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[MadMojoMonkey]

So the combinations are 12W, 12W1L, 12W2L, 11W3L, 10W3L, ..., 1W3L 0W3L

In each of these combos the final result has to be a W when 12W and a L when 3L so can be thought of as a combination of all results except the last.

Each of these can be worked out using (total outcomes)!/ [(W outcomes)!(L outcomes)!]

Example - 8W3L can be thought of as a combo of 8Ws and 2Ls so 10!/(8!2!)

Each outcome has a probability of (x)^W * (1-x)L where x is your win%

By multiplying the # combos with the probability of each happening we get how likely that result is.

edit

Then I assume we can work out our avg # of wins just by doing # wins * how likely for every result.

I didn't see this when I posted my initial response to your question.

This reasoning is spot on.

Your result of 10!/(8!2!) is equivalent to my notation of C(10,8). This is the binomial coefficient.
By symmetry, C(10,8) = C(10,2); just like 10!/(8!2!) = 10!/(2!8!)

To restate my solution here, using your notation of W for number of wins and L for number of losses:

For cases where you lose 3 matches
EV(W,L=3,x) = C(W+3-1,2)*X^W*(1-X)^3

For cases where you win 12 matches
EV(W=12,L,x) = C(12+L-1,L)*X^12*(1-X)^L

For all the "inner" cases, your EV of being in any state of 0 <= W <= 11 and 0 <= L <= 2
EV(W,L,x) = C(W+L,L)*X^W*(1-X)^L
or, by symmetry of the binomial coefficient
EV(W,L,x) = C(W+L,W)*X^W*(1-X)^L

The only notable difference is the "-1" in the binomial coefficient for the "end states" that is not present in the "intermediate states."

I still find that you need x > 81.3% to get a 50% shot at winning 12 matches.

Of course, this all assumes that all matches are of equal difficulty and your winrate taken as an aggregate of all matches is acceptable as an estimator as the Arena progresses to different matches. I think this is not too bad a model, since the arena matches seem to be always against players of roughly your rank +/- 1, therefore roughly equal skill, no matter how many W/L you have.

[Savy]

I also made an excel sheet!!!! I think it becomes beneficial to play arena rather than spending gold on other things somewhere between the 4 and 5 wins per arena mark and that 7 wins guarentees you enough gold that you always have enough gold to enter another arena whilst getting other rewards too.

Of course, this all assumes that all matches are of equal difficulty and your winrate taken as an aggregate of all matches is acceptable as an estimator as the Arena progresses to different matches. I think this is not too bad a model, since the arena matches seem to be always against players of roughly your rank +/- 1, therefore roughly equal skill, no matter how many W/L you have.

This is actually why I think it's got faults as a model (I thought I'd commented on this somewhere but I had not, was on another forum).

First of all you get to draft a new deck every arena of which some are better and some are worse. I think it's fair to assume that this averages out and can be ignored but I don't have any real explanation for why I think this.

What I don't think averages out is how likely you are to win each game. Assuming your deck can be considered an average deck that deck gets less and less likely to win after every game because the average deck you are playing against and the average player you are playing against (these can be combined into it gets harder) whilst your deck and skill level don't change. So there would be a decrease in likelihood of winning after every win you have*.

I imagine (no strong reason) that this means your results are squished up a bit more and drop down quicker whilst hopefully not having a mental effect on your results. Worth noting that I think the lower your win rate the bigger this effect will be.

That all being said (sorry bit boring) for what I wanted this for it does a sufficient job (how can I say this when I don't actually know?) for what I want it to do (the reason is it's of 0 real importance to anything I'm doing).

*I think it's done on wins so there's no bias in if you've just started you're going to play a 0-0 rather than a 0-2 compared to if you're 0-2 yourself.