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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%. |
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