Introduction:
Magic the Gathering: Arena is a online collectible card game where you build a deck of cards and battle other players based on the physical trading card game of the same name. In the context of this question, you only need to know that in this game, there is an event called "Jump In!" in which you are given two packets of cards that will then become your deck to play with, and you get to keep these cards in your collection once chosen. Even more specifically, you only need to know that in these two packets, each will have one valuable card, a rare or mythic, which is what we are primarily interested in. More specific details of the mechanics are below.
Mechanics:
The specific mechanics of the Jump In! event in MTG Arena are that each time you enter, you are given a choice between three packets of cards, twice. Each packet will give, among other less valuable cards we will ignore, one rare or mythic card with known probability, usually 80% and 20% between a rare or mythic, respectively. Some packets only have a single possible rare or mythic to receive, so 100% chance. Rarely, a packet will have three possible, 40/40/20, or 60/30/10. Other distributions are possible. These numbers and the specific card reward are all known in advance. Once you have received four of a specific card, you can no longer receive that card, but can still be presented with it as a reward, resulting in no gain from selecting that packet.
What I want:
What I want to know is how to determine which packet to choose in order to maximize the total number of rares/mythics I receive in the long run before I have four copies of the rares/mythics from too many packets and the expected return drops below about one rare/mythic per event at which point the event is no longer cheaper than just buying a single booster pack of cards. (each event will, in the beginning, guarantee two rares/mythics, one per packet)
Example:
I am tracking how many of each card I have, but complicated situations arise where I am unable to determine which choice is best. For instance, if one packet has a single reward for which I already have one, and the other has two at 80/20 probability, and I already have two of the 80% card, which choice is correct?
Example 2:
What if I have three of the 20% card, and one of the 80%? That is still four cards possible to eventually receive from that packet, compared to only three in the other packet, but a 20% chance that I'll pull the fourth card and spoiling this packet for all future events. Choosing the first packet in that case at least guarantees no immediate spoilage, but is that statistically correct?
Optional, additional complexity to consider:
Oh, and to introduce more complexity that it might be best to ignore: The selection of packets in the second round will be influenced by the choice made in the first round. Each packet has one or two of five colors associated with it, and once chosen, the second selection of three will only contain compatible packets of colors. For instance, if you select a white packet then no dual-color packets that don't contain white will be available to choose from in the second round. If you chose a black/green packet, then only black, green, or black/green packets would be available. Trying to factor that in would probably be a nightmare, I'm just including it to be complete. Information on the complete pool of packets and their colors are known.
More Info:
There's plenty information about how Jump In! works to be found online if I've missed anything. I am also available to expand or explain anything that isn't clear.
Extraneous complaining:
I've tried so hard to understand this problem and work it myself, but I just can't get it. I'm really just winging it with my choices, and I hate that.
Hello fellow Planeswalker! There are a couple layers to this problem. To start with probability of outcome, there is a formula you need, so if you don't know at least high school Statistics you may have a rough time with this one:
Let r=number of successes/desired outcomes
Let n=number of attempts
Let p=chance of success
Let q=chance of failure
Then: $$P(Event)=nCr*p^{r}*q^{n-r}=\frac{n!}{r!(n-r)!}*p^{r}*q^{n-r}$$
Now, just fill it out to calculate the odds. From your example above, let's find out what the odds are that you'll pull 3 rares without hitting that 4th mythic at 80-20:
Let r=3 (3 rares in a row without a mythic)
Let n=3 (we don't want any mythics along the way)
Let p=0.8 (80% chance of rare)
Let q=0.2 (20% chance of mythic)
Then:
$P(3\;rares)=_3C _3*0.8^{3}*0.2^{3-3}=\frac{3!}{3!(3-3)!}*0.8^{3}*0.2^{3-3}$
$P(3\;rares)=\frac{1}{0!}*0.8^{3}*0.2^{0}$
$P(3\;rares)=\frac{1}{1}*0.8^{3}*1=0.8^3=0.512$
So, in this case you would have a 51.2% chance to pull three rares in a row from the 80-20 chance. Or, to put it another way, you have a 48.8% chance to spoil the pack by hitting that 4th mythic within the next three packs. Note that even if you already have three copies of a given mythic, you are still more likely to hit three rares than to get the 4th one.
Now, let's compare this to the other situation you posed, a pack with two rares and no mythics:
Let r=2 (What if we got 2 more rares)
Let n=2 (in a row, with no mythics?)
Let p=0.8 (80% chance of rare)
Let q=0.2 (20% chance of mythic)
Then:
$P(2\;rares)=_2C _2*0.8^{2}*0.^{2-2}=\frac{2!}{2!(2-2)!}*0.8^{2}*0.2^{0}$
$P(2\;rares)=\frac{1}{0!}*0.8^{2}*0.2^{0}$
$P(2\;rares)=\frac{1}{1}*0.8^{2}*1=0.8^2=0.64$
So, you have a 64% chance to spoil the pack by pulling the next two rares in a row without pulling a mythic. Note that these odds are worse than the previous ones because of the addition of only a single rare.
So, in summary, always pick the pack for which you have the fewest rares. This will give you the most mythics in the long term. I haven't exactly proven this properly, but it does seem to be the trend, and it lines up with my expectations as well.
As for the secondary pack selection, that is a much more complex mathematical question. In lieu of solving it, might I suggest a way to not have to do the math? It seems to me that you could start up a list (or better, an excel spreadsheet) of your Rares/Mythics from Jump Start sorted by color and use that to track which colors you have the greatest completion of. That would enable you to try to pick packets from colors that you have the least completion, which is essentially the conclusion the maths would bring you to as well. Counting is much easier than calculating!