Why is this 4850 so cheap?

Darkrage

Senior member
Dec 15, 2008
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price looks good, and don't the 4850's preform better than the 9800gtx? I say go for it.
 
Apr 20, 2008
10,067
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Welcome Darkrage!

But yeah, i'm contemplating spending a little more cash for this..

Choices Choices Choices..
 

Denithor

Diamond Member
Apr 11, 2004
6,298
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Here's a MSI 4850 for $130AR with free shipping (making it $8-10 cheaper than the one you found).

And here's an Asus 4850 with aftermarket cooling for $135AR + FS.

Or here's an EVGA 9800GTX+ clocked at 756/2246 for $145AR + FS.
 

SeductivePig

Senior member
Dec 18, 2007
681
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Originally posted by: Denithor
Here's a MSI 4850 for $130AR with free shipping (making it $8-10 cheaper than the one you found).

And here's an Asus 4850 with aftermarket cooling for $135AR + FS.

Or here's an EVGA 9800GTX+ clocked at 756/2246 for $145AR + FS.

I found a 9800gtx+ for $130 AR.

But I'm very interested in that MSI 4850.. do you know what it actually looks like? They are saying in reviews that the card pictured is completely different..

Do you think it looks like this? http://img.neoseeker.com/v_ima...rticleid=2157&image=10
 

SlowSpyder

Lifer
Jan 12, 2005
17,305
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That Asus card for $135 after rebate is probably the best buy. It is overclocked form the factor to 680MHz on the core and 2100MHz on the memory.
 

AzN

Banned
Nov 26, 2001
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They had deal back few days ago 4850 for $115 after rebate.

You have to watch for some of these fans on 4850 and 4830. Lot of them have only red and black wires. That means the fan isn't variable.
 

kmmatney

Diamond Member
Jun 19, 2000
4,363
1
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Originally posted by: Azn
They had deal back few days ago 4850 for $115 after rebate.

You have to watch for some of these fans on 4850 and 4830. Lot of them have only red and black wires. That means the fan isn't variable.


This makes sense - however the third wire should just be for monitoring the fan speed - you can still adjust the voltage to the fans with just 2 wires. You should still be able to set fan speeds, as a percentage of full voltage versus temperature, I would think.