Jump to content


Photo

How is Avg found?


  • Please log in to reply
2 replies to this topic

#1 stage4mtbk

stage4mtbk

    Member

  • Silver Members
  • PipPip
  • 13 posts

Posted 01 February 2012 - 11:58 AM

Hello, how is the avg price found? I can't figure out if it's based on time (6 months/year) or the past 10/20 sales etc. Just wondering. Thanks

#2 Bobby

Bobby

    Master Member

  • Root Admin
  • PipPipPipPip
  • 478 posts
  • Gender:Male
  • Location:At my computer
  • Interests:Being with my girls.
  • Collection:Was a heavy Pre-war collector, mostly Hall of Famers type cards.

Posted 01 February 2012 - 12:07 PM

Hello, how is the avg price found? I can't figure out if it's based on time (6 months/year) or the past 10/20 sales etc. Just wondering. Thanks



Based off the sales during the past year if no card in the grade has sold in that time frame then it is the last sales record. The average depends on how many sales records their are and more then 3 the high and low are disregarded and the remainder are averaged.
Bobby Binder
VintageCardPrices

#3 patriots74

patriots74

    Advanced Member

  • Limited Members
  • PipPipPip
  • 47 posts

Posted 03 February 2012 - 07:18 PM

Based off the sales during the past year if no card in the grade has sold in that time frame then it is the last sales record. The average depends on how many sales records their are and more then 3 the high and low are disregarded and the remainder are averaged.


Why would the High and Low be disregarded when more than 3? The listed sales will only include auctions correct?, BIN are being tossed. So an auction is an auction.

I am just curious, since its basically just splitting hairs, and would probably avg roughly the same, I mean data is data correct?

It may be better to consider that if your gonna toss out the high and low, then completely remove that data and not make it visible... Because it is confusing to people... Or just leave it as is and avg the sales...


Thanks in advance for your response.
1951 Bowman




1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users