Wow values just beat me - I have been writing a very similar post. Well I guess there is no need to do so now... Still I will copy it below in case anybody wants to go through the math for the general case or use it as an exercise for high school math classes....

The arena format says that you play until you accumulate 3 losses

or 9 wins, whatever comes first. This is an arena "match" that consist of between

3 and 11 single "games". In an arena game, either one player wins and the

other loses, or both lose (this is called a "draw", and happens rarely when both

heroes are at 0 or lower health, for instance if a card like Hellfire is played

when both heroes have 3 or less health). We will define the following variables:

N: maximum number of losses allowed in Arena (has the value of 3 currently)

M: maximum number of wins allowed in Arena (has the value of 9 currently)

p: probability of winning a game (also equal to the fraction of games won)

Note that p would be equal to 1/2 if there were no draws, and in reality is

very close but slightly lower then 1/2 since (as far as I can judge from

watching streams) draws are pretty rare. Note that the probability of losing

a game is 1-p.

Now we will evaluate the probability (or fraction) of Arena matches ending in

a result A:B, where A are the wins and B the losses (for instance if A=7 and B=3,

you have accumulated 7 game wins and 3 game losses in your Arena run).

To evaluate the probability of getting an A:B result, we have to distinguish

two different situations: a game with N losses (i.e. B=N) where we know that the

last game played was a loss, and a game with strictly less then N losses, where

we know the last game played was a win.

1) games with N losses (B=N). In this case we have B-1 losses spread among

A+B-1 games, since we know the last game is a loss, and there are A+B games

in a match with A wins and B losses. The different number of ways we can

combine B-1 losses in a set of games of size A+B-1 is given by the binomial

coefficient C(A+B-1,B-1) where C(n,k)=n!/(k!*(n-k)!).

The probability of ending up A:B is therefore given by the probability of achieving

A wins and B losses (this being equal to p^A *(1-p)^B ) multiplied by the different

number of ways of achieving that A:B result. So we have:

Probability of achieving A:B (when B=N) = C(A+B-1,B-1)*p^A*(1-p)^B =

= (A+B-1)!/(A!*(B-1)!)*p^A*(1-p)^B

2) games with strictly less then N losses (B<N). In this case the last game is a win,

and we have B losses spread among A+B-1 games. The different number of ways

we can combine B losses in a set of games of size A+B-1 is given by the binomial

coefficient C(A+B-1,B). Therefore we get:

Probability of achieving A:B (when B<N) = C(A+B-1,B)*p^A*(1-p)^B =

=(A+B-1)!/((A-1)!*B!)*p^A*(1-p)^B

Let us evaluate the formula, and start with the assumption

that p=1/2 (i.e. draws never happen). The results is as follows:

WIN LOSS Fraction of games with this result if Prob(win)=50%

9 0 0.20%

9 1 0.88%

9 2 2.20%

8 3 2.20%

7 3 3.52%

6 3 5.47%

5 3 8.20%

4 3 11.72%

3 3 15.63%

2 3 18.75%

1 3 18.75%

0 3 12.50%

For instance, only about 5.5% of games end in a result of 6 wins to 3 losses.

About 9% of games achieve 7:3 or better.

Here it should be noted that while exceptional players will be more likely

to end up in the 9 wins bracket, it is a basic fact that the 91% majority of

players will get less then 7 wins.

What is the effects of draws? We do not know what is the percentage of games that

are drawn, but we could try setting p=0.495 (corresponding to 1% of games that end

in draws) and we would obtain:

WIN LOSS %

9 0 0.18%

9 1 0.81%

9 2 2.05%

8 3 2.09%

7 3 3.38%

6 3 5.30%

5 3 8.04%

4 3 11.60%

3 3 15.62%

2 3 18.93%

1 3 19.12%

0 3 12.88%

The effect is small, but it can be noted that there are fewer games with many wins

and more games with 2 or less wins. The table can easily be recomputed for p=0.49

or p=0.45.