Randomly choosing a number and then the dealer picks $20$ from the same set

Question

I am reading the following problem:

In the game of one-spot keno, a card is purchased for $\$1$. It allows a player to choose a number from $1$ to $80$. A dealer then chooses $20$ numbers at random. If the player's number is among those chosen, the player is paid $3.2$ but does not get to keep the $1$ paid to play the game. Find the expected value of a $\$1$ bet. Describe what it means

My approach:

Outcome	Gain or Loss	Probability
Win	$\$3.2 - \$1 = \$2.2$ (winning minus cost for the ticket)	$\frac{1}{80}$
Lose	$\$-1$	$\frac{79}{80}$

My reasoning is that the probability that the person picks the winning number is $1$ out of the $80$ available. The probability to lose is $79$ out of $80$.
The expected value is: $(2.2 \cdot \frac{1}{80}) - \frac{79}{80} = -\$0.96$

which means that on average the person is expected to lose -$\$0.96$ per game in the long run.

The answer though says: -$\$0.2$

I think that because I am not using the $20$ numbers as part of my formula is the error here, although I don't really understand how to include them since, the person first picked a number between $[1,80]$ and then the dealer picked $20$ numbers.

Could someone help me understand this?

Answer 1

Say the player has purchased a card numbered $n \ (1 \leq n \leq 80$).

Now dealer chooses $20$ numbers out of $1-80$. We need to find probability that number $n$ is in those $20$ cards.

As the card numbered $n$ must be in one of the $20$ numbers chosen, the probability is simply,

$\displaystyle \frac{20 \choose 1}{80 \choose 1} = \frac{1}{4}$

Alternatively look at it this way. Favorable outcomes are when the dealer chooses number $n$ and chooses $19$ numbers out of rest $79$.

$\displaystyle \frac{79 \choose 19}{80 \choose 20} = \frac{20}{80} = \frac{1}{4}$

So expected gain is $ \displaystyle \frac{1}{4} \times 3.2 + \frac{3}{4} \times 0 - 1 = -0.2$

So in fact the player is expected to lose $ \$ 0.20$ playing the game.

Answer 2

For you to win the bet the dealer's pick must include $1$ winning number (the one you picked) and $19$ other numbers. The probability of that is $$\frac{\binom11\binom{79}{19}}{\binom{80}{20}}=\frac14$$ This may be derived in a simpler way, since for the purposes of calculating the probability of winning it does not matter whether you or the dealer moves first – neither action influences the other. By letting the dealer roll the winning numbers first, and then you choosing a number, the probability of winning is clearly $\frac{20}{80}=\frac14$. From there the expected gain can be computed to be negative $20$ cents as stated.

Answer 3

If the dealer picks $20$ out of the $80$ numbers, then one-fourth of the numbers have been selected. Each number is equally likely to be selected, meaning that the probability of the dealer selecting your number is $1/4$.