A Budgetary "What Should I Do" Question

[SzczerbiakManiac]

Do those numbers change significantly if instead of 100 packs it's 120? Cause that's how many I'll be getting since I get 20 additional free. It's not going to change my decision (I'll get the 100+20 pack offer), I'm just curious.

[Alex]

Originally Posted by SzczerbiakManiac

Do those numbers change significantly if instead of 100 packs it's 120? Cause that's how many I'll be getting since I get 20 additional free. It's not going to change my decision (I'll get the 100+20 pack offer), I'm just curious.

Yes, you're odds of getting random URs with 120 packets are:

Odds of:
0 random URs: 30.29%
1 random UR: 36.14%
2 random URs: 21.68%
3 random URs: 8.67%
4 random URs: 2.6%

So with 120 you'll have about a 1 in 3 chance of getting the 2 extra random URs you'll want to trade and get all the desired URs.

By the way, the math here is relatively straightforward if you want to track it down and do any other similar problems. Look up Poisson Distribution for a fuller explanation but the formula is:

X = (λ^k)(e^-λ)/(k!)

Where:

X = The odds of a specific number of outcomes in the given number of tries.
λ = The expected number of outcomes in the given number of tries (so if the odds are 1 in 100, you'd expect 1.2 of the outcome in 120 tries).
k = The number of of desired outcomes you're interested in (2, 3, 4, or 5 in this case)
e = The mathematical constant (using 2.71828 will do well enough for government work)
k! = the factorial of the number of desired outcomes. Essentially multiply that number by every lower integer down to 1 (so 2! = 2x1; 3! = 3x2x1; 4! = 4x3x2x1, etc.)

Or, you can follow this link and replace the variables with your values and click submit to have Google do the math.


ETA: To reiterate one big caveat, these numbers really do rely on the URs being randomly distributed through all the packets. I do not know anything about the methods here but "random distribution" tends to be very difficult to achieve (if they even tried for true randomness over the entire population) and the form of failure can make those numbers dramatically wrong. For example, if they group the packets into groups of 100 and then put one UR into a random packet from each group your distribution curve would essentially be limited to either 0, 1, or 2 packets when you buy 100.

[SzczerbiakManiac]

Originally Posted by Alex

Odds of 2 random URs: 21.68%

So with 120 you'll have about a 1 in 3 chance of getting the 2 extra random URs

A 1 in 3 chance of getting two URs? What am I missing? It looks like the odds are closer to 1 in 5.

[Alex]

Originally Posted by SzczerbiakManiac

A 1 in 3 chance of getting two URs? What am I missing? It looks like the odds are closer to 1 in 5.

That's a 21% chance of getting exactly 2 URs. An 8% chance of getting exactly 3 URs, etc.

So a total chance of about 33% chance of getting 2 or more URs.