Searching for a probability distribution appropriate for my task

Question

I'm making a game (not important), but I'd like to have real probability distribution function (instead of classical dice notation). I like the normal distribution, but I would like to also shift the odds so to speak to a given side. My crude drawing (not necessarily accurate):

I've tried to shift the function along x-axis, but that isn't it. Because sum must be always 100% (1.0) obviously. So the question is.

Is there some way to make normal distribution behave like this? If not, is there some other parametric function (or distribution), that allows me to do it?

regards,
Kate

Update: Solution in Wolfram Mathematica (for anyone interested in this):

Manipulate[ListPlot[Table[{k, PDF[BetaBinomialDistribution[a, b, 50], k]}, {k, 0, 50}], Filling -> Axis], {a, 1, 3, 0.1}, {b, 3, 1, 0.1}]

Answer 1

I am making a separate answer here because while my suggestion using a beta distribution remains applicable, the added information that the eventual distribution will be discrete makes another approach possible.

I suggest that a binomial distribution with appropriate parameters is suitable and easy to simulate. Recall that if $X \sim {\rm Binomial}(n,p)$ is a binomially distributed random variable with parameters $n$ and $p$, where $n$ indicates the number of trials and $p$ is the probability of "success" for a single trial, then the probability mass function is $$\Pr[X = x] = \binom{n}{x} p^x (1-p)^{n-x}, \quad x = 0, 1, 2, \ldots, n.$$ The mean of the distribution is ${\rm E}[X] = np$, and the variance is ${\rm Var}[X] = np(1-p)$. The selection of parameters is dictated by the desired support. For instance, if the range of generated values is to be from $[1, 8]$, then we choose $n = 8-1 = 7$. Then if the probability mass is to be concentrated on $Y = 6$, we want to choose $p$ such that $np = 6-1 = 5$, or $p = 5/7$. Then we simulate $X \sim {\rm Binomial}(7,5/7)$ and calculate $Y = X + 1$.

To perform the actual simulation, we can simply generate $n$ independent and identically distributed Bernoulli trials, each of which has probability $p$ of "success," then count the number of successes. In the above example, we would generate $7$ random reals $x_1, x_2, \ldots, x_7$ on $[0,1]$, and count trial $i$ as a success if $x_i \le p$.

Now, the remaining issue is that of controlling variance. The binomial model has two parameters but our parameter selection procedure does not allow us to choose the variance and therefore get finer control over the shape of the distribution. To address this, we could go the cheap route and introduce a "pseudoparameter" $k$, which we multiply by $n$. Then let $n^* = [kn]$ be the resulting rounded integer, and bin the simulated values appropriately. This isn't really ideal, however. A better solution would be to use the beta-binomial distribution. But that is probably best left to another post. For more information, see the wikipedia page http://en.wikipedia.org/wiki/Beta-binomial_distribution

Answer 2

From the picture that you've drawn, it seems that you want the distribution to have finite support--in other words, you want the range of the random values that you generate to be in some fixed interval. In that case, shifting a normal distribution is not going to work, as you noted, because then you "chop off" a big chunk of probability mass if you shift the mean.

To correct this, you could renormalize the normal density, but to be honest, this is added computational complexity, and I think you would be better served by dealing directly with a more appropriate choice of parametric distribution. Because your model seems to need finite support, a natural choice would be the beta distribution, whose probability density function is defined by $$f_X(x) = \frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)} x^{a-1} (1-x)^{b-1}, \quad 0 \le x \le 1,$$ where $a,b > 0$ are parameters that you set that determine the shape of the distribution. To pick where the distribution has the "hump," you want to use the mode, which is $\frac{a-1}{a+b-2}$. So for example, say you want the highest density to occur at $x = 2/3$. Then you would solve the equation $$\frac{2}{3} = \frac{a-1}{a+b-2},$$ which would give you $a = 2b-1$. That doesn't uniquely determine $a$ and $b$ yet. The next decision you want to make is how narrow the peak at $2/3$ will be. The general rule is that the larger the values of $a$ and $b$, the more narrow and tall the peak, and the greater the probability that the random number generated will be close to the corresponding $x$-value. So if you want less variance, choose something like $b = 6$, $a = 11$. If you want more variance, choose $b = 2$, $a = 3$.

If you want the peak of the distribution to be at $x = 1/2$, then obviously choose $a = b$.

Also, note that when $a = b = 1$, you get a uniform distribution, which is nice.

It should be pretty obvious that if you need your random number to be distributed on some other interval besides $[0,1]$, then you can simply apply an appropriate linear transformation $y = (d-c) x + c$ to get it in the interval $[c,d]$.

Finally, $a$ and $b$ are not required to be positive integers. If they are positive integers, then the gamma functions in the front reduce to factorials, since for positive integer values of $z$, $\Gamma(z) = (z-1)!$; this can be computationally easier. But $a$ and $b$ can be any real number, if you need them to be.