Hearthpwn

Quote from Boozor >>

Is it correct to say that 50% of the population will have a 0-3 / 1-3 / 2-3 result then ?

odd are with you, you have a 50/50 chance to win first, 50/50 in case of a loss, and the same onwards. Average is about 4-6, but it's also about luck and what you get to draft. Combo and easy opponents play a good part in it as well. I can say i learned something about this.

Ex: combos like the nerubian 6 cost card 4/4 which gets cheaper + master of evolution.

Quote from CranberrySawce >>

What is the average number of victories in an arena run?
This is a nice statistical problem! The answer is 2.99157714844. Here is the demonstration.

Let's make the following assumptions:

1) A player is always matched up against another player
   with the same number of victories and losses.
2) No one ever 'retire' from their run.

Because of assumption 1, the probability that a player wins their next game at any stage of the run is 1/2. Indeed, for a given number of victories and losses, exactly half of the players will win and half will lose.

So here is how we can compute the probability of finishing the run with N victories.

Let C(n,m) be the number of different ways of picking 'n' elements among a set of 'm' elements. This is

    C(n,m) = m! / ( n! * (m - n)! )

Then, the probability of losing 'n' games among 'm', (only for n < 3) is

    P(n,m) = C(n,m) / 2^m       ; for n < 3

As long as n < 3, you could have lost your games in any order. But for n = 3, the last game must be a loss. Therefore, the probability of losing 3 games among 'm' is

    P(3,m) = P(2,m-1) / 2

Hence, the probability of finishing the run with 'N' victories (only for N < 12) is

    V(N) = P(3,N+3) = P(2,N+2) / 2      ; for N < 12

In the case that you finish the run with 12 victories, we must add the probabilities of finishing with any number of losses from 0 to 2 and that you win your last game. Therefore

    V(12) = ( P(0,11) + P(1,12) + P(2,13) ) / 2

Now, we can compute the probability of finishing the run with any number of victories:

Victories   Probability (%)
    0     :    12.5
    1     :    18.75
    2     :    18.75
    3     :    15.625
    4     :    11.71875
    5     :    8.203125
    6     :    5.46875
    7     :    3.515625
    8     :    2.197265625
    9     :    1.3427734375
   10     :    0.8056640625
   11     :    0.47607421875
   12     :    0.64697265625

And the average number of wins is

    A = sum_(N=0,12)   V(N) * N = 2.99157714844

This result is exact under the two assumptions we made.

The first assumption is reasonable as long as there is a large number of players doing arena runs at any time. However, if you match up players with different number of wins and losses, the probability of each outcome is no longer 1/2.

Also, there may be some players who retire after realizing that they crafted the worst deck in the world. This would then lower the average number of victories, because these players deny other players from easy victories.

There ya go! You should no longer feel bad about those 3-win arena runs!

Quote from DynamischerDiskord >>

Quote from Boozor >>

Is it correct to say that 50% of the population will have a 0-3 / 1-3 / 2-3 result then ?

 What do you mean by population? 

Some comments about the calculation:

-Another underlying assumption was that all the population finish their arena run. In reality, the population is dynamic; some new players come in before the other finish their run. However, for a very large number of players, this should not make a significant difference in the win distribution.

-The probabilities listed above are the actual distribution of the population. The "population" is just a large number of players that meet the criteria (perfect match up, no retirement, everyone finishes).

-Yes, there are more players that finish with 12 wins than 11 wins. This is because 12 is the maximum number of victories, so you count the players that finish with any number of losses in the same bin.

-Suppose you increase the maximum number of victories above 12, then the average win will get closer and closer to 3 as you increase the limit. This makes sense; if you completely remove the maximum number of victories, then in the end, there will be exactly the same total number of victories and losses in the population.

This is why I rather buy Boosters :p

Well, it's a 2 player game, so the average winrate is exactly 50%. 

Quote from Boozor >>

Quote from DynamischerDiskord >>

Quote from Boozor >>

Is it correct to say that 50% of the population will have a 0-3 / 1-3 / 2-3 result then ?

 What do you mean by population? 

 Sample population. So players that meet Cranberry's original criteria.

People that retire before completing their arena makes it not so simple. Because it means some loses are not met with a win, therefore average result is logically under 3.