probability

[CAMELSMOM]

Experiment: I deal you one card from a standard deck of 52 and then you toss a single 6sided die
The Payout: If you get an ace and a 5 you get $25
If you get a heart and an even number you get $20
You cost: How much should you pay in order for this game to be fair? Assume you do not get your original money returned to you in the chance that you win.

So far I have this
P(ACE AND 5)=4/52*1/6= .01282
p(HEART AND EVEN #)= 12/52 * 3/6=.125

I am getting stuck when you throw the money in there

 

[soroban]

Hello, CAMELSMOM!

Experiment: I deal you one card from a standard deck of 52 and then you toss a single 6-sided die.
. . If you get an Ace and a 5, you get $25.
. . If you get a Heart and an even number, you get $20.
How much should you pay in order for this game to be fair?

$\displaystyle \text{There are: }\:52\cdot6 \:=\:312\text{ possible outcomes.}$

$\displaystyle \text{To get an Ace and a 5, there are: }\:4\cdot1\:=\:4\text{ ways.}$
. . \(\displaystyle P(\text{Ace \& 5}) \:=\(\text{win \$25}) \:=\:\frac{4}{312}\)

$\displaystyle \text{To get a Heart and an even number, there are: }\:13\cdot3 \:=\:39\text{ ways.}$
. . \(\displaystyle P(\text{Heart \& even}) \:=\(\text{win \$20}) \:=\:\frac{39}{312}\)

$\displaystyle \text{You win in: }\:4+39 \:=\:43\text{ outcomes.}$
. . $\displaystyle \text{You lose in the other: }\:312 - 43 \:=\:269\text{ outcomes.}$
$\displaystyle P(\text{lose }x\text{ dollars}) \:=\:\frac{269}{312}$


$\displaystyle \text{For a fair game, the expected value is \$0.}$

$\displaystyle \text{We have: }\;(25)\!\left(\frac{4}{312}\right) + (20)\!\left(\frac{39}{312}\right) - (x)\!\left(\frac{269}{312}\right) \;=\;0$

$\displaystyle \text{Multiply by 312: }\;100 + 780 - 269x \:=\:0 \quad\Rightarrow\quad 269x \:=\:880$

, . $\displaystyle x \:=\:\frac{880}{269} \:=\:3.271375465$


$\displaystyle \text{Therefore, you should pay about }\$3.27\text{ per game.}$


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Reality Check:

$\displaystyle \text{If the "house" (casino) is establishing the rules,}$
. . $\displaystyle \text{it would charge you \$3.28 per game.}$

$\displaystyle \text{If you paid only \$3.27 per game, your expected value would be:}$
. . $\displaystyle (25)\!\left(\frac{4}{312}\right) + (20)\!\left(\frac{39}{312}\right) - (3.27)\!\left(\frac{269}{312}\right) \:=\:+0.37$

$\displaystyle \text{That is, you could expect to }win\text{ an average of 37 cents per game.}$


$\displaystyle \text{And, of course, the casino would never allow that . . .}$