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Can't figure out the math to this iphone game..

Blogs > FalconPunch
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FalconPunch
Profile Blog Joined November 2010
United States127 Posts
Last Edited: 2013-03-14 02:00:00
March 14 2013 01:48 GMT
#1
Hey guys,

There's this iphone game I've been playing that keeps bothering me because I don't know if the math makes sense to me.

Here's the problem:

For $5, I have the chance of getting one of the following items:

Common - 55%
Uncommon - 30%
Rare - 10%
Epic - 5%

So the way I see it.. for $5, I have a 5% chance to get the epic (best items). So to obtain an epic, it will cost me on average $100 ($5/0.05). By this math, I consider these items to be "worth":

Common - $9.09
Uncommon - $16.67
Rare - $50
Epic - $100

First off, is this a good way to determine what these items are worth? Also keep in mind that every time I spend $5, I will get an item guaranteed. When I think about it this way, wouldn't I consider a Common to be worth $5? But if that's the case, how do I determine the worth of the other items using this logic? My mind gets blown to pieces at this point as I don't know how much an item is worth! Help!

[EDIT] Actually there is a part 2 to this problem. I was going to save it for later, but might as well do it now.

So let's look at only the epic items, since that's what everyone wants to get.

There are 4 epic items in this pack. When you are awarded an epic, you basically have a 25% chance to get the specific epic you want. So on average, to get a specific epic, you would need to spend $400 right? ($100 to get an epic, divide by 0.25 to get the specific epic). But the thing is, at $400, you will have gotten 1 of each epic on average. So can I still say that the specific epic is worth $100 (part 1), or is it worth $400 (part 2), or somewhere in-between?

*
You only live twice.
AgentW
Profile Blog Joined July 2012
United States7471 Posts
Last Edited: 2013-03-14 02:05:51
March 14 2013 02:05 GMT
#2
Here's my input, and I'm still thinking about this so I may edit it:

Common would be worth $5 if that was all you were looking to get out of it, but since you're trying to get something more (one of the better things), you're overpaying by whatever $5 minus your supposed cost of the Common to roll the dice may be. Also, your first calculation is incorrect, as common can't be worth more than $5, given that it is the least desirable outcome.
Who's the bigger scrub, the scrub, or the scrub who loses to him?
Muirhead
Profile Blog Joined October 2007
United States547 Posts
March 14 2013 02:08 GMT
#3
For the first, question, an epic item is worth $100 only if the other items are worth 0. What you actually know is the equation
$5 = 0.55*c+0.3*u+0.1*r+0.05*e
where c is the average value of a common item, u is the average value of an uncommon item, etc.

This equation tells you nothing about the relative values of c,u,r, and e, but you can see for example that if only epic items are worth anything than e must be $100.

As to the second question, assuming everything not epic is worthless, what you know is that the average value of all 4 epic items is $100. The value of each given epic item cannot necessarily be deduced from that. If for example all 4 epic items are equally valued across the community, and trading is allowed, then each would be worth $100.
starleague.mit.edu
FalconPunch
Profile Blog Joined November 2010
United States127 Posts
March 14 2013 02:14 GMT
#4
I wouldn't consider everything else useless. To put it simply, I would estimate each item is worth the chance of obtaining it.
So since an epic is 10x harder to get than a common (roughly), an epic is 10x stronger than a common. Likewise, an epic is about 2x stronger than a rare. etc.
You only live twice.
Muirhead
Profile Blog Joined October 2007
United States547 Posts
March 14 2013 02:15 GMT
#5
I guess the main point is that you do not actually give us enough information to determine the values, but you do give us some constraints that give upper and lower bounds, and those are the numbers in your post.
starleague.mit.edu
Muirhead
Profile Blog Joined October 2007
United States547 Posts
Last Edited: 2013-03-14 02:21:00
March 14 2013 02:18 GMT
#6
On March 14 2013 11:14 FalconPunch wrote:
I wouldn't consider everything else useless. To put it simply, I would estimate each item is worth the chance of obtaining it.
So since an epic is 10x harder to get than a common (roughly), an epic is 10x stronger than a common. Likewise, an epic is about 2x stronger than a rare. etc.


What you are saying then is that the ratio of the value to c to e should be 5 to 55, and similar such constraints. I do not know if this is actually true (just because a weapon has 10x better stats doesn't mean it's exactly 10x more valuable), but if so you can use that information to calculate the values precisely.

You just need to solve the system of 4 variables in 4 unknowns
$5 = 0.55*c+0.3*u+0.1*r+0.05*e
0.05*e=0.55*c
0.05*e=0.3*u
0.05*e=0.1*r

It will tell you that the value of an epic under your assumptions is $25.
starleague.mit.edu
Muirhead
Profile Blog Joined October 2007
United States547 Posts
March 14 2013 02:21 GMT
#7
I would guess in real life an epic is worth more than $25, because a weapon with 10x better stats is worth more than 10x the $.
starleague.mit.edu
Aerisky
Profile Blog Joined May 2012
United States11038 Posts
March 14 2013 02:22 GMT
#8
Whoa haha, nice breakdown and pretty interesting hm... MIT OP :D
Jim while Johnny had had had had had had had; had had had had the better effect on the teacher.
Fishgle
Profile Blog Joined May 2011
United States1943 Posts
March 14 2013 02:22 GMT
#9
question, is the chance of getting an Epic each time you open a pack 5%, or do 5% of the packs have an Epic? because that changes the math quite a bit
aka ChillyGonzalo / GnozL
FalconPunch
Profile Blog Joined November 2010
United States127 Posts
March 14 2013 02:29 GMT
#10
On March 14 2013 11:22 Fishgle wrote:
question, is the chance of getting an Epic each time you open a pack 5%, or do 5% of the packs have an Epic? because that changes the math quite a bit


Hmm, maybe I confused it a little bit by calling it a pack. Essentially, pay $5 to get 1 card out of a "pack" of 20 cards. The chances are as mentioned in OP.
You only live twice.
FalconPunch
Profile Blog Joined November 2010
United States127 Posts
March 14 2013 02:32 GMT
#11
On March 14 2013 11:15 Muirhead wrote:
I guess the main point is that you do not actually give us enough information to determine the values, but you do give us some constraints that give upper and lower bounds, and those are the numbers in your post.


Not sure what kind of information to give you. I can give you the items' stats, but I don't think you'll understand them necessarily unless you play the game..
You only live twice.
Fishgle
Profile Blog Joined May 2011
United States1943 Posts
Last Edited: 2013-03-14 02:35:48
March 14 2013 02:33 GMT
#12
On March 14 2013 11:29 FalconPunch wrote:
Show nested quote +
On March 14 2013 11:22 Fishgle wrote:
question, is the chance of getting an Epic each time you open a pack 5%, or do 5% of the packs have an Epic? because that changes the math quite a bit


Hmm, maybe I confused it a little bit by calling it a pack. Essentially, pay $5 to get 1 card out of a "pack" of 20 cards. The chances are as mentioned in OP.

so then, you don't get an Epic 1 in 20 packs? rather it's a new 5% each time you open one? Because if that's the case, you only have a 75% chance of having at least one Epic after 20 5$ packs.

edit: it might be even less. 13% chance of having at least one. I really don't know math very well. srry
aka ChillyGonzalo / GnozL
run.at.me
Profile Joined December 2011
Australia414 Posts
March 14 2013 02:43 GMT
#13
What's the game called so I download it
MidnightGladius
Profile Blog Joined August 2007
China989 Posts
Last Edited: 2013-03-14 03:05:09
March 14 2013 03:02 GMT
#14
Assuming the draws are independent (and if this is online, they almost certainly are), and you're only concerned with getting Epics, you can treat this as Bernoulli trials with p=0.05, q=0.95.

Pr(at least 1 Epic in 20 draws) = 1 - Pr(0 Epics in 20 draws) = 1 - ((20 ncr 0)*(0.05)^0*(0.95)^20) = 64%

In order to get your chances up to 95%, you'd need to open roughly 60 packs.

EDIT: Going on to your question about relative values... it really depends on how the player community values individual items, and how much trading goes on. In trading card games, you see a lot of highly sought-after commons/uncommons and "worthless" rares.

Can you provide us with any more details?
Trust in Bayes.
Cambium
Profile Blog Joined June 2004
United States16368 Posts
Last Edited: 2013-03-14 03:27:52
March 14 2013 03:03 GMT
#15
This is actually a really good question, and it's related to game theory, in particular utility theory. Here's a very mathematical approach to what you are asking:

=============

Assuming the participants are logical, we'd naturally have

u(A) >= u(B) >= u(C) >= u(D)

where the function u() defines the utility of the item to the players, and that is our constraint. Utility essentially means happiness, meaning, how happy you are with items A, B, C, or D. Where A is epic, B is rare, etc.

The equation we are trying to solve is:


u($5n) <= 0.05n * u(A) + 0.1n * u(B) + 0.3n * u(C) + 0.5n * u(D)

where n is the number of times you play the game.

We often assume u($) to be a linear mapping between utility and money, as it is true for small quantities, and makes our calculations a lot easier as we use $ as the worth of the items. This slightly simplifies our equation to

u(5$) <= 0.05 * u(A) + 0.1 * u(B) + 0.3 * u(C) + 0.5 * u(D)


So essentially, we have one equation, four unknowns, and one constraint. This is precisely why the 'value' of these items are entirely subject, and is therefore driven by the players' own utility functions.

For example, if you value 'epics' much much more than everything else, i.e. u(A) >> u(B)

You would thusly have

u(5$) <= 0.05 * u(A)
20*u(5$) <= u(A)

which reads: you would rather have item A than to have $100.

Furthering my example, I can show you why it's incorrect, or rather, illogical to assume B is worth $50:

Since we had assumed our players are logical:

u(A) >= u(B)


We make the best effort to maximize B's worth without altering our initial constraint, and therefore:

u(B) = u(A) >> u(C)

We then have:

u(5$) <= 0.05 * u(A) + 0.1 * u(B)
u(5$) <= 0.15 * u(B)
u(B) >= u($33.33)

Or B is at most worth roughly $33, and that's under the assumption that you value A and B equally.
When you want something, all the universe conspires in helping you to achieve it.
FalconPunch
Profile Blog Joined November 2010
United States127 Posts
Last Edited: 2013-03-14 09:02:16
March 14 2013 09:01 GMT
#16
Just to quickly answer a few questions before I go to bed.

What's the game called so I download it


Underworld Empire.

Can you provide us with any more details?


Of course, but I don't know what kind of details are relevant. And to answer your paragraph above it, this is not a trading card game and trading is not allowed. It's an rpg game where you basically buy (with real and fake money, real money is of course better) items and heroes that make your character stronger.

I don't have time to follow all the math formulas right now, but I'll take a look at them tomorrow. And thanks for all the responses!

One last thing. I think I really confused everyone by using the word "pack". Essentially there are ~20 (for the sake of this argument, let's just say 20) items at the shop. 4 epics, 4 rares, 6 uncommons, and 6 commons. When you go to buy an item at the shop, it tells you the drop rates for the items and they are:

Common - 55%
Uncommon - 30%
Rare - 10%
Epic - 5%

The chance of getting an epic is not 4 out of 20 (20%), the chance of obtaining an epic is 5% as stated. On average, you need to buy 20 items (at $5 each) to obtain 1 epic, assuming the above drop rates are true.


Ahh, I was thinking about it some more and my head exploded again! Can't think about it anymore till tomorrow!
You only live twice.
alQahira
Profile Joined June 2011
United States506 Posts
March 14 2013 14:55 GMT
#17
Well, the real question is can you sell items to other players? Then there would be a market value which would answer all our questions.
razorsuKe
Profile Blog Joined July 2008
Canada1823 Posts
March 14 2013 16:08 GMT
#18
On March 14 2013 23:55 alQahira wrote:
Well, the real question is can you sell items to other players? Then there would be a market value which would answer all our questions.


w0rd, any given item is only worth what someone is willing to pay for it.
EntusGalleries.com - CJ Uniform Sale
MidnightGladius
Profile Blog Joined August 2007
China989 Posts
March 14 2013 17:23 GMT
#19
Hold on, let me make sure this is clear. The store doesn't actually sell you items, it sells you $5 tickets for a chance at one of N items? And you can't trade or sell off duplicates of items you get?

Math aside, that seems like a really sketchy way to design a market :/
Trust in Bayes.
Aerisky
Profile Blog Joined May 2012
United States11038 Posts
March 14 2013 18:10 GMT
#20
That's how a lot of these games work nowadays lol. Basically gambling for the items. Not a particularly honest way of going about business but it does bring in the dough.
Jim while Johnny had had had had had had had; had had had had the better effect on the teacher.
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