I see my initial post as an exercise in analytical thought utilising certain mathematical concepts useful and pertinent to poker. My stakes hardly matter - what does matter is practicing and learning to think in ways that are appropriate to poker. I posted it deliberately in the beginner's forum because I figure any poker player who isn't a beginner is familiar with the principles of unexploitable play and has his own way of producing random or pseudo-random numbers on the spot - I thought beginners might find this type of thing useful or entertaining to consider. It's flexing the mathematical and analytical muscle in a way that is not (let's be clear) immediately useful, but still revolves around poker concepts.

Additionally writing it down was a way for me to develop the idea in my own head, and having written it I saw no particular reason not to share it. I've worked a little more with it and come to the conclusion that this pseudo-random number generator has a number of significant problems which make it unusable for the intended purpose. I'll be happy to continue the braindump if anyone has any interest, but as that may not be the case I won't pursue that. And as badgers is correctly pointing out it's not the main thing I should busy my head with if I'm trying to improve my play - and it really isn't the main thing anyway. This was just a stray thought.

In a completely unrelated thought exercise I saw an example of the outcome of an odds calculation and thought it was interesting - especially for beginners who may not be aware of combining odds in that way. I was vaguely aware that such odds maths should be possible but I hadn't seen an example of it or put any serious thought to figuring it out.

An assertion is made that a raise can have been made with a hand that has you dominated where you are a 10-1 underdog, or a semi-bluff where you are a 3-1 favourite, and since the pot is giving you 2-1 odds you should call if the villain semi-bluffs a third of the time. Where'd he get that from I asked myself - I discovered the answer by adding up the odd components associated with me winning vs the odd components associated with me losing: 10 + 1 + 1 (me losing) to 1 + 3 + 2 (me winning) => 12 to 6 or 2 to 1. That must be where the observation comes from that he must semi-bluff a third of the time (2 to 1 odds against) for me to have a chance of winning. I could be very wrong in assuming this is where the assertion comes from, but from how I'm seeing it, it fits.