A Budgetary "What Should I Do" Question
 

Once a year I participate in a live-action D&D-like game. (It's not a LARP though.) Part of the fun involves the use and collection of tokens which can aid a player within the game. Most tokens represent items your character wears and/or uses like armor & weapons. Tokens come in four rarities: common, uncommon, rare, and ultra-rare. The rarer the token, the greater its relative power/usefulness. The full set of tokens changes every year and are never sold again (though there are after-market opportunities, such as eBay.) Tokens come in packs of 10 and cost a dollar per pack. Each pack has 7 common, 2 uncommon, and 1 rare. However, 1% of the token packs contain one ultra-rare in place of a common. Some ultra-rare tokens form sets. If all the items in a set are used an additional bonus is granted.

I have a budget of $100 to spend on tokens. The company that runs the game/sells the tokens has a special deal. If you buy 25 packs, you get an extra 5 packs free and your choice of an ultra-rare. (You might get an additional UR by random chance in the packs.) If you buy 100 packs, you get 20 free packs, four URs of your choice, and a special ultra-rare (one might call it an ultra-ultra-rare) that is only available via this offer. This UUR is part of a seven-year set that, when complete four years from now, will be über-powerful. I have the first two parts of this set. (FYI, it's called The Rod of Seven Parts, a famous D&D artifact.)

Each year there are 20 different URs. This year there are three which are part of three different sets that I "need" to complete. (I have all the other parts of the sets.) There are three other URs that are way freakin' cool that I also covet, not to mention segment three of the Ro7P.

After this year's tokens are shipped in April (we're in the pre-order phase right now), tokens will start showing up on eBay. Current-year URs generally sell for $10-$20 (since for $25 you can pick what you want and get a bunch of less commons to boot.) Previous year's UR prices are a lot more volatile and often go for more money, depending on usefulness. Ro7P segments have also been seen on eBay and while they too range in price, tend to be spendy.

So here's my dilemma. Do I:

• Blow my entire budget on the 100 pack offer, select four URs, guarantee myself segment three of the Ro7P, but miss out on the other two URs that I want. (but very likely get at least one random UR which I may be able to trade for one I want)
• Buy 25 packs, pick one UR and hope I can get the other six tokens I want on eBay.
• Buy 50 packs, pick two URs and hope I can get the other five tokens I want on eBay.
• Buy 75 packs, pick three URs and hope I can get the other four tokens I want on eBay.
• Don't pre-order anything and hope I get really lucky on eBay.

This sounds like a really complicated word problem. :(

Do you use the tokens after the weekend or do they just sit in a box? And how much more would you spend buying the remainder off ebay. That's how I would make the decision for myself.

Is this a real world situation or a problem presented just to see a mathematical solution?


Since you give an eBay price of $10-$20 for the eBay purchases won't these two options almost certainly exceed your $100 budget?

Option 1: $50 + $50 to $150 (over budget unless you get lucky on all auctions)
Option 2: $75 + $30 to $60 (over budget unless you get the auctions at less than your estimated minimum.

So if you're estimates are solid and your budget firm those don't seem options likely to lead to happiness. Since you can't recover from failing to obtain the three R07Ps and you're also in progress on the UUR and you don't indicate any way of getting that other than qualifying to buy it I would say the best option is to just buy the 100 packs.

You're guaranteed continuing your R07P sets.
You're guaranteed continuing your UUR set.
You have a fair chance of getting another UR (though while "probably" is correct it is only a 63.4% chance of getting one randomly when purchasing 100).

The other options either do not guarantee much and have the potential to be wasted saving if eBay costs are higher than projected.

The tokens can be/are used year after year. Some of them are turned in when used (e.g., a potion that is "drunk") but things like weapons & armor are permanent. However, during the year they just sit in a box.
It's very much a real situation.The 75 pack option was just included for completeness sake, but I had contemplated the 50. I hadn't done the math on average eBay prices, thanks for pointing that out.

By the way, if you want a sense of your odds of getting extra random URs by buying packets. Assuming they are truly randomly added to packets (since your purchase will almost certainly be consecutive packets)

For 100 packets:
36.6% chance you get no random URs.
37% chance you get one.
18.5% chance you get two.
6.1% chance you get three.
1.5% chance you get four.
Less than 1% chance you get more than four.

So add to option one the fact that you also have a 25% chance that you'll get enough extra random URs that you can try to trade for the two other ones you want (though aren't so critical) but didn't get through your purchase.

Alex

You have WAY too much free time on your hands.

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.

I think I just LARPed in my pants.

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.

Someone pass me a mop. My brain has melted and is dripping out of my ears.