Predicting isotope patterns in mass spectrometry
The first step is to write all the possible combinations of H atoms: $$\mathrm{H^1H^1~~ H^1H^2~~ H^1H^3 ~~H^2H^1 ~~H^2H^2~~ H^2H^3~~ H^3H^1~~ H^3H^2 ~~H^3H^3}$$ Now it is important to note that each of these parings have an equal probability of forming and exist in equal quantities in the mixture. Their molar mass are respectively: $$2, 3, 4, 3, 4, 5, 4, 5, 6$$ So there is $1$ way to make $2$, $2$ ways to make $3$, $3$ ways to make $4$, $2$ ways to make $5$ and only $1$ way to make $6$. Since each way has an equal probability of occurring, then the ratio of the molar masses should be in the ratio -  $~1:2:3:2:1$. 

Hence for peaks of mass = $2, 3, 4, 5, 6$ ratio of heights = $1:2:3:2:1$

The reason why there is no peak for mass $1$, I am not too sure. However I think it is because if the hydrogen molecule was to break up, it would form a proton and a H radical. Both of these are extremely reactive and unstable and will probably just get lost in the machine and never actually get recorded. This could make sense as the mass spectra for $\ce{HCl}$ or $\ce{HBr}$ doesn't include a peak for mass $1$. Also in mass spectra for hydrocarbons, such as heptane, the smallest mass peak recorded is usually 29 which is the ethyl group. Therefore any thing lower than that (such as the methyl group) is probably too unstable.