Re: What is the expectation of this game?
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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.
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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%.
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