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Let's assume a imaginary lottery where each draw, we are allowed to buy the following tickets:

Price   Probability    Payout        EV
$1      1/600 000      $1,200,000    2
$1      1/300 000      $600,000      2    
$1      1/200 000      $400,000      2
$1      1/100 000      $200,000      2

If we play the lottery 1,000,000 times (buying those 4 tickets each time), how can we calculate the probability of losing money?

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Let random variable $X$ represent our (gross) win if we buy one ticket of each kind. Then $X$ has mean $8$. We can also compute the variance of $X$. For example, let $W_1$ be our winnings on the first type of ticket. Then $E(W_1^2)=\frac{(1200000)^2}{600000}=2400000$. Thus $W_1$ has variance $2400000-4$. Do the same for the other types of ticket, and add up. (We are asssuming the four lotteries are independent.) Thus $X$ has variance $4800000-16$.

Now let $Y$ be our total gross winnings on the $1$ million trials. This has mean $8000000$ and variance $4.8\times 10^{12}$. Our outlay has been $4000000$, so we want the probability that $Y\lt 4000000$.

Now comes the crossing of the fingers part. Assume that $Y$ has a distribution close enough to normal for our computations to be not too far off.

Then we want the probability that a normal with mean $8000000$ and variance $4.8\times 10^{12}$ is less than $4000000$. This is the probability that $Z\lt -4/\sqrt{4.8}$, where $Z$ is standard normal. That turns out (tables) to be about $0.039$.

Remark: The (very) weak point of the calculation is the normal aapproximation. The greatest contributor to the weakness is the $1200000$ prize. So I would suggest redoing the computation as follows. It is easy to calculate to calculate the probability $p_0$ that we win no big prize, the probability $p_1$ that we win exactly one, and so on. For practical calculation we only need these two, and the probability $p_r=1-p_0-p_1$ that we win two or more big prizes.

Calculate the probability $l_0$ that we lose money given that we got no big prizes, the probability $l_1$ that we lose money given that we got exactly one big prize. These can be estimated by the same sort of normal approximation as before, but the approximation is substantially more reliable.

For all practical purposes, we will lose money only if we win no big prize or only one, and the smaller prizes are not enough to cover our big prize losses. The probability we lose money is well approximated by $p_0l_0+p_1l_1$.

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  • $\begingroup$ Thanks for youre reply! Gonna take some time trying to process it. $\endgroup$ – Martin Wiklund Jul 11 '13 at 6:08

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