Suppose your probability of winning if your number comes up if $\frac{1}{38}$. If so, you win $\$35$, otherwise you lose a dollar. Use the Poisson distribution to show that if you play 70 times then the approximate probability you will have won more money than you have lost is larger than $\frac{1}{2}$

My thoughts are:

Let's say you win k games over the 70 games. You will win more money than you lose if 35k-(70-k) > 0 or k $\ge 2$. So, P(X $\ge$ 2) = 1 - [P(X = 0) + P(X = 1)] = 1 - [$\frac{\lambda^0}{0!}e^{-\lambda} + \frac{\lambda^1}{1!}e^{-\lambda}$] = $1-(1 + \lambda) e^{-\lambda}$

However, I'm not getting a prob > $\frac{1}{2}$ if I use $\lambda = \frac{1}{38}$

Is $\lambda$ supposed to be $\frac{1}{38}$ or $\frac{70}{38}$?

  • $\begingroup$ Based on my research, looks like it does need to be 70/38. Lambda needs to be the average for the interval being considered, which in this case would 70/38 games. However, since I'm a newbie at this, hopefully someone more qualified can comment or add to the answer. $\endgroup$ – EggHead Oct 31 '13 at 3:18
  • $\begingroup$ You are correct. You provide a per game rate, then extend to an interval of 70 games, so you get 70/38. Your calculation will show P(making money) > .5 $\endgroup$ – user76844 Oct 31 '13 at 14:47
  • $\begingroup$ $e^{-\lambda}+\lambda e^{-\lambda} = (1+\lambda)e^{-\lambda}$ $\endgroup$ – Alex Nov 3 '13 at 18:07

Based on Eupraxis's comment, if we use: $\lambda = \frac{70}{38}$
$P(X \ge 2) = 1-\frac{108}{38}e^{-\frac{70}{38}} = 0.5495$


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