Lucky Lottery Probability
A lottery draws 6 balls at random (without replacement) from an urn containing 70 balls labeled 1 through 70. Your ticket also has 6 distinct numbers, chosen uniformly at random from 1 to 70.
You win if at least 5 of your 6 numbers match the 6 drawn balls (order does not matter).
- Find the probability $p$ of winning.
- Report $p \times 10^6$, rounded to the nearest hundredth.
Hints
- The number of matching numbers $N$ follows a Hypergeometric distribution -- identify the three parameters: total population (70), number of 'successes' in the population (6 balls that match your ticket), and number of draws (6).
- The win probability is $P(N=5) + P(N=6)$. Use the Hypergeometric PMF: $P(N=k) = \binom{6}{k}\binom{64}{6-k} / \binom{70}{6}$.
- Compute $\binom{70}{6} = 131,\!115,\!985$. The numerator is $\binom{6}{5}\binom{64}{1} + \binom{6}{6}\binom{64}{0} = 384 + 1 = 385$. Divide and multiply by $10^6$.
Worked Solution
How to Think About It: This is a sampling-without-replacement problem. You have 70 balls, 6 of which match your ticket ('successes') and 64 of which do not ('failures'). The lottery draws 6 balls. The number of matches $N$ follows a Hypergeometric distribution -- the natural model whenever you sample without replacement from two groups.
Quick Estimate: The number of ways to pick 6 from 70 is $\binom{70}{6} \approx 131$ million. The '5 match' outcome requires choosing 5 from your 6 winning numbers and 1 from the 64 non-winners: $\binom{6}{5}\binom{64}{1} = 6 \times 64 = 384$ ways. The '6 match' outcome has $\binom{6}{6}\binom{64}{0} = 1$ way. Combined: 385 ways out of roughly 131 million -- a probability around $3 \times 10^{-6}$. So $p \times 10^6 \approx 3$.
Formal Solution:
Let $N$ = number of ticket numbers that match the drawn balls. Since we draw without replacement from a population of 6 'matching' and 64 'non-matching' balls:
$$N \sim \text{Hypergeometric}(\text{population}=70,\; \text{successes}=6,\; \text{draws}=6)$$
The PMF is: $$P(N = k) = \frac{\binom{6}{k}\binom{64}{6-k}}{\binom{70}{6}}$$
Win condition: $N \geq 5$, i.e., $N = 5$ or $N = 6$.
$$P(N = 5) = \frac{\binom{6}{5}\binom{64}{1}}{\binom{70}{6}} = \frac{6 \times 64}{\binom{70}{6}} = \frac{384}{\binom{70}{6}}$$
$$P(N = 6) = \frac{\binom{6}{6}\binom{64}{0}}{\binom{70}{6}} = \frac{1}{\binom{70}{6}}$$
Now compute $\binom{70}{6}$: $$\binom{70}{6} = \frac{70 \times 69 \times 68 \times 67 \times 66 \times 65}{720} = 131,\!115,\!985$$
So: $$p = P(N \geq 5) = \frac{385}{131,\!115,\!985} \approx 2.9363 \times 10^{-6}$$
$$p \times 10^6 \approx 2.94$$
Answer: $p \times 10^6 \approx \mathbf{2.94}$.
Intuition
The Hypergeometric distribution is the go-to model whenever you sample without replacement from a finite population with two types (success/failure). It differs from the Binomial in one key way: draws are not independent, because each ball removed changes the composition of the remaining population. For large populations relative to the sample size, the Binomial is a good approximation (the dependence becomes negligible), but for lottery-scale problems where 6 of 70 balls are 'successes,' you need the exact Hypergeometric.
Lottery probabilities are a good reference point for calibrating your sense of small numbers. Roughly one in 340,000 for a 5-match, one in 131 million for the jackpot -- both are far smaller than they feel. This is a recurring theme in risk management: people are bad at intuiting probabilities below $10^{-4}$, which is why tail risk is chronically mispriced. When you work with rare event probabilities in practice (option tail pricing, credit default modeling, scenario analysis), always compute the number of standard deviations from the mean and cross-check against historical base rates.