Expected Value

[jbone]

Suppose you are offered a bet based upon the roll of three standard dice. These dice
are fair, so for any given die, the odds of a specic number being rolled is 1/6. In order
to participate in this bet, you must pay $1 up front. You select a number from 1 to
6, and if this number comes up on all three dice, then the payout is $4. If this number
comes up on two of the three dice, then the payout is $3. Finally, if your number comes
up on only one of the dice, then the payout is $2. How much money can you expect to
win or lose on average from playing this game?

I worked it out to get:

E(w)= (1/216)*4+(15/216)*3+(75/216)*2+(125/216)*-1
E(w)= .34

Which seems fine, but my professor made a comment at the end of class insinuating that these "gambles" always had a negative expected wealth (i.e. gambling is dumb).

Is this what you got?

 

[Deleted member 4993]

suppose you are offered a bet based upon the roll of three standard dice. These dice
are fair, so for any given die, the odds of a specic number being rolled is 1/6. In order
to participate in this bet, you must pay $1 up front. You select a number from 1 to
6, and if this number comes up on all three dice, then the payout is $4. If this number
comes up on two of the three dice, then the payout is $3. Finally, if your number comes
up on only one of the dice, then the payout is $2. How much money can you expect to
win or lose on average from playing this game?

I worked it out to get:

E(w)= (1/216)*4+(15/216)*3+(75/216)*2+(125/216)*0 -1
e(w)= .34

which seems fine, but my professor made a comment at the end of class insinuating that these "gambles" always had a negative expected wealth (i.e. Gambling is dumb).

Is this what you got?

e= 0.921296 - 1 = - 0.08

 

[soroban]

Hello, jbone!

You ignored the $1 ante, a very common error.

Suppose you are offered a bet based upon the roll of three standard dice.
These dice are fair, so for any given die, the odds of a specic number being rolled is 1/6.
In order to participate in this bet, you must pay $1 up front.
You select a number from 1 to 6.
If this number comes up on all three dice, then the payout is $4.
If this number comes up on two of the three dice, then the payout is $3.
Finally, if your number comes up on only one of the dice, then the payout is $2.
How much money can you expect to win or lose on average from playing this game?

I worked it out to get:

\(\displaystyle E(w) \:=\: \frac{1}{216}(4)+\frac{15}{216}(3)+\frac{75}{216}(2)+\frac{125}{216}(-1) \:=\:0.34\)

Since you already paid $1 to play,
. . your winnings are: $3, $2, $1 and -$1, respectively.

 

[JeffM]

Although this is the identical mathematical answer as everyone else has given you, I find it most intuitive to express it as follows, starting with the ante and showing the payoff for every possibility thereafter.

\(\displaystyle E(w) = \left(- 1 * \dfrac{216}{216}\right) + \left(4 * \dfrac{1}{216}\right) + \left(3 * \dfrac{15}{216}\right) + \left(2 * \dfrac{75}{216}\right) + \left(0 * \dfrac{125}{216}\right) = \dfrac{-216 + 4 + 45 + 150 + 0}{216} = \dfrac{-17}{216} < 0.\)