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#5
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636 Re: What is the expectation of this game?

[ 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%.