorem: $P(\text{Saturday} \mid \text{party happened}) = P(\text{party on Saturday}) \,/\, P(\text{party happened})$, where $P(\text{party happened}) = 0.30 + 0.28 = 0.58$.
Worked Solution
How to Think About It: This is a clean Bayes' theorem problem with a small event space. You have two independent weather draws and a deterministic rule (party on the *first* sunny day). Once you learn the party happened, you need to update your beliefs about which day it was. Before any math, think about it: Saturday has a 30% chance of being sunny, but if Saturday is sunny the party is guaranteed to be Saturday -- it's the first sunny day. Sunday only gets the party if Saturday fails. So Saturday has a slight edge in the posterior even though Sunday is the sunnier day.
Quick Estimate: P(party on Saturday) = P(Saturday sunny) = $0.30$. P(party on Sunday) = P(Saturday rainy) $\times$ P(Sunday sunny) = $0.70 \times 0.40 = 0.28$. Total probability the party happens = $0.30 + 0.28 = 0.58$. So the posterior for Saturday is roughly $0.30 / 0.58 \approx 0.517$. Just above half -- Saturday has a slight edge because it gets "first dibs."
Approach: Apply Bayes' theorem directly. Define the events and compute the conditional probability.
Formal Solution:
Let $S$ = Saturday is sunny, $U$ = Sunday is sunny. We are given:
$$P(S) = 0.30, \quad P(U) = 0.40$$
with $S$ and $U$ independent. Define:
- $A$ = party held on Saturday = $\{S\}$ (Saturday is sunny, so party is Saturday regardless of Sunday)
- $B$ = party held on Sunday = $\{S^c \cap U\}$ (Saturday rainy, Sunday sunny)
- $H$ = party happened = $A \cup B$
Compute the probabilities:
$$P(A) = P(S) = 0.30$$
$$P(B) = P(S^c) \cdot P(U) = 0.70 \times 0.40 = 0.28$$
$$P(H) = P(A) + P(B) = 0.30 + 0.28 = 0.58$$
We can verify: $P(H) = 1 - P(\text{both rainy}) = 1 - 0.70 \times 0.60 = 1 - 0.42 = 0.58$. Consistent.
By Bayes' theorem:
$$P(A \mid H) = \frac{P(A)}{P(H)} = \frac{0.30}{0.58} = \frac{30}{58} = \frac{15}{29}$$
Answer: The conditional probability the party was held on Saturday is $\dfrac{15}{29} \approx 0.5172$.
Intuition
The key insight is that "first sunny day" creates an asymmetry between the two days even though the weather is independent. Saturday gets priority -- if it's sunny, the party is automatically Saturday, no matter what Sunday looks like. Sunday only gets the party if Saturday fails. This means Saturday's share of the posterior is simply its unconditional sunny probability ($0.30$), while Sunday's share is discounted by the probability that Saturday was rainy ($0.70 \times 0.40 = 0.28$). The result is that Saturday has a slight edge ($15/29 \approx 52\%$) despite being the rainier day.
This pattern shows up constantly in sequential decision problems and market microstructure. Whenever outcomes are resolved in a fixed order -- first limit order at a price gets filled, first signal triggers a trade -- the earlier position has a structural advantage even if its individual probability of "activating" is lower. In Bayesian terms, conditioning on "something happened" shifts weight toward whichever event had the cleanest path to occurring, which is usually the one that came first in the sequence.