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A bet gives 50% chance of winning 200U\$ and 50% chance of losing 100U\$. I understand that the expected value after having played 200 rounds of the same bet independently is 10.000U\$=(50%*200-50%*100U\$)*200. However, I do not know how to calculate the probabilities of : 1) The total winning is indeed 10.000U\$ 2) There is no winning at the end of the 200 rounds 3) The final outcome is a loss of 2000U\$ Thanks for explaining how the calculations of the probabilies of those outcome work. Thanks in advance

2 Answers 2

1

Let $X$ be the number of wins and $Y$ the sum won (or lost if negative!) after $n=200$ rounds. Since each bet $X_i\in\{0,1\}$ is an independent Bernoulli trial with success (and failure) probability $p=\frac12$ and outcome $Y_i=100(3X_i-1)=200$ or $-100$, the sums are Binomially distributed $ \eqalign{ X &= \sum_{i=1}^n X_i = \operatorname{Binom}\left(n=200,p=\frac12\right) \\ Y &= \sum_{i=1}^n Y_i = \sum_{i=1}^n 100\left(3X_i-1\right) =100\left(3X-n\right)=300X-20000 \\ } $ and your exact probabilities for parts a & c are given by $ \mathbb{P}[X=k]={n\choose k}p^k(1-p)^k=\frac{n!}{k!\,(n-k)!}p^n $ for $k=100$ (for $Y=10$K) and $-20$ (for $Y=-2$K), since $X=\frac13\left(\frac{Y}{100}+200\right)=\frac{Y}{300}+\frac{200}{3}\,.$ For cumulative probabilities, since $n$ is large, we can use the fact that the distributions of $X$ and $Y$ approach the normal distribution, $X$ with parameters $\mu=np=100$ and $\sigma^2=np(1-p)=50$: $ \eqalign{ X &\approx \mathcal{N}\left(\mu=100,~\sigma^2=50\right) \implies\\\\ X&=\sigma Z+\mu=5\sqrt2\,Z+100 \\\\ Y&=100\,\left(3X-200\right)=100\,\left(15\sqrt2\,Z+100\right) \\\\ Z&=\frac{X-100}{5\sqrt2}=\frac{Y/100-100}{15\sqrt2} } $ where $Z\sim\mathcal{N}(0,1)$ is a standard normal variate. So for example, $ \mathbb{P}[X \le x]= \mathbb{P}\left[Z \le z=\frac{x-100}{5\sqrt2}\right] $ $ \mathbb{P}[Y \le y]= \mathbb{P}\left[Z \le z=\frac{y/100-100}{15\sqrt2}\right] $ which one can look up from the cumulative normal function, $\Phi(z) =\frac12\left[1+\operatorname{erf}\left(\frac{z}{\sqrt2}\right)\right] =\frac1{\sqrt{2\pi}}\int_{-\infty}^ze^{-t^2/2}dt$ using the appropriate normalized $z$-score. I think the question in part b is that there are no winnings, i.e. that the cumulative win is nonpositive: $Y\le0$, the probability of which you can accurately estimate using the cutoff $y=0$ and $z$-score from the above formula of $z=-\frac{100}{15\sqrt2}=-\frac{10\sqrt2}{3}\approx-4.714$, to get a probability of about $1.214\times10^{-6}$.

Finally, if you find yourself unable to calculate an exact probability as in part a or c because you can't calculate the binomial coefficient (but you do have access to the CDF of the normal distribution either from a table or a calculator), you can even use the (continuous!) normal approximation to estimate the exact (discrete) probability that, for example, $X=k$ by computing the $z$-scores of $x=k\pm\frac12$ and taking the difference of the standard normal CDF at these endpoints: $ \mathbb{P}[X=k]\approx \Phi\left(\frac{k+\frac12-100}{5\sqrt2}\right)- \Phi\left(\frac{k-\frac12-100}{5\sqrt2}\right) $

0

You need to translate each question into a count of how many wins are required, then you can find the probability of that many wins from the binomial distribution. Let us take $W$ for the number of wins, $L$ for the number of losses. So for the first, to win exactly \$10,000 you have $200W-100L=10000, W+L=200$. This gives $W=100, L=100$. What is the chance that you get exactly $100$ heads? For the second, do you mean the result is exactly $0$ or that the result is less than (or less than or equal) to $0$? The chance of exactly $0$ is $0$, as you would have to have $\frac {400}3$ losses.