[Fr. Frank]

see what percentage 1/4 inch is for the 15 1/4....double that since it's blade area

Actually, that's pretty close. To simplify it greatly, since the propeller diameter is not full engagement, ie not a perfect disk, you take the percentage diameter difference and multiply it by the percentage of area of a disk the prop blades actually cover , usually between 70 and 80%and then multiply that times Pi. That's your effective differential of thrust.
The hub on a V6 outboard, (except Honda), is approximately 4.25 inches, giving an effective disk area on a 15.25" prop of 11". Increasing that by .25 inches gives you an thrust differential of of .25 x 11 x .75 x 3.1415 = 5.4% thrust differential.
In other words, an increase in diameter from 15 to 15.25", assuming all other factors were unchanged, would give the effect of a same diameter increase in pitch of about 5 1/2 %. If your propeller was 20" pitch, it would be like adding an inch of pitch to 21"

This greatly over-simplified process does not take into account things such as rake, pitch/rake progression, blade shape, propeller ventilation, slippage, angle of thrust, surface tension, depth of thrust, etcetera.

I used to have an MS Excel macro that would calculate all that, but have lost that program.