« And the winner is… | Home | How Do We Beat the Bitch? »
I Have Met the Enemy And He Is Me
By Corvus | December 4, 2007
In a rush to be done with the Witcher chapter of my life yesterday morning, I scanned through the 60 comments in the giveaway thread and tried to sort through the comments of people who did and did not want the game. I pulled out 16 names and ended up not including Red Bull, the very first commenter.
Sheesh!
Then, not thinking clearly, I grabbed two 8 sided dice to make my random selection. And here’s where the “problem” with being completely open about process comes in… people immediately point out that I’m an idiot. Not only does that remove the possibility of coming up with a ‘1′, but rolling two 8 sided die means certain numbers, like 10, stand a statistically higher chance of being rolled than others, like… well, all of them. These are all things I knew… but, for whatever reason, didn’t think about in regards to the giveaway.
Have I mentioned that my entire RPG system has been designed to make math as non complex as possible? Well, I perhaps there’s more than one reason for that.*
Anyway… hardly an auspicious beginning to my week, not to mention for my thoughts about giving away games as a regular feature. My self critical voice kicked into high gear and pestered me for the rest of the day. I do pretty well at avoiding that particular album, but it’s still there in the mix, waiting to jump to the top of the queue and play havoc with my brain.
So, this morning, I put Red Bull at the top of the list and produced two more random numbers using Python’s randint method.** This method (randint(a,b)) generates a random integer where a < = X <= b. Here's a screenshot:
I’m using the same list as yesterday, with Red Bull added at the top and bookamp scratched out. This means that Rampant Coyote at position 8 and Red Bull in position 1 are the consolation prize winners. I’ll be contacting them via email (and in order) to see which prize they’d like. Once they’ve made their selections, I’ll post which prizes they selected.
Now, if you’ll excuse me, I have to leap immediately into action. I somehow managed to completely forget that I have a dental appointment today, which is going to eat up a few hours of productivity and further vex my plans for the week.
*The other being that during play, I want the math not to intrude on the storytelling aspect of the experience
**Python?! I thought you were using Java these days, Corvus. Hm… yeah, there’s a post about that lurking under the surface of my awareness too.
Tagged:game giveaway, the witcher. | 5 Comments »
December 4th, 2007 at 8:40 am
I’m often surprised at how many programmers do not pick up on the bell curve function in multiple dice operations.
My cRPG designs often use bell curve distributions, produced by adding multiple random numbers together in the style of a multi-die random evaluation. Yet I often have to explain to programmers that we cannot substitute a single random number generation step – because I want to use the bell curve to give me more or less frequent results.
Tangential to your post, but I thought I’d share.
December 4th, 2007 at 9:39 am
Well then. I was using this as evidence that I had not enough programming in my background, but you’ve just knocked the legs out from under that idea.
That actually makes it easier to chalk it up to temporary oversight. Which helps. Thanks!
December 4th, 2007 at 10:34 am
You’ve been very good about the whole thing. It would have been very easy to say “I’m giving a game away for free, I’ll do it whatever way I feel like!” Instead, you did your best to make it right.
December 4th, 2007 at 10:38 am
Thanks, Red Bull. In person I usually come across as so confident that people believe I’d be unwilling to admit when I’m wrong or when I’ve dropped the ball.
The fact of the matter is, I usually love it when I’m called on things. How else would I grow?
Oh, and as soon as RC gets back to me with his prize choice, I’ll be sending you an email with the list of prizes, RB!
December 16th, 2007 at 1:22 pm
“… but rolling two 8 sided die means certain numbers, like 10, stand a statistically higher chance of being rolled than others, like… well, all of them.”
Except for 9, which is the number that is most likely to be rolled. Much like 7 is the one most likely to be rolled if you roll two 6 sided dice. You take the lowest possible number and add it to the highest possible number to get the number with the most possible combinations. At least for rolls with two of the same die.
I hope you love being called on this small, unimportant thing.
(Hey, it’s still on the front page! It’s not too late!)