Die Numbers

[zookeeperz 470]

Hi folks I am assuming this subject has probably been broached before . How to ascertain  How many of each die number was allocated to each coin. I was trying to think of a logical formula .

Example for 1873 sixpence there were a total of 90 different die numbers used that are recorded. So mintage of 4,594,733 /90 = 50,000 (rounded to make it easier) different die combinations of which 79 die numbers were used on reverse A again rounded for ease of explanation that equates to 82% of the total die numbers struck are reverse A  which is 40,000 (simplified) Which leaves 10,000 coins struck between reverse B with 6 die numbers and reverse C with 5 die numbers So something like 5,500 for rev B and 4,500 for rev C. The only problem is they would of had to use every die number the same amount of times throughout the mintage year but did they? Who is to say the used more of one number than another Which I suspect would account for the missing die numbers?  Any thoughts ? it's a very complex formula to try and fathom  and probably impossible. Only anecdotal evidence would be a strict keeping of all sales records for every die number and that would still leave those coins in private collections.

[Nick 329]

You just cannot tell how many coins each reverse die produced (other than by estimates based upon previous sales).  During that mid-Victorian period they struggled to get quality steel and therefore some dies would have broken having struck just a few coins and others perhaps as many as 100,000.

[1949threepence 3,367]

Impossible to tell, but I wouldn't mind betting that if you built up a big enough collection of Victorian shillings, for example, you might see a trend towards some die no's having been used much more than others. 

No idea whether that's ever been tried.