Below are two "free-body diagrams" of 2 pinned bridges, to help understand the effect the various forces and assumptions. For both, the "free-body" is defined as including the bridge, saddle, most of the pins, and bit of string between the saddle and the bottom of the bridge (since all these are inside the free body, any forces inside them are internal and don't show up on the diagram). The free-body does not include any of the top, or the bit of string and ball-end below the bottom surface of the bridge.
I assumed that the string tension force is the same at both ends; I think this is fairly reasonable after the guitar has been played a while, but it isn't a critical assumption.
The left diagram shows the forces that would occur if everything were perfectly rigid. The string tension tries to tip the bridge counterclockwise; this concentrates a compression force on the front edge (Nfront). To understand whether the bridge could tip or not without glue (but with a shear stop), first set Nrear=0, then look at the clockwise and counterclockwise torques acting around the black dot at the bottom left:
Clockwise torques = T*b (Nfront and Shear go right through the black dot, and we're not relying on Nrear at all)
Counterclockwise torques = T*h.
So it will tip if T*h > T*b, or if h > b. Nice and simple!
Unfortunately, as Chuck noted, a real bridge, top, and bridge plate are not rigid -- they flex. The right diagram is much more complex, with the surface forces replaced by unknown pressure distributions (the diagram shows linear but that's a simplification). Of particular importance is the new force in blue: "R", which I just thought of tonight while making the diagrams. To understand R, first go back to the
rigid case and think about the string's ball-end force that is applied to the
bottom of the bridgeplate (not in the diagrams): this force would be reacted all the way around to Nfront and Nrear, which is dandy to prevent tipping. But with a flexible bridgeplate, most of that ball-end force applied to the plate would be reacted
right back through the plate and top to the bridge, just in front of each pin hole. So, R*b mostly cancels the benefit of T*b holding the bridge down. Bummer. From this, it would appear that pinned bridges probably only have a small advantage over pinless, as far as tip prevention goes.
If the origin of "R" seems fuzzy, think of some sheetmetal with a hole, a bolt through the hole, and a nut on the other side. Now tighten the bolt and nut to (pointlessly) clamp the sheetmetal. The bolt shaft is in tension (like the string), the nut below the sheetmetal is pushing on the bottom of the sheetmetal (like the ball-end). That nut force is transmitted right through the sheetmetal to the head of the bolt. Unless the sheetmetal is inches thick, that compressive reaction force is very localized around the bolt shaft itself. Similarly, unless the bridgeplate is very thick, most of the ball-end force will be transmitted back to the bridge via a small spot by the pin hole.
