# Example for a simple game of chance where the average value converges fast to the expected value

I want to introduce the concept of expected value (for a discrete random variable, finite case) in a high school class. My first idea was to start with a simple game of chance and let the students play it. Suppose there are 25 students. If everybody plays the game once I have 25 outcomes which I could take to calculate the mean value of profits for this game. Then I want to discuss it and go over to calculate the probabilities of the game and then introduce the concept of expected value.

Throw a fair dice twice (= two random integers $i$ and $j$ between 1 and 6). If $i = 4$ and $j = 4$ you win 4\$, if exactly one of them is 4 you win 2\$ if $i \neq 4$ and $j \neq 4$ you lose 1\$. The expected value for this game turns out to be -0,03\$.

However if you simulate this by playing the game 25 times you never end up with a mean value close to this, even if I let each student play it two times, such that I get 50 values it is most times far away from the mean value. I simulated it with a little python script and noticed that you have to play it at least 1000 times to come close to some extent to the expected value, so this example converges too slow for my purpose.

So is there another game (of similar complexity) which converges very much faster and which is such that I can play it with the students as described above before discussing it?

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Joke answer: toss a two-headed coin. More serious: convergence will always be slow if the variance is not tiny, and the only way to a tiny variance is close to the two-headed coin. – André Nicolas Dec 1 '12 at 17:47

Consider the weak law of convergence of the occurence frequencies $k_n$ of an event $A$ to its probability in $L^2$--fomulation: $$\|k_n - P(A)\|_2 = \frac{1}{\sqrt n}\cdot \|1_A-P(A)\|_2,$$ where $$\|1_A-P(A)\|_2 = \sqrt{P(A)\cdot (1-P(A))}$$ is the standard deviation of the atomic experiment, that gives $1$ by occurence of $A$ and $0$ otherwise. That means that the relative (!) error $$\frac{ \|k_n - P(A)\|_2 }{P(A)}= \frac{1}{\sqrt n}\cdot \sqrt{\frac{1-P(A)}{P(A)}}$$

That means $L^2$--convergence is faster, the closer $P(A)$ is to $1$. So in most trials you will observe better convergence (that means faster convergence of the relative errors) with events that are more likely.

Similarily, to find a fast convergent example of the arithmetic means of successive samples of a random variable $X$, you have to keep $$\frac{\|X-E(X)\|_2}{E(X)}=\frac{\sigma(X)}{E(X)}$$ as small as possible.

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Thanks. Do you have some simple examples which answer my question? Why do you need $L^2$ convergence in the discrete, finite case? – user51379 Dec 1 '12 at 19:57

As you say, the expected value is $-\dfrac{1}{36} \approx -0.02778$. This is rather small compared with the dispersion of the outcomes.

The variance of a single game is $\dfrac{2915}{1296} \approx 2.24923$, so the variance of the mean of $25$ games is $\frac{1}{25}$ of this, about $0.08997$, and the standard deviation of the mean of $25$ games is the square root of that, about $0.29995$; the standard deviation of a single game is just under $1.5$.

With the standard error of the sample more than ten times the absolute value of the expected value, it is not surprising if the sample often seems not to give a good estimate of the expectation. Even looking at the mean of $2500$ games, the standard deviation would be bigger than the absolute value of the expected value.

This is one of the reasons that casinos attract customers despite the small negative expectation. A substantial number of customers can walk away having beaten the house, even if they play several games.

If you want apparently faster convergence, you should start with a more biased game. If each game is independent then convergence will still be proportional to $1/\sqrt{n}$, but the constant of proportionality will be different. A example will look as if it has better convergence if $\frac{\sigma(X)}{|E(X)|}$ is smaller. In your particular example, for a single game, it is close to $54$.

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I did something similar to this just today.

Roll three 6-sided dice. If at least one 6 is rolled, you win \$1. If not, you lose \$1. This is a nice one because you may be able to "convince" students that the game is fair: the probability of getting a 6 on one die is 1/6, there are 3 dice, and $3 \times 1/6 = 1/2$. But in fact the probability of winning is only about 42%.

I had 13 groups and asked each of them to play 10 times, and aggregated the 130 total results. (I believe there were 52 wins, about 40%.) With this much data the standard deviation is about 4.3%, so it's very unlikely you will be anywhere close to 1/2, and you can with high confidence reject the null hypothesis that the game is fair. This kind of parallelizing is a nice way to gather a lot of data fast.

This is similar to the carnival game chuck-a-luck and its relative sic bo. There are other payoff schemes possible, which generally still result in a quite unfavorable game.

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