Question

In Win1, the player picks a single letter from A-J and a single digit 0-9. If both the letter and digit match the letter and the digit picked on that day, the player wins $150. In PBall, the player picks a single letter A-T and a single digit from 0 to 9. If both the letter and the digit match the letter and the digit picked on that day, the player wins$280.--Which lottery game has a better expected value, and what is the value?(A)Win1 has a higher expected value=$0.50.(b)-PBall has a higher expected value,=-$0.60. (C)Win1 has a higher expct. value= 0.50$(D)PBall has higher expt. value= 0.60$

Answer 1

im confused @cwrw238

Answer 2

i litter dont know what to plug in for what

Answer 3

litteraly

Answer 4

can u help me @ankit042

Answer 5

@SithsAndGiggles how would i plugg it in

Answer 6

i see but how would i find one of the answers @SithsAndGiggles @mathstudent55 @mathmale

Answer 7

Oh hold on, I made a few mistakes. First, I mixed up the winnings for Win1 (150) and PBall (280).

You have a \(\dfrac{1}{100}\) chance of getting any one pair matching letters/digits. Since there are 10 pairs you can get (and 90 other losing pairs), you actually have \(\dfrac{10}{100}\) chance of winning.
\[\begin{align*}E(\text{Win1})&=\sum xp(x)\\
&=0\underbrace{\left(\frac{1}{100}+\cdots+\frac{1}{100}\right)}_{\large90\text{ non-matching pairs}}+150\underbrace{\left(\dfrac{1}{100}+\cdots+\frac{1}{100}\right)}_{\large10\text{ matching pairs}}\\
&=\frac{1500}{100}\\\\
&=15
\end{align*}\]
For PBall, there are 20 letters and ten digits. As with Win1, I'm assuming pairs like A-1 and B-2 are winning pairs. This means there are \(20\times10=200\) total pairs, only ten of which allow you to win, so the chance of winning is \(\dfrac{10}{200}\), or \(\dfrac{1}{200}\) per winning pair.
\[\begin{align*}E(\text{PBall})&=0\underbrace{\left(\frac{1}{200}+\cdots+\frac{1}{200}\right)}_{\large190\text{ non-matching pairs}}+280\underbrace{\left(\dfrac{1}{200}+\cdots+\frac{1}{200}\right)}_{\large10\text{ matching pairs}}\\
&=\frac{2800}{200}\\\\
&=14
\end{align*}\]
So Win1 has the better expected winnings by a dollar.

The answer choices you're given don't make sense. For one thing, you can only have a negative expectation if the player pays to play, or is supposed to pay if he/she DOESN'T get a winning pair.

Second, the expectation can't be less than 1. The product of the winnings and the probability of the winnings will always exceed 1, unless there's a typo, or we're missing some information, or you made a mistake transcribing the question.