Average wins for arena is slightly less than 3 wins. The average for forumers would be higher than that because only more experienced/better/hardcore gamers would visit this site. (Many with less than 3 win average are are usually new players).

The average number of wins for an Arena run is exactly 3, because for every win someone else has to lose, and vice versa. Therefore, the expected value of wins after 3 losses is also 3.

This is if you're referring to the average as the arithmetic mean. As the link above shows, the mode for wins per Arena run is actually tied at 1-3 and 2-3. Since an Arena run ends at 3 losses or 12 wins, the curve of the distribution of wins is skewed to the right.

The average number of wins for an Arena run is exactly 3, because for every win someone else has to lose, and vice versa. Therefore, the expected value of wins after 3 losses is also 3.

This is if you're referring to the average as the arithmetic mean. As the link above shows, the mode for wins per Arena run is actually tied at 1-3 and 2-3. Since an Arena run ends at 3 losses or 12 wins, the curve of the distribution of wins is skewed to the right.

Actually, i was thinking the same, but i am not so sure anymore, if the average is exactly 3 wins. It would be true, if every Arena run would end with 3 losses. But since you can have less losses, i think that the average is less than 3 wins. If you compare a 12-2 run with 12-0 run, then in both cases the number of wins is the same, but the number of wins feed to the System is different, which effects the Overall win rate.

4 wins is not profit and means you have to spend 7 games when you could just crack a pack for 100 gold. just sayin

If you are only counting card gain yes but that's not all you get. Just getting two wins gets you a pack and a chance at 40-60 gold or 50-60 dust. Also a small chance for an extra pack. Standard pack for of duplicates DE's for 40 dust. Worse case scenario your down 10 extra gold on the run compared to just getting a pack. Best case you gain a pack and a half worth of dust. That's at just two wins which shouldn't be hard to get to. If you get a good deck and get to 4-5 it's quite a bit better than just buying a pack.

2 Wins does not guarantee 40-60 gold or 50-60 dust. It guarantees 20-30g and 10-30 dust. Also I do not believe you can get an extra pack for 2 wins you need at least 3.

Basically if you can average 3 wins in Arena its better than buying packs as long as you dont mind investing the time in playing those games (Arena is fun why wouldnt you want to play :)). Also this is assuming your main goal is OG packs.

The average number of wins for an Arena run is exactly 3, because for every win someone else has to lose, and vice versa. Therefore, the expected value of wins after 3 losses is also 3.

This is if you're referring to the average as the arithmetic mean. As the link above shows, the mode for wins per Arena run is actually tied at 1-3 and 2-3. Since an Arena run ends at 3 losses or 12 wins, the curve of the distribution of wins is skewed to the right.

Actually, i was thinking the same, but i am not so sure anymore, if the average is exactly 3 wins. It would be true, if every Arena run would end with 3 losses. But since you can have less losses, i think that the average is less than 3 wins. If you compare a 12-2 run with 12-0 run, then in both cases the number of wins is the same, but the number of wins feed to the System is different, which effects the Overall win rate.

I did some quick math, and you're right. Because you can end a run before reaching 3 losses, the expected value of wins per Arena run is slightly less than 3. (2.9916 wins, to be more precise). If there was no win limit, then the expected value would be 3 wins.

The median number of wins in Arena will always remain constant, though, regardless of what the win limit is. For any win limit at 3 or above exactly half of Arena runs will end up below 3 wins, and half will end up with 3 or more.

There are also retired arena decks and people that quit the game entirely, in the middle of an arena run, to factor in. Regardless, the average is close enough to 3.

For the person saying arena isn't worthwhile, it's easily the fastest way to build your collection for free. It's also not very hard to learn to perform better than that 3 average. I was averaging around 4 in my first couple weeks of playing Hearthstone. Haven't played much arena the past couple months, so who knows how I'd do now. Hoping to jump back in and push my average to 5 in the next couple months.

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:

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!

To be clear - 65.625% of players win 0-1-2-3 games ?

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

Thanks !

Yes, as long as all players are matched with other players with the same number of win and losses, exactly 50% of the population will finish with less than 3 wins, which makes perfect sense. These numbers are both probabilities and the actual (theoretical) distribution.

Correct me if I'm wrong but the pack you win in Arena is from a random expansion right? So if you always Open classic packs is much better to buy them with Gold and increasing your chances of a legendary (with the pity counter) than opening random packs from the arena, especially when the other expansions don't have as many good cards as classic.

I also average 3 wins, so I don't see it worth it at all unless you can average 5 and love playing arena because it's really time consuming

Correct me if I'm wrong but the pack you win in Arena is from a random expansion right? So if you always Open classic packs is much better to buy them with Gold and increasing your chances of a legendary (with the pity counter) than opening random packs from the arena, especially when the other expansions don't have as many good cards as classic.

I also average 3 wins, so I don't see it worth it at all unless you can average 5 and love playing arena because it's really time consuming

It is not random anymore. Now the pack is always from the last expansion.

Hello ,

what's the general average winrate in arena (i'm consistent with 4-5 wins min) ?

And what's the winrate per class after WOG ?

thanks,

I am at 4.3 wins on average in my latest 37 runs (that is when i started to keep track on them)

Average wins for arena is slightly less than 3 wins. The average for forumers would be higher than that because only more experienced/better/hardcore gamers would visit this site. (Many with less than 3 win average are are usually new players).

http://hearthstone.gamepedia.com/Arena#Statistics

The average number of wins for an Arena run is exactly 3, because for every win someone else has to lose, and vice versa. Therefore, the expected value of wins after 3 losses is also 3.

This is if you're referring to the average as the arithmetic mean. As the link above shows, the mode for wins per Arena run is actually tied at 1-3 and 2-3. Since an Arena run ends at 3 losses or 12 wins, the curve of the distribution of wins is skewed to the right.

Thanks mate :D

sounds like not worth doing based on rewards

4 wins is not profit and means you have to spend 7 games when you could just crack a pack for 100 gold. just sayin

“Name dropping is not a good thing to do. Robert DeNiro told me that." - Bob Saget

Minion (1)Ability (29)It was numerous amount of times said that arena run is better than a bougth pack. Take as granted.

Just gettig to 2 wins gives you a pack and a chance at

2 Wins does not guarantee 40-60 gold or 50-60 dust. It guarantees 20-30g and 10-30 dust. Also I do not believe you can get an extra pack for 2 wins you need at least 3.

Basically if you can average 3 wins in Arena its better than buying packs as long as you dont mind investing the time in playing those games (Arena is fun why wouldnt you want to play :)). Also this is assuming your main goal is OG packs.

I did some quick math, and you're right. Because you can end a run before reaching 3 losses, the expected value of wins per Arena run is slightly less than 3. (2.9916 wins, to be more precise). If there was no win limit, then the expected value would be 3 wins.

The median number of wins in Arena will always remain constant, though, regardless of what the win limit is. For any win limit at 3 or above exactly half of Arena runs will end up below 3 wins, and half will end up with 3 or more.

There are also retired arena decks and people that quit the game entirely, in the middle of an arena run, to factor in. Regardless, the average is close enough to 3.

For the person saying arena isn't worthwhile, it's easily the fastest way to build your collection for free. It's also not very hard to learn to perform better than that 3 average. I was averaging around 4 in my first couple weeks of playing Hearthstone. Haven't played much arena the past couple months, so who knows how I'd do now. Hoping to jump back in and push my average to 5 in the next couple months.

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!

^ Wow nice.

To be clear - 65.625% of players win 0-1-2-3 games ?

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

Thanks !

## Twitch Arena Stream: https://www.twitch.tv/boozor / Youtube Chanel: https://www.youtube.com/c/boozortv

#1 Arena Leader Board Player in North America - August 2018 and April 2020#2 NA Nov 18, #2 Asia July 19, #2 NA Feb 20, #2 NA June 20It would be nice to show the distribution in a graph.

both the cumulative and the discrete.

think people often retire, but get 12 more than they retire.

Often you just get such a useless deck that losing 50 gold is worth not spending the time to get 1 - 3

Which skews the distribution.

This would not hold.

but the other conclusion would probably be true.

Correct me if I'm wrong but the pack you win in Arena is from a random expansion right? So if you always Open classic packs is much better to buy them with Gold and increasing your chances of a legendary (with the pity counter) than opening random packs from the arena, especially when the other expansions don't have as many good cards as classic.

I also average 3 wins, so I don't see it worth it at all unless you can average 5 and love playing arena because it's really time consuming

And here is the graph of the distribution (attached)

Minion (1)Ability (29)