probabilityy

[spardawg]

Problem: In a carnival game, the theoretical probability of winning is 4/25. The cost to play is $3. Winning players recieve $15. Given that the prize value remains $15, change the cost of playing to make this a fair game.

I said "The price to play should be 50 cents because the average win or expected value is 60 cents and the cost to play is $3." Would this be right? It's 10 cents less to play than the average winnings are...I'm just making sure. Thank you

 

[arthur ohlsten]

Probability of winning [P] = 4/25=.16
Probability of losing [Q]= ,84

amount you win=$15
expected gain=$15[.16]=$2.40

expected loss for a fair game=bet[Q]
$2.40 = [.84] x
x= $2.85

 

[soroban]

Hello, spardawg!

I don't understand your reasoning at all . . .

In a carnival game, the theoretical probability of winning is 4/25.
The cost to play is $3. Winning players recieve $15.
Given that the prize value remains $15, change the cost of playing to make this a fair game.

I said: The price to play should be 50 cents . ??
because the average win or expected value is 60 cents . How did you get this?
and the cost to play is $3. . So you subtracted $2.50 ?

Would this be right?
It's 10 cents less to play than the average winnings are. . Where did you get the 10?

Arthur has the right approach.


Have you ever met the Expected Value Formula?


. . \(\displaystyle \begin{array}{ccc}\text{Event} & \text{Prob.} & \text{Result} \\ \hline \\[-3mm] \text{Win} & \frac{4}{25} & +\$15 \\ \\[-3mm] \text{Lose} & \frac{21}{25} & -\$3 \\[-3mm] \\ \hline \end{array}\)

\(\displaystyle E \;\;=\;\;\left(\frac{4}{25}\right)(+15) + \left(\frac{21}{25}\right)(-3) \;\;=\;\;\frac{60}{25} + \frac{63}{25} \:\;=\;\:-\frac{3}{25} \;\;=\;\;-12\rlap{/}c\)

. . You can expect to lose an average of 12 cents per game.



For a fair game, we want \(\displaystyle E = 0.\)

. . \(\displaystyle \begin{array}{ccc}\text{Event} & \text{Prob.} & \text{Result} \\ \hline \\[-3mm] \text{Win} & \frac{4}{25} & +15 \\ \\[-3mm] \text{Lose} & \frac{21}{25} & -x \\[-3mm] \\ \hline \end{array}\)

\(\displaystyle \text{We have: }\:\frac{4}{25}(15) + \frac{21}{25}(-x) \:=\:0 \quad\Rightarrow\quad \frac{60}{25} - \frac{21x}{25} \:=\:0 \quad\Rightarrow\quad 60 - 21x \:=\:0\)

. . . . . . . . . . . \(\displaystyle 21x \:=\:60 \quad\Rightarrow\quad x \:=\:\frac{60}{21} \;=\;2.85714\hdots \;=\;\$2.86\tfrac{5}{7}\)