0
$\begingroup$

I am recalling a game that people played in a night market in my town long time ago. I am going to do some math to find out what's the chance of winning the game with my limited knowledge of statistics. Here is the game description and my math. It charges 1 dollar to play the game. There is 500 covered boxes. Some boxes has cash prized and others has nothing. Let me assume the following prize structure

  • 3 boxes contain \$10
  • 2 boxes contain \$20
  • 1 box contains \$100
  • 30 boxes contain \$4
  • 20 boxes contain \$5
  • 10 boxes contains \$8
  • 434 boxes have nothing inside

From my text, I find the expectation value as $$ \overline{x} = \frac{3\times 10 + 2\times 20 + 1\times 100 + 30\times 4 + 20 \times 5 + 10 \times 8 + 434\times 0}{500} = 0.94 $$

My understanding is if I play the game $N$ times, each game cost \$1, I will get $0.94 N$ dollars back in average. Now let's consider the standard deviation as

$$ \sigma = \sqrt{\frac{3(10-0.94)^2 + 2(20-0.94)^2 + 1(100-0.94)^2 + 30(4-0.94)^2 + 20(5-0.94)^2 + 10(8-0.94)^2 + 434(0-0.94)^2}{500}} = 4.96 $$

Assuming millions of games played, my book said there will be about 96% of chance to get the outcome falls into $(0.94-2\times4.96, 0.94+2\times 4.96)$. The range is $(-8.89, 10.86).$ It is quite confusing what does that range tell. If I played the game many times, most times I will get zero but some times I will get \$100, \$20 etc. but they are all positive so I don't understand what does the negative number (-8.89) really tells.

My second question is if I have 1000 people to play the game, each play it 10,000 times. I think each people will get in average \$9400 back but actually, someone may get more than \$9400 and some will get less. I want to know how does that mean value distribute. I have a feeling that the mean value will be in normal distribution also. If this is correct, will the standard deviation of the mean value distribution related to $\sigma$ we calculated above?

$\endgroup$
  • $\begingroup$ The distribution of the expected gain is not normal... I think a binomial is more adapt for the problem. $\endgroup$ – N74 Jan 12 '17 at 21:35
  • $\begingroup$ Thanks for your comment. I wonder if the definition of the standard deviation only works for normal distribution or not. If SD definition is general (so work for binomial), what is the significance to have such big SD with small mean value. $\endgroup$ – user1285419 Jan 12 '17 at 23:34
  • $\begingroup$ The difficulty is with the "96%" confidence interval, which assumes a normal distribution. If $Z$ is standard normal, then $P(|Z| < 2) \approx .955.$ That's the 'source' of the '2' in your formula. $\endgroup$ – BruceET Jan 13 '17 at 2:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.