minnesotajoe
• Posts: 168
Joined: Dec 18, 2010
February 15th, 2013 at 2:38:00 AM permalink
A set of baseball cards contains 300 cards 1-300.
I currently have 226 of the 300 cards.
Therefore, I am 74 cards short of the set.... obviously.

I'm considering buying a box to help complete the set.
Each box contains 116 cards that are completely random, no one card has higher likely hood than another. There are also no duplicates per box.

If I were to purchase a box of 116 cards:

Question 1
How many of the 74 cards I need to complete the set should I expect to receive from a statistical stand point?

Question 2
What is the probability that of the 116 cards I purchase that I will receive ALL 74 cards I need to complete the set?

Question 3
What is the probability that of the 116 cards I purchase that I will NOT receive a single card of the 74 that I need to complete the set?

Thanks.
boymimbo
• Posts: 5994
Joined: Nov 12, 2009
February 15th, 2013 at 6:26:08 AM permalink
(1) If there are 116 cards in the set, the odds of any one card of being included in the set is 74/300. I would think the expected value of cards you would receive would be 116/300*74 = 28.6.

(2) ZERO. But I don't know how to calculate that.

(3) ZERO (224/300)^116 = 521,549,364,692,073 : 1
----- You want the truth! You can't handle the truth!
minnesotajoe
• Posts: 168
Joined: Dec 18, 2010
February 15th, 2013 at 8:32:58 AM permalink
Thanks. I recently cut back significantly on casino gambling, so now I'm "gambling" on sports cards and storage units (I did storage auctions before the tv shows!!! lol).
Long story short, I can buy a box for about 130\$ that will give me:
2 autograph cards (which vary between .99 USD and literally 2,500 USD with average being 15 USD each)
2 refractor cards which have same value as autos
2 insert cards which are more 3 USD to 70 USD average 10 USD each
116 base cards. -- which if I get the set is worth about 80 USD.. plus I can sell the duplicates.

If I were to get "the worst box distributed" bare minimum I would get ROI is 30\$. I'm debating if it is worth a gamble. Just going for the set alone, not worth it because box costs more than set... but the other six cards having shot at big money.. plus selling duplicates.. Right now debating.

... Also if my box was by chance the 521,549,364,692,073 : 1 shot.. wow
AZDuffman
• Posts: 14070
Joined: Nov 2, 2009
February 15th, 2013 at 9:20:39 AM permalink
Quote: boymimbo

(1) If there are 116 cards in the set, the odds of any one card of being included in the set is 74/300. I would think the expected value of cards you would receive would be 116/300*74 = 28.6.

(2) ZERO. But I don't know how to calculate that.

(3) ZERO (224/300)^116 = 521,549,364,692,073 : 1

My stats knowledge is very rusty but isn't 2 more or less figured the same as 3 then subtracted from 1 or something like that? It looks like a classic "figure the opposite result probability then take the remsinder" kind of problem.
All animals are equal, but some are more equal than others
Ibeatyouraces
• Posts: 11933
Joined: Jan 12, 2010
February 15th, 2013 at 9:27:06 AM permalink
deleted
DUHHIIIIIIIII HEARD THAT!
ThatDonGuy
• Posts: 6406
Joined: Jun 22, 2011
February 15th, 2013 at 10:08:11 AM permalink
Quote: minnesotajoe

A set of baseball cards contains 300 cards 1-300.
I currently have 226 of the 300 cards.
Therefore, I am 74 cards short of the set.... obviously.

I'm considering buying a box to help complete the set.
Each box contains 116 cards that are completely random, no one card has higher likely hood than another. There are also no duplicates per box.

If I were to purchase a box of 116 cards:

Question 2
What is the probability that of the 116 cards I purchase that I will receive ALL 74 cards I need to complete the set?

There are 4.048 x 1085 possible different boxes of cards.
Of these, 9.091 x 1045 of them have all 74 cards you need.
The probability is 1 in 4.452 x 1039.
Quote:

Question 3
What is the probability that of the 116 cards I purchase that I will NOT receive a single card of the 74 that I need to complete the set?

Remember those 4.048 x 1085 possible different boxes of cards?
5.281 x 1066 have none of the 74 cards you need.
The probability is 1 in 7.664 x 1018.
minnesotajoe
• Posts: 168
Joined: Dec 18, 2010
February 15th, 2013 at 11:54:27 AM permalink
Quote: Ibeatyouraces

I used to just go to card shows or to the Ebay Marketplace. Some sellers even have their own store where you can singley buy the individuals you need.

Yea, I saw that too. One person is selling lots of 20 cards each for 11.25 USD with shipping included. It would cost me about 45\$ to complete a set that I could sell for 80\$ :/

On eBay with cards, I try to find people selling lots of 10-100 cards. I look at the best cards, see what they are selling for, then bid such that I could profit 25+ USD by selling them individually (plus get positive feedback).
That and I prospect. I will load up on minor league baseball player autograph cards... then hope they break out.

My thing is, I figure if I buy a box that I will get 25-30 cards towards my base set... the rest of the duplicates I can sell individually (rookies) or as team sets (all Bulls, Lakers, etc) and the box I am guaranteed those 6 SP cards. Some of the autograph cards are selling for serious money.
minnesotajoe
• Posts: 168
Joined: Dec 18, 2010
February 15th, 2013 at 11:56:50 AM permalink
Quote: ThatDonGuy

There are 4.048 x 1085 possible different boxes of cards.
Of these, 9.091 x 1045 of them have all 74 cards you need.
The probability is 1 in 4.452 x 1039.

Remember those 4.048 x 1085 possible different boxes of cards?
5.281 x 1066 have none of the 74 cards you need.
The probability is 1 in 7.664 x 1018.

You're good. I am pretty sure that I will purchase a box in the future. I will post how it works out for me.
onenickelmiracle
• Posts: 8277
Joined: Jan 26, 2012
February 16th, 2013 at 11:49:41 PM permalink
From what I remember from almost 20 years ago, baseball card packs are not really random but only appear to be random. Donruss used to have Diamond Kings inserts and I figured out which pack to buy from the box to get one every time if the box wasn't disturbed.
Gambling on baseball cards is not worth it. The problem with it is the dealers and speculators always pay a lot less than you and once they decide to dump them, you'll never get it back.
I am a robot.
AlanMendelson
• Posts: 5937
Joined: Oct 5, 2011
February 17th, 2013 at 3:38:59 AM permalink
Hold it. All of the card makers have what are called "short prints" which have a limited number. The short prints are mostly the sought after "relic cards" which include pieces of uniform, bats, balls, bases, and of course autographs.

Each year the card companies release a list of their "short print cards."

Read this just for a taste of the problem: http://www.talkingchop.com/2012/11/18/3661292/the-death-of-the-baseball-card-collector

And with companies/services like PSA that grade and rate cards for condition, you have a big problem that one man's vision of a well centered, well printed, sharp cornered card may not be someone else's vision of a well centered, well printed, sharp cornered card.

And most of all, remember that card collecting operates on the "greater fool theory."

You would be better off trying to buy the individual cards you need at below market prices than taking a gamble on boxes and cases off the shelf.
JB
• Posts: 2089
Joined: Oct 14, 2009
February 17th, 2013 at 4:07:09 AM permalink
Quote: minnesotajoe

Question 1
How many of the 74 cards I need to complete the set should I expect to receive from a statistical stand point?

74 * 116/300 = 28.61333

Quote: minnesotajoe

Question 2
What is the probability that of the 116 cards I purchase that I will receive ALL 74 cards I need to complete the set?

The Excel formula =HYPGEOMDIST(74,74,116,300) will calculate this. The result is approximately 1 in 4,452,714,598,049,282,523,719,800,081,875,225,005,019. Don't count on this happening.

Quote: minnesotajoe

Question 3
What is the probability that of the 116 cards I purchase that I will NOT receive a single card of the 74 that I need to complete the set?

The Excel formula =HYPGEOMDIST(0,74,116,300) will calculate this. The result is approximately 1 in 7,664,287,003,599,031,939. Don't count on this happening either.