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.

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

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

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

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.

Quote:minnesotajoeA 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 10

^{85}possible different boxes of cards.

Of these, 9.091 x 10

^{45}of them have all 74 cards you need.

The probability is 1 in 4.452 x 10

^{39}.

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 10

^{85}possible different boxes of cards?

5.281 x 10

^{66}have none of the 74 cards you need.

The probability is 1 in 7.664 x 10

^{18}.

Quote:IbeatyouracesI 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.

Quote:ThatDonGuyThere are 4.048 x 10

^{85}possible different boxes of cards.

Of these, 9.091 x 10^{45}of them have all 74 cards you need.

The probability is 1 in 4.452 x 10^{39}.

Remember those 4.048 x 10^{85}possible different boxes of cards?

5.281 x 10^{66}have none of the 74 cards you need.

The probability is 1 in 7.664 x 10^{18}.

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.

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.

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.

Quote:minnesotajoeQuestion 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:minnesotajoeQuestion 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:minnesotajoeQuestion 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.

Wikipedia page on the hypergeometric distribution