Constraint on the probability of an event

Question

I have following exercise. The casino is testing a new machine. The player puts 1\$. With a probability of 32/200 the game returns 3 \$, with a probability of 1/200 the game returns 100\$, with a probability of 167/200 returns 0. The casino has found that the machine has lost 10,000\$ after a million games. Show constraint on the probability of an event.

I am not sure how to do this task. I decided to use Chebyshev inequality. Let $X$ - player's expected earnings $$E[X] = 32/200 * 2 + 99 *1/200 + 167 * (-1) = -1/50$$ $$E[X^2] = 32/200 * 2^2 + 99^2 *1/200 + 167 * (-1)^2 = 1262/25$$ $$Var[X] = E[X^2] - E[X]^2 = 1262/25 - (-1/50) = 50.4796$$ Chebyshev inequality: $$\Pr(|X-\mu |\geq k\sigma )\leq {\frac {1}{k^{2}}}.$$ And i'm not sure how to show constrain on the probability. I've tried transform the equation: $$\Pr(|X-\mu |\geq k )\leq {\frac {\sigma ^{2}}{k ^{2}}}$$ After substitution: $$\Pr(|X-\mu |\geq k )\leq {\frac {50.4796^2}{10000 ^{2}}}$$ I am pretty sure that my reasoning is incorrect because I did not use the number of games played. I would ask for advice. thank you

Answer 1

Side comment: In your computations $167$ should be $167/200$ and in the third line your $(-1/50)$ should be $(-1/50)^2$, but I think these are just typos since your final results are correct...

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Your $X$ represents a player's earnings in a single game. You can then use your computations to find the analogous quantities for $Y = X_1 + \cdots + X_{1000000}$ where each $X_i$ is a single independent game.

You can check that $E[Y] = 10^6 E[X_1] = -20,000$ and $\text{Var}(Y) = 10^6 \text{Var}(X_1) = 50,479,600$ where $E[X_1]$ and $\text{Var}(X_1)$ are the quantities you have already computed.

Then we have $$P(Y \ge 10000) = P(Y - E[Y] \ge 30000) \le P(|Y-E[Y]| \ge 30000) \le \frac{\text{Var}(Y)}{30000^2}.$$

Answer 2

• The expected earnings and variance you have calculated are for a single game. Thus, the Chebyshev inequality $$ P\left(|X-\mu|>10000\right)\le \left(\frac{50.4796}{10000}\right)^2\approx2.5\times10^{-5} $$ isn't very informative, because the most a player can win in a single game is $\$100$, so $\ P\left(|X-\mu|>10000\right)\ $ is not merely bounded above by $\ 2.5\times10^{-5}\ $, it is precisely zero.
• The random variable you're interested in is not merely the casino's loss in a single game but its total loss $\ L=\sum_\limits{i=1}^{10^6}X_i\ $ over a million games, where $\ X_i\ $ is the amount won by the player of the $\ i^\text{th}\ $ game. As shown in angryavian's answer, $\ E(L) = -20,000\ $, and $\ \text{Var}(L)=50,479,600\ $.

You can get a much tighter bound on the quantity you're really interested in, $\ P(L\ge10000)\ $, rather than $\ P(|L-20000|\ge10000)\ $, by using the normal approximation to the distribution of $\ L\ $ instead of the Chebyshev inequality. This gives \begin{align} P(L\ge10000)&=P(L+20000\ge30000)\\ &\approx P\left(L+20000\ge4.22\sqrt{50479600}\right)\\ &\approx1-\mathcal{N}(4.22)\\ &=\mathcal{N}(-4.22)\\ &\approx 1.22\times 10^{-5}\ , \end{align} much smaller than the bound of sbout $\ 0.056\ $ given by the Chebyshev inequality. In the above calculation $\ \mathcal{N}\ $ denotes the standard normal cumulative distribution function.