The Massachusetts state lottery game, Cash WinFall, used to have a way that anyone with enough money and time could stand a good chance of getting rich, and it is reported that an MIT computer scientist did just that. In this game, a player picks 6 numbers from the range from 1 to 46. If he matches all 6, then he could win as much as $2 million, but the odds of that payout don’t justify a bet, so let us ignore the possibility of winning this jackpot. Nevertheless, there were times when matching just 5 of the 6 numbers in a $2 lottery ticket would pay $100,000. Suppose in this scenario that you were able to bet $600,000.
(a) What is the expected amount that you would win?
(b) Derive a bound on the probability that you would lose $300,000 or more in this scenario, that is, that you would have 3 or fewer of the 5 of the 6 winning tickets.
The Correct Answer and Explanation is :
Let’s break this down carefully.
(a) Expected Amount Won
Each Cash WinFall ticket costs $2, and you are betting $600,000, meaning you are purchasing:
[
\frac{600,000}{2} = 300,000
]
tickets. You win $100,000 if you match 5 out of 6 numbers. The probability of this happening can be computed using combinatorial counting:
[
P(\text{5 out of 6}) = \frac{\binom{5}{5} \binom{40}{1}}{\binom{46}{6}}
]
[
= \frac{1 \times 40}{9,366,819} = \frac{40}{9,366,819} \approx 4.27 \times 10^{-6}
]
The expected number of winning tickets is:
[
300,000 \times 4.27 \times 10^{-6} = 1.281
]
Each such winning ticket earns $100,000, so the expected winnings are:
[
1.281 \times 100,000 = 128,100
]
Thus, the expected amount won is $128,100.
(b) Probability of Losing $300,000 or More
To lose at least $300,000, you must win 3 or fewer of these 5 out of 6 matches.
Let ( X ) be the number of such wins. Since we are playing 300,000 independent trials with probability ( p = 4.27 \times 10^{-6} ) per ticket, ( X ) follows a Binomial distribution:
[
X \sim \text{Binomial}(300,000, 4.27 \times 10^{-6})
]
By normal approximation:
[
\mu = 1.281, \quad \sigma^2 = 300,000 \times p \times (1 – p) \approx 1.280
]
Thus, ( \sigma \approx 1.13 ). Using the normal approximation, we compute:
[
P(X \leq 3) \approx P\left( \frac{3 – 1.281}{1.13} \right) = P(Z \leq 1.52)
]
Using the standard normal table:
[
P(Z \leq 1.52) \approx 0.935
]
Thus, there is about a 93.5% chance that you will lose $300,000 or more in this scenario.
Now, let’s generate an image illustrating the probability distribution of winnings.
The image illustrates the probability distribution of the number of $100,000 winning tickets in a Cash WinFall scenario where 300,000 tickets are purchased. The binomial probability distribution (blue bars) is approximated by a normal curve (red dashed line). The green dotted line represents the threshold of 3 winning tickets, beyond which a loss of $300,000 or more occurs. As seen, the probability mass is concentrated around 1-2 tickets, aligning with our calculated 93.5% chance of significant loss.