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Lets says I have a lottery game where the ticket costs $1 and has the following probability/prize distribution:

  • 0.3 -> \$1
  • 0.2 -> X
  • 0.5 -> \$0

If X = \$1, then the expected value is: 0.3(\$1) + 0.2(\$1) = \$0.50

If X = FREE_TICKET, then I've calculated (and confirmed) via sampling that the EV is either:

  • 0.3(\$1) / (0.3+0.5) = 0.375
  • 0.3(\$1) + 0.2 * \$1 * 0.3/(0.3+0.5) = 0.375

Obviously, a FREE_TICKET isn't the same as \$1, the price of a ticket.

However, I'm stumped when calculating the EV of the table for the following two cases:

  • X = 2 FREE_TICKETS
  • X = \$5 + FREE_TICKET

How do I calculate them?

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You have the expected value as $Y = 0.3\cdot 1 + 0.2\cdot X + 0.5\cdot 0$. If, in addition, $X = 2Y$, or $X = 5 + Y$, you can solve for $X$ and $Y$ simultaneously. –  mjqxxxx Nov 28 '12 at 18:24
    
Thanks, you're right, it's just basic algebra. –  AlexD Nov 29 '12 at 14:31

1 Answer 1

up vote 1 down vote accepted

My guess: EV = 0.3 + 0.2 * (2 * EV) in the first case EV = 0.3 + 0.2 * (5 + EV) in the second

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1  
I tried that too, but it's wrong (doesn't match my random sampling results) –  AlexD Nov 28 '12 at 18:35
    
@AlexD: Programming errors happen, at least when I am doing the programming. –  André Nicolas Nov 28 '12 at 19:10
    
My program was fine, but I simply misread GaTTaCa's solution. It was right after all. –  AlexD Nov 29 '12 at 14:32

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