What is the expectation of this game?

I've been playing a game on jackpotjoy.com all day called "Lucky Dice." In the game you throw three dice, and I bet 10 points on one, 10 points on two and 10 points on three.

So if none of the die come up one/two/three I lose the 30 points I bet.

if any one die comes up one/two/three I get 20 points back (so I lose 10 overall)
if any two die are the same and come up with one/two/three I get 30 points (so I lose nothing overall)
if all three are the same and come up with one/two/three I get 100 points (so I get 80 points overall)

so for example if the die roll as ...
one, one, five. I get thirty points but I bet thirty points so overall I get nothing.
if they rolled as four, five six I get nothing (so my net gain is -30)
if they rolled as one, two, three I get 20 back for each (net gain is 30)
if they rolled as one, one, three I get 50 back, (so my net gain is 20)

ANyway, enough examples. The reason I'm going on about it is it seems to have really good return, like a really high positive expectation. And given that I've been playing all day I really am quite surprised to be still winning.

Does soembody familiar with basic probability want to calculate expectation for my strategy?

Looks like 8 possible outcomes, each equally likely. You are betting on coin flips, for all intents and purposes. The outcomes are distributed as follows:

-30 (1, -30)
-10 (3, -30)
0 (3, 0)
+80 (1, +80)

-30 - 30 + 0 + 80 = 20
20/8 = 2.5

Your total EV for each play should be +2.5 points, if I'm not mistaken.

Wait, you only get +70 points if all three hit with a 100-point return, right? Not +80? So your EV is +1.25

so it is worth playing?

[BruceZ]

[ QUOTE ]
Looks like 8 possible outcomes, each equally likely. You are betting on coin flips, for all intents and purposes. The outcomes are distributed as follows:

-30 (1, -30)
-10 (3, -30)
0 (3, 0)
+80 (1, +80)

-30 - 30 + 0 + 80 = 20
20/8 = 2.5

Your total EV for each play should be +2.5 points, if I'm not mistaken.

[/ QUOTE ]

The outcomes are not even close to equally likely, and the overall EV is -1.53 points out of 30, or -5.1%, almost as bad as double-zero roulette.

You also didn't include the net gains of +20 and +30 that he mentioned explicitly. There is also a +10. I'm assuming that for 3 lows you get back 100 for a net gain of +70, not +80 as the OP said.

In the following, X stands for a 4-6. Where the number 1 appears alone, this stands for any number 1-3. 11 stands for any pair 11,22, or 33. 111 includes 222 and 333. Where 12 appears, this is any 2 different numbers 1-3. These can occur in any order. These are all the combinations with their probabilities, which sum to 1:


XXX: (1/2)^3 = 1/8

1XX: (1/2)^3 * 3 = 3/8

11X: 3/6 * 1/6 * 1/2 * 3 = 1/8

12X: 3/6 * 2/6 * 1/2 * 3 = 1/4

112: 3/6 * 1/6 * 2/6 * 3 = 1/12

123: 3/6 * 2/6 * 1/6 = 1/36

111: 3/6 * 1/6 * 1/6 = 1/72


EV =

1/8*(-30) +

3/8*(-10) +

1/8*(0) +

1/4*(+10) +

1/12*(+20) +

1/36*(+30) +

(1/72)*(+70)

=~ -1.53, and -1.53/30 =~ -5.1%.

Right, I completely misread the OP. Sorry.