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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?

  • 0
    Thanks, you're right, it's just basic algebra.2012-11-29

1 Answers 1

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

  • 0
    My program was fine, but I simply misread GaTTaCa's solution. It was right after all.2012-11-29