2 for 1 pricing
BloodyIron
Join Date: 2009-11-09 Member: 69321Members, Reinforced - Shadow
Join Date: 2009-11-09 Member: 69321Members, Reinforced - Shadow
Comments
Except that your comic is a perfectly reasonable request. Companies take returns, what's the difference between returning it, rebuying it, or just asking for the difference? Less paperwork for the business.
I prefer reading about it on the forum to be perfectly honest
You can complain to forumgoers who have no power whatsoever, or you can directly ask the developers themselves, who have all the power in the world. They'll even answer you.
Gee.... what should we do....
Here's a suggestion:
Say that X = $35, the price of the game. If you bought the game today you'd be getting 2 copies of the game for X, or paying X/2 for each. In your situation, if you bought another copy of the game today, you'd have 3 copies of the game for 2 * X or 2/3 * X each. If you bought 100 copies, you'd pay X * (101) / 201. With n copies, it's X * (n + 1) / (2n + 1). The limit as n approaches infinity is X/2, or the same price you would have paid if you had only waited a few days!
So clearly the solution to your problem is to buy as many copies as you possibly can while the holiday special lasts.
In all seriousness, your situation is something we always consider when approaching an offer like this. Will the people that bought the game yesterday feel cheated? Our conclusion is that with Steam constantly have nutzoid deals, people are used to this sort of thing and they won't. And of course the alternative is to not ever have deals which seems like a bad solution. But if you feel differently, you can e-mail us and we'll put your case before our Review Board for examination.
Say that X = $35, the price of the game. If you bought the game today you'd be getting 2 copies of the game for X, or paying X/2 for each. In your situation, if you bought another copy of the game today, you'd have 3 copies of the game for 2 * X or 2/3 * X each. If you bought 100 copies, you'd pay X * (101) / 201. With n copies, it's X * (n + 1) / (2n + 1). The limit as n approaches infinity is X/2, or the same price you would have paid if you had only waited a few days!<!--QuoteEnd--></div><!--QuoteEEnd-->
You've been programming to much for today I think there Max! Better be careful, or you'll start speaking binary <img src="http://members.home.nl/m.borgman/ns-forum/smileys/biggrin.gif" border="0" class="linked-image" />
Say that X = $35, the price of the game. If you bought the game today you'd be getting 2 copies of the game for X, or paying X/2 for each. In your situation, if you bought another copy of the game today, you'd have 3 copies of the game for 2 * X or 2/3 * X each. If you bought 100 copies, you'd pay X * (101) / 201. With n copies, it's X * (n + 1) / (2n + 1). The limit as n approaches infinity is X/2, or the same price you would have paid if you had only waited a few days!
So clearly the solution to your problem is to buy as many copies as you possibly can while the holiday special lasts.<!--QuoteEnd--></div><!--QuoteEEnd-->
My calculus teacher would love you.
There have been plenty of games on steam I have bought where one week later or a day later it goes on sale for half off. Same thing with computer hardware... it goes on sale without you knowing and you may buy something before the sale.
It sucks but if they gave you the deal we all should get the deal. I bought the game a month ago and didn't know about the holiday sale. What's the difference between one day and one month in this situation?
By selling all 200 copies at $28, your net profit will be (28*200) - (35*101) = $2065. More than enough to buy a brand new rig that can run NS2 at the highest graphics settings.
Say that X = $35, the price of the game. If you bought the game today you'd be getting 2 copies of the game for X, or paying X/2 for each. In your situation, if you bought another copy of the game today, you'd have 3 copies of the game for 2 * X or 2/3 * X each. If you bought 100 copies, you'd pay X * (101) / 201. With n copies, it's X * (n + 1) / (2n + 1). The limit as n approaches infinity is X/2, or the same price you would have paid if you had only waited a few days!<!--QuoteEnd--></div><!--QuoteEEnd-->
Thank you for the bad memories...
I hate that Calc 1/2 has to be a part of the CS degree :(
On topic: To me this is just one of those things. You got what you paid for, you were fine with the price and amount of copies prior to the holiday special. If lenience is given, where does it stop? 1 day before? 2? so on...
It does suck, I've had it happen to me, as I'm sure everyone else has. however, I would much rather have these deals and get miss them from time to time, than to never have deals offered.
I got a great deal on it... Bought it over a year ago when it was first offered so feels like a free game now =)
By selling all 200 copies at $28, your net profit will be (28*200) - (35*101) = $2065. More than enough to buy a brand new rig that can run NS2 at the highest graphics settings.<!--QuoteEnd--></div><!--QuoteEEnd-->
Dangit! Stop stealing mah entrepreneurial ideas!
"Healthy competition" is what "they" call it these days :P
I will make two friends very happy
Say that X = $35, the price of the game. If you bought the game today you'd be getting 2 copies of the game for X, or paying X/2 for each. In your situation, if you bought another copy of the game today, you'd have 3 copies of the game for 2 * X or 2/3 * X each. If you bought 100 copies, you'd pay X * (101) / 201. With n copies, it's X * (n + 1) / (2n + 1). The limit as n approaches infinity is X/2, or the same price you would have paid if you had only waited a few days!
So clearly the solution to your problem is to buy as many copies as you possibly can while the holiday special lasts.<!--QuoteEnd--></div><!--QuoteEEnd-->
<!--quoteo(post=1818205:date=Dec 22 2010, 12:42 PM:name=senor_hybrido)--><div class='quotetop'>QUOTE (senor_hybrido @ Dec 22 2010, 12:42 PM) <a href="index.php?act=findpost&pid=1818205"><{POST_SNAPBACK}></a></div><div class='quotemain'><!--quotec-->Based on Max's calculations, if you buy an additional 200 copies now at the price of 100 copies, you can sell those extra copies later at 80% of the original price, ie $28, and make a profit.
By selling all 200 copies at $28, your net profit will be (28*200) - (35*101) = $2065. More than enough to buy a brand new rig that can run NS2 at the highest graphics settings.<!--QuoteEnd--></div><!--QuoteEEnd-->
I love this thread.
Well done....
Well done... On getting got.