I've attempted to solve this problem, but I'm not 100% sure I'm correct.


Consider a game played with the following extra game trigger conditions:

There is a 0.001 probability of triggering 10 extra games, a 0.0015 probability of triggering 5 extra games, and a 0.002 probability of triggering 3 games. Each of these extra games has a 0.005 probability of retriggering 15 extra games, and a 0.008 probability of retriggering 5 extra games. Assuming all games are independent, what is the expected number of games played?

My attempted solution (includes first game):

N(expected) = 1 + 0.001(10)[0.005(15) + 0.008(5)] + 0.0015(5)[0.005(15) + 0.008(5)] + 0.002(3)[0.005(15) + 0.008(5)] = 1.0027025.

Am I missing something?


There are a few missing terms. For example, for the $10$ extra games triggered situation, we need $$(0.001)(10)[1+(0.005)(15)+(0.008)(5)].$$ A similar change is needed in the other cases.

  • $\begingroup$ I considered that, but I thought I'd be double counting the (0.001)(10) case along with the (0.0015)(5) and (0.002)(3) cases. $\endgroup$ – user161676 Jul 4 '14 at 20:10
  • $\begingroup$ You were counting the expected number of first generation games ($1$) and the expected number of third generation games, but not counting at all the second generation games. For example, there is a probability $0.001$ that we get to play $10$ second generation games. That already makes a contribution of $0.01$ to the expected number of games, bringing us beyond the number $1.0027\dots$ you obtained. $\endgroup$ – André Nicolas Jul 4 '14 at 20:15

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