Small statistics correction. The chance of you winning is ~63%.
The possibility of not winning is (999/1000)^1000, which is chances of not winning times the giveaways entered.
Not that it really matters though ;)

Another small statistics correction. The chance of winning in only ONE giveaway out of a thousand giveaways with a thousand participants in each one is 100%. Calculated from (1/1000) + (1/1000) ....... + (1/1000) = 1000*(1/1000) = 1.

I don't know what (999/1000)^1000 calculates, but you don't use multiplication when calculating odds for independent draws.

If you want to calculate the odds by the chance of NOT winning, it would go like this:
1 - ((1/1000) for each giveaway) = 1 - ((1/1000)*1000) = 1 - 1 = 0%.

Lastly, this is a calculation based on average observation. Which means that you will not win on the 1000th giveaway each time you enter 1000 giveaways. It means that if you entered x*1000 giveaways you would win on average x games. (those x games can be on any of the x*1000 giveaways, even the x last ones)

Coin flip example.
I flip one coin twice. The possible results are HT, HH, TT, TH, with T being tails and H being heads. The chance of getting only one heads results is 2/4, aka 1/2, and not (1/2) + (1/2) = 1.

Coin flip continues. The chance of me getting at least one heads result on two coin flips is 3/4. That is calculated by finding the chances of not getting the desired result each time, raised to the number of times, and subtracting that from 1. So in this case the chance is 1-((1/2)^2)=3/4

This does mean that flipping a coin 10 times leaves you with a very small chance of not getting a heads result (less than 1/1000), but it is still possible. Going by the multiplication you are referencing, the chance of not getting a heads result is 1 - ((1/2)*10)=-4
I believe you do see the problem with the above calculation.

Been a while since I did Statistics (and I didn't much care for it even then), but are you sure that is correct?
By my understanding, the odds of winning ONE and only ONe giveaway out of a thousand giveaways with a thousand participants each is:

odds of winning 1 specific giveaway out of the 1000 AND losing the other 999 giveaways + odds of winning ANOTHER specific giveaway out of the 1000 AND losing the other 999 giveaways.... (and so on for a total of 1000 sums)

so it'd be:
(Odds of winning 1 specific giveaway out of 1000 AND losing the other 999 giveaways) x 1000
(Odds of winning 1 specific giveaway out of 1000 x losing the other 999 giveaways) x 1000
(1/1000 x (999/1000)^999) x 1000
~=37% chance of winning 1 and only 1 giveaway out of 1000 giveaways with 1000 entries each.

Assuming that the indipendent draws have the same probability p to be won, the chances to win k giveaways out of n entered can be calculated using the Binomial distribution.

Wow! I have studied all kinds mathematics, calculus and statistics for about 10 years and I really hoped I was done with it.

If it's an unbearable hassle for you and you believe the odds are not worth it - don't do it. The games aren't that expensive anyway. Just enjoy the raffles and don't put so much thought into it :P

Wow, there was some interest in this game after all!

Grynn, you're the lucky winner! Your key has been mailed to you :D

Hitman: Contracts went to mariganza. Enjoy!

Someone please, invite me to GOG.com Gift Club.

My Steam profile: http://steamcommunity.com/id/ZX_Kuroi

You should have read the first post BEFORE posting that...
http://www.gog.com/forum/general/steamgiftscom_gog_gift_club_v3_giveaways/post1

You should also need to read the rules:
http://www.gog.com/forum/general/steamgiftscom_gog_gift_club_v3/post1

Nice giveaway!

http://www.steamgifts.com/giveaway/QaSXG/realms-of-arkania-blade-of-destiny

Hope someone is interested in this. :)

Ohhh, good one. I already have this, but it's a fun game! Thanks for the giveaway!

Looks good and +1 for you.