General Forum

 
The question is inspired by the dim experiences of completing sets with newly found lots.

A 100 card set is missing 3 cards.

You get a lot of 20 random cards from the set, no duplicates.

What is the percentage chance all 3 three missing cards are in the 20 card lot ? 

What size lot is needed to have at least a 50% chance of filling the set ?

Guesses and calculations both welcome.

I could answer  this when I was in HS and had math classes on combinations and permutation.   55 years later, I remember zilch.

0.6% of getting all 3.  I think the calculation is you need 3 / 100 and you have 20 / 100 so you multiply the 2 fractions getting 0.006.  It is either that or it is 20/100 x 19/100 x 18/100 given that for each correct card, you have one less in the random lot to get another (I beleive this is the right way to calculate it).  This gives you a 0.0684% chance.

80 cards for a 50% chance based on the second calcualtion.

It has been 40 years since I did this kind of thing.

The short answer is "VERY LOW" percentage chance if the 20 card lot is truly randomized.

There are over 1.3E+39 possible combinations of 20 cards from the 100 card set. Here's your formula:

P(n,r)=P(100,20)=100!(100−20)!

Then you take the probability of hitting the specific 3 cards from that, and, well, let's just leave it at "very low" probability.

However, with a group of 23 people you have approximately a 50% pro(Click here if you doubt it)bability that two have the same birthday... but that is more random than two specific days in a year. And more random than three specific cards even if the pool is only 100, as opposed to 365. Seems to me like you have a 20% chance of pulling one of the random cards if each card has one percent chance of being pulled and there are no duplicates. How the math works after that I don't know.:)

How that applies to this...im not sure but I thought it was connected. 

Hmmm...let's see...(hits figures on calculator)...and if you add...(writes down new figure)...and divide by...(calculator again)...you've got...

Ah, screw it! (throws calculator, paper, and pencil into trash. Goes to Sportlots and buys the three cards)

The answer is RED. its always red

It's been a while since I have tried my hand at one of these problems, but I think the solution is 0.705% chance of getting all three cards.

That's based on 20/100 to get the first one, then 19/99 to get the second one and 18/98 to get the third.

(20×19×18) / (100×99×98) = 0.00705.

You are in the right church but the wrong pew. The birthday paradox is that you are internally seeking any match, so the likelihood increases as you add more people. In this instance we are looking for specific cards, so it is sort of the inverse of the birthday paradox.