Question

In a lottery game, a player picks six numbers from 1 to 25. If the player matches all six numbers, they win 40,000 dollars. Otherwise, they lose $1.

What is the expected value of this game? $

Answer 1

Answer: around -$0.77

Explanation:

To calculate the expected value of this lottery game, we need to find the probability of winning the game and the probability of losing the game.

Probability of winning:

There are a total of C(25, 6) possible combinations of choosing 6 numbers out of 25, where C(n, k) represents the number of combinations of choosing k items from a set of n items. The formula for combinations is:

C(n, k) = n! / (k!(n-k)!)

C(25, 6) = 25! / (6!19!) = 177100

So, there are 177,100 possible combinations. Since there's only 1 winning combination, the probability of winning is:

P(Winning) = 1 / 177,100

Probability of losing:

Since there's only 1 winning combination, the probability of losing is:

P(Losing) = 1 - P(Winning) = 1 - (1 / 177,100) = 177,099 / 177,100

Expected value:

Now, we can calculate the expected value (EV) by multiplying the probabilities of each outcome by their respective values and then summing up these products:

EV = P(Winning) * Value(Winning) + P(Losing) * Value(Losing)

EV = (1 / 177,100) * $40,000 + (177,099 / 177,100) * (-$1)

EV ≈ ($40,000 / 177,100) - ($177,099 / 177,100)

EV ≈ $0.2256 - $1

EV ≈ -$0.7744

The expected value of this lottery game is approximately -$0.7744. This means that, on average, a player would lose about 77.44 cents per game.