Thought I should wait, but then decided otherwise.
The DT lottery algorithm programmed into the UR mainframe is becoming a bit predictable. If you draw a frequency distribution curve with the already awarded prices, we can find most of them huddled around two magic numbers.
I have found two. 33 and 94. Meaning if you are placed around these two places in DT, you stand a better chances of winning a prize in DT, the former being better (less standard deviation in the case of latter).
Any thoughts on this?
Hmmm, I was going to run it through spss last week, but will move that to next week due to busy period ( to try and unraffle the pattern)
few questions that I am wondering about are concerning the method:
- When the prices have been distributed, you see the best rank of the winner displayed, right? does this mean everyone that ends up in the top 150 (regardless of the amount of times a day) only participates with his best position? or gets one extra chance for every time he hits top 150? (if the latter is true, the prices won do not necessarily have to be won on the displayed rank let's say position 33 if that is simply the best rank the winner of the draw, he could had won it on a worse rank in a other dt that day).
- if the latter is true about the previous question, so ending up in top 150 a day multiple times does give you more "lottery ticket" can a player that plays 10 dts a day win multiple prices? or are his tickets withdrawn when he wins the first price?
I understand that by ending multiple times in top 150 he always enlarges his chances to win a price regardless by which of the anwsers of the last question is correct, but their is a huge difference in the amount of chance you gain by playing more dt's across the both explanations (option 1; more times in top 150 -> more tickets -> more chance | option 2: more times in top 150 -> only 149 other players that have a chance in the price draw -> very slight increase of chance of winning)