OP, I commend you on going to the effort here, but unfortunately, nobody cares.

The people who understand and agree already know.

The people who don't already know are choosing not to.

They'd rather say

"I watch so and so on a stream and he only paid for one arena entry and now it pays for itself"

or

"If you're good enough you can win, it's all about skill"

Somehow, the fact that you can get absolutely horrible cards to start with, let alone bad draws, let alone any number of factors that go into winning or losing, they're convinced that if you try hard enough, you'll always win.

But in the end, it's their loss.

Casinos make millions out of stupid people.

Arena is just a casino disguised as a free to play game.

This post couldn't be any more true, the people who don't recognize this are just ashamed of how much money they've spent or simply like being taken advantage of and feel like they actually matter to Blizzard and are not another penny in the bank

It's not quite the same thing - right now it's possible for people to gain gold by running arena if they're good enough (at the expense of the people they play against) *but* with a sufficiently large population and a functioning MMR it will not be anymore, because you would be playing against players who are just as good at arena as you are.

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.

Otherwise, good detailed post.

09/02/2013 10:53 AMPosted by GraoggI would be interested if you also took into consideration the value of rewards

Agreed, the value of dust, packs and gold given have to be considered in order to truly evaluate the ROI of arena. In order to do that however, it will take a bit longer on the math side since the rewards are not fixed, and not linear. Additionally, you cannot disenchant every card (iirc) so some type of representative average has to be estimated/calculated for that as well.

With all of that said, I think with just the current math and general observation we can say with a high confidence rating that the Arena play will still 'not pay for itself' technically, even if it comes just a few percentage points from doing so.

To be fair, I suspect it has to be that way. This is a business that needs to make money for Blizzard and their mathematicians probably figured out these numbers long before Hearthstone was even in production. Thus, how close to that self-sustaining curve Arena is for players (on average) is the only real question. It never 'should' be self sustaining or better (on average) because the other modes/pricing of the game would have to disproportionately change to support the revenue stream.

With that said, I have gone 6-3 and received zero gold, so there is no guarantee that even 6 wins will help pay for the next Arena entry.

And what Dobby said is true - if your opponent gets a huge selection of spells then it is very hard to beat them. I once fought someone with 1 Blizzard and 3 Flamestrikes (that I saw through 18 cards, he could have had even more). I had no chance.

http://us.battle.net/hearthstone/en/forum/topic/9572767831#1

https://docs.google.com/document/d/1nFZNcBRTNNwKi2_VZ3irT0JjuDmTeCYL0rXydHEkEDc/edit?usp=sharing

Use that to figure out your chances of breaking even, cuz you did very little real math in your original post.

IIf four exist, and they go 9-0, 0-3, 0-3, and 0-3, three of them lose greatly while one wins moderately.

With all this math you don't like to go into details about this amount do you? You would much rather stay at vague assessments of the rewards for those brackets, well here is some numbers.

I have seen something like 10+ reward openings of 9 win brackets. Those that included pure gold was all around 300-350 gold on top of the 1 guaranteed pack in one of the boxes. Since a pack has a direct value of 100 gold that means the value of 9 wins is 400-450 gold or more minus the 150 gold you used to enter = 250-300 gold paid out.

What is this

*great*loss the 0-3 bracket gets that you talk of? Well normally it's 1 pack and roughly 15 gold worth. That means the 0-3 players paid 150 gold andmultiply that with the 3 players you say it takes to even out the 9-0 player and you get a total of roughly 105 gold combined.

So all in all a setup with a 9-0 and 3 x 0-3 will result in Blizzard losing around 150-200 gold worth of rewards every time people decide to play the arena and choose not to just buy packs... those bastards right?

stop treating this !@#$ like statistics class.

if you played lebron james 1 on 1 basketball, 1st to 20, 10 games...

your win % isnt going to be 50%. if you're wanting to win out in arena, maybe you have to play better than opponents, have a better understanding of deck types, know about pick orders etc etc etc.

do things that will increase your win% to above 50%, its pretty simple.

If you "do things that increase your win% to above 50%", the MMR system will increase the difficulty of your opponents until you get back down to 50% win.

Unless you happen to be the 1 best player in the world *rolleyes*

stop treating this !@#$ like statistics class.

if you played lebron james 1 on 1 basketball, 1st to 20, 10 games...

your win % isnt going to be 50%. if you're wanting to win out in arena, maybe you have to play better than opponents, have a better understanding of deck types, know about pick orders etc etc etc.

do things that will increase your win% to above 50%, its pretty simple.

If you "do things that increase your win% to above 50%", the MMR system will increase the difficulty of your opponents until you get back down to 50% win.

Unless you happen to be the 1 best player in the world *rolleyes*

Not in the arena *rolleyes*

really?

REALLY?

There is not any actual math in this post just a bunch of conclusions you have came up with.

**assumption**. The math is right but as stated before. you forgetting other values. Such as card packs, dust and gold. Nice try though.

09/02/2013 09:27 AMPosted by Valuespeople should stop claiming it does.

Is that what this thread is about?

Either way, it's more profitable than any of the other game modes.

(really not scharcasim)

Sep 2, 2013
-1

I don't know how much sense a static MMR would make in arena considering the high variance in decks that you get. I think it might try to match you against opponents with similar records in their current arena run though, as I often get walkovers at the start and stronger opponents deeper in, but that's just my experience.

This is just wrong. All that 9% means is that an average player who you expect to win 50% of the time in every game would get 7+ wins about 9% of the time. That says nothing about how well the top 9% of the playerbase would do. That's really dependent on how well the top 9% do against the rest. How often does a top 9% player win against an average player? In a low skill game it might only be 55%, and it would be very difficult for anyone to consistently get 7 wins.

Real math, as in making up totally arbitrary estimates of win rates and running a simulation.

I love how math gets used on forums. Just make long posts with some random math in it to make it look like you know what you're talking about, and basically no one will actually read it, but suddenly you're credible.

So what does this mean? This means that if YOU are a good enough player, YOU won't go 9-0 0.22% of the time. It means that 0.22% of ALL PLAYERS will go 9-0. Roughly 9% of ALL PLAYERS will go 7-3 or better.

In this scenario, players reasonably COULD become so good that they can pay for their own arena admission: you just need to be in the top 9% of the playerbase.

This is just wrong. All that 9% means is that an average player who you expect to win 50% of the time in every game would get 7+ wins about 9% of the time. That says nothing about how well the top 9% of the playerbase would do. That's really dependent on how well the top 9% do against the rest. How often does a top 9% player win against an average player? In a low skill game it might only be 55%, and it would be very difficult for anyone to consistently get 7 wins.

You want some real math? I did an in-depth analysis of the Arena awhile back with a simulation that has realistic probabilities of reaching each level for different sections of the populace:

http://us.battle.net/hearthstone/en/forum/topic/9572767831#1

Real math, as in making up totally arbitrary estimates of win rates and running a simulation.

I love how math gets used on forums. Just make long posts with some random math in it to make it look like you know what you're talking about, and basically no one will actually read it, but suddenly you're credible.

What I have been saying however, is that at the current price structure it's probably better to just burn through endless arena events then buy packs in bulk since even going 0-3 is giving you a pack ($1.25 to $1.50 of your entrance fee) plus some amount of gold and dust that is going to get you fairly close to $2 depending on how much you value arcane dust.

4 wins = 1pack,, 40 gold, 5 dust. Mind you these values have been known to change some as 2 victories has never resulted in the same results...

7 wins = 1 pack 50 dust and 50 gold + gold card

8= 1 pack 200 gold and 50 dust.....

This is just what I have been able to see on streams and what I have obtained in my time in arena.

The amounts vary per win, but increase at intervals. If you win none you get pretty crap, if you win 1 you get crap, if you win 2 you win crap but not as bad, 3 also not as bad crap, 4 less crappier 5 less crappier, 6 mid range 7 mid range, 8 can play again, 9 play again.....

So yeah pretty much only the top players can use arena to farm...And by top I mean those with a greater understanding of card interaction and who are extremely lucky. Or those who lucked out on three or more of the same same damage spell and exploit the hell out of it.