This is however just a calculation of single player average wins using combinatorics, basically finding the highest point on the probability bell curve. It didn't really adress the entire system and its two limits of 2.25 an 3 as we approach infinity.
The 50/50 (p=0.5) case does properly address the entire system. It's not finding the highest point on the probability bell curve, it's performing a weighted average over all possible exit conditions. As you said earlier, the 9-win condition reduces the average number of wins ever so slightly, because it truncates the infinite series. You're right that if we remove that condition, we should approach the limit of 3 (essentially the original conjecture - that since 3 losses is the only way to remove a deck from the system, the average number of wins has to be 3).
I'm not sure where the 2.25 comes from, but I'll admit I haven't read the entirety of the thread; just came in here at the request of someone on twitter.
I think you guys are severely underestimating the value of arcane dust. Some one equated a pack to the amount of dust require to craft those cards: this is wrong. Early on it may be the case, but after you have all the cards and are looking to craft golden cards, often a pack is worth the dusted value of the 4 commons 1 rare you get. That's 40 dust--yeah, a pack is worth 40 dust. At minimum, mind you, but that needs to be factored into the equation.
Also don't forget to factor in the 10 gold per 3 wins into play mode. If you consider arena an average of 3 runs (which I know it's slightly less than) that requires arena to yield an additional 10 gold.
There is another factor to consider as well: an average of 3 wins in arena does NOT mean you get, on average, the rewards for winning 3 times (unless you get 3 wins every time). I'm not sure what the math is, but it would be based on the rewards given at all 10 levels, based on their frequency, averaged.
EX: you get 2, 3, and 4 wins. 2 wins = 120 value; 3 wins = 145 value; 4 wins = 165 value. that's an average of ~143, not 145. (These numbers are completely made up)
On average: 0, 1, 2 wins are achieved most often, 3, 4, 5 are achieved less often, and 6+ are achieved the least often. You would need to know the average yield of each reward level and the weight of each level's frequency to be achieved. Frequency will change over time: as the playing field evens out over time, there will be less 6+ wins and more 3/3 wins.
Finally there's another thing to consider: in play mode, a new player can achieve a deck which is very powerful based on online recommendations (find the flavor of the day cookie-cutter deck) and have a better chance at winning. In arena, that player would need to know everything that goes into making a good arena deck every time he plays. This coupled with play mode's focus on new packs (which are better for new players) would make play mode better for new players and, respectively, make arena mode potentially better for experienced players.
Well yes, its exactly what it does. Assigning a given probability of winning each match, in this case 50%, and running a number of trials, in this case I believe you chose one million, to approach the 'true' outcome. I do apologize, I know the math but am not fluent in the english part of mathematical expressions. By 'true' outcome I mean the average outcome regarding of number of matches played as well as the number of wins. A weighted average over all possible conditions is in fact also apex of a probability bell curve, there is no contradiction here. You are most likely to play 5.9 games and have 2.95 wins. A million trials is certainly good enough for an estimate, but this can be increased if you want higher accuracy in form of lower variance and standard deviation.
It's not the apex. If you have a probability density function (i.e. probability of winning N games), the probability weighted average is not necessarily the same as the apex of that curve. This is only the case in specific circumstances, in particular symmetric PDFs. Based on your usage of "probability bell curve" it's clear this is what you're assuming, but there's no guarantee that the PDF for number of games won looks like a bell curve.
In fact, it intrinsically does *not* in this case - both the 3-loss exit condition and 9-win condition introduce asymmetric skew. For example, what is your probability of winning 1 or 2 games? 13 games? If it were a bell curve, it would be nonzero.
Here's what the PDF actually looks like:
This is most certainly not a bell curve, and the average (nearly 3) is certainly not the apex of the distribution.
Note that this is not a Monte-Carlo simulation - the "one million games" thing was brought up by someone else. This is an analytical (i.e. exact) solution to the problem using simple combinatorics.
Well, it's not the integration of this chart, strictly speaking. For one, the chart isn't continuous; but if we integrate discretely (or more simply, cumulative sum) you still don't get a bell curve. It would be a monotonically increasing function that starts at 0 at x=0 and goes up to 1 at x=9:
That said, I do see what you're saying. You're answering the question, "given M trials (decks played to completion), what is the probability of getting an average of 2.95 wins?" That should indeed look like a normal distribution (bell curve) thanks to the Central Limit Theorem, unless I'm mistaken.
I found this page searching Google for Arena win rate math. Theck, if you're still there (or if anyone knows how to contact you)... I have some questions.
This chart assumes each match is a coinflip (50% win chance for each participant). That's a fair assumption for the overall population because the Arena tries to match players by win rate - as I understand it if you went 0-2, they will try to give you an opponent who has also gone 0-2 also, so even though they're both worse players, they should be worse by the same amount.
What would be more useful is actually charts modelling win rates for a characters with a known skill levels. This is practical because then we can look at the charts and say, hm, if you're at the 30th percentile of skill, you'll win on average X games, which isn't worth it, but if you're at the 80th percentile of skill, you'll actually manage to break even with Y games most of the time. Let me give you an example of what I mean, why it's different to the 50% win rate assumption you use for the entire population...
Say we assign players a skill rating between 0 and 1, and when two players play, the player with the higher skill rating wins. Say skill rating is distributed normally in the population with mean 0.5 and standard deviation 0.165, truncating at 0 and 1.
The first player you fight is drawn from a population with mean 0.5 and standard deviation 0.165
If you lost the first match, you'll be matched with a weaker opponent. Presumably we can find out exactly how much weaker with some maths, but I don't know it. I just plugged the numbers into Excel (10,000 samples, two players match, look at the distribution of skill of the losers coming out of game 1 and going into game 2). A person who has lost one game is drawn from a population with mean 0.405 and standard deviation of 0.135 - this is a significantly weaker choice of opponent compared to your first game. A person who has lost two games in a row is drawn from a population with mean 0.327 and standard deviation 0.115.
The maths would look something like this: "you" are a person with a skill distribution of mean 0.6 and standard deviation 0.05 (you're a 60th percentile player, and the standard deviation is small because it's only the variance in the decks you build and hand draws). The first match you fight is against a person drawn from a population with mean 0.5 and standard deviation 0.165. If you lose it, you then go on to fight a weaker person with mean 0.405 and standard deviation 0.135. If you won, however, you get given a person who also went 1-0 - this person is drawn from a population with mean 0.595 and standard deviation 0.135.
And so on. This has some stabilizing effect on the win rate forcing it to tend to a mean of some sorts, reducing the probability of outliers. The shape of the graph would be very different from the chart you show with a flat 50% win rate. The sum of every person's individual chart though, would sum up to the entire population chart which you showed.
I vaguely remember some methods for adding / subtracting / using conditional formulas on PDFs but it's all forgotten to me now...
it's simple...pack is 100, arena is 150. As long as you make 50 gold per arena run, you got your packs worth of gold value. Anything over 50 is profit compared to getting a pack.
At 9 wins you get 150g + 1 pack + 1 pack/Golden card + 1 random + 1 random
150g is worth one arena ticket or 1.5 packs
1 pack is worth at least 40 dust
1 Golden card is worth at least 50 dust
At 9 wins you are guaranteed at least 140 dust, but have a very high chance of getting more dust.
Gold on the other hand cannot be converted from anything, gold can be changed into stuff but stuff cannot be changed into gold.
This limits the least amount of gold to 150 for 9 wins.
What we need to determine is the pool of stuff the 2 random slots can be.
We know they include 1 pack, 1 random golden expert card, but we do not know the amounts of gold or arcane dust possible for maximums and minimums per level of rewards.
You are all too smart to get over a very stupid invalid assumption: This is NOT a closed system.
1. New decks are entering arena all the time.
2. Bad decks/players are getting wiped out below 3 wins.
3. Long streakers are getting wiped out some of the time.
4. Even longer streakers are getting fewer and more far between.
It is a very fluid, very dynamic system, and therefore cannot assume to be closed regardless of your guys' statistical analysis. One can, however, capitalize on the population fluctuations to some degree, if they have all hours of the day available to invest in Hearthstone.
Wow, you must've digged down deep to find this thread! =)
Anyway, does anyone know what current 12 wins arena reward system looks like? How many wins do you need for a guaranteed card or pack? I've got 12 wins only once and was awarded 1 pack, 1 normal rare card, 90 gold, 150 gold, 235 gold. If I remember correctly, you get
5 chests only for 12 wins,
4 chests for 7-11 wins,
3 chests for 3-6 wins,
2 chests for 0-2 wins
You can get a card if you get only 3 chests, maybe even less. So I guess 1st chest is a guaranteed card pack. 2nd and 3rd can be X gold or Y dust or a card. 4th and 5th can be dust, gold, card or a card pack? Also not sure if a number of wins affects the probability of getting a golden card instead of normal.
Whoa, I forgot I made this lol. I suppose it's relevant again since the gold-per-win was upped (since I made the thread) and arena wins were increased to 12 possible, making a lot of the previous math invalid/outdated (although the general picture is basically the same).
I'm not sure if this has been posted yet.. But you don't break even if you get 50 gold from arena. Don't forget that you would get 10g per 3 wins if you invested that time into constructed, unless you hit your gold cap on a daily basis. So if you've earned 50 gold in a 5/3 run, you've actually lost about 17.7 gold.
According to this site http://www.arenamastery.com/ you have a chance to get a second card pack only at 12 wins (about 30% chance) and a golden card only at 8-12 wins (~50% chance), while you can get a normal card in any of 0-12 outcomes (10-30%, depending on # of wins).
I'm not sure if this has been posted yet.. But you don't break even if you get 50 gold from arena. Don't forget that you would get 10g per 3 wins if you invested that time into constructed, unless you hit your gold cap on a daily basis. So if you've earned 50 gold in a 5/3 run, you've actually lost about 17.7 gold.
Old thread, with old posts and outdated info. This was made with the old "max 9 win" arena system.
The thread been necroed (I see no reason for it), but I'm sure some people won't realize that and will read the first few posts nonetheless (I did, and then I found something "off" about the numbers).
Since this thread got necroed and information provided is way outdated, I'm closing it here. If you have something to add, just use one of the current ones.
Rollback Post to RevisionRollBack
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The 50/50 (p=0.5) case does properly address the entire system. It's not finding the highest point on the probability bell curve, it's performing a weighted average over all possible exit conditions. As you said earlier, the 9-win condition reduces the average number of wins ever so slightly, because it truncates the infinite series. You're right that if we remove that condition, we should approach the limit of 3 (essentially the original conjecture - that since 3 losses is the only way to remove a deck from the system, the average number of wins has to be 3).
I'm not sure where the 2.25 comes from, but I'll admit I haven't read the entirety of the thread; just came in here at the request of someone on twitter.
I think you guys are severely underestimating the value of arcane dust. Some one equated a pack to the amount of dust require to craft those cards: this is wrong. Early on it may be the case, but after you have all the cards and are looking to craft golden cards, often a pack is worth the dusted value of the 4 commons 1 rare you get. That's 40 dust--yeah, a pack is worth 40 dust. At minimum, mind you, but that needs to be factored into the equation.
Also don't forget to factor in the 10 gold per 3 wins into play mode. If you consider arena an average of 3 runs (which I know it's slightly less than) that requires arena to yield an additional 10 gold.
There is another factor to consider as well: an average of 3 wins in arena does NOT mean you get, on average, the rewards for winning 3 times (unless you get 3 wins every time). I'm not sure what the math is, but it would be based on the rewards given at all 10 levels, based on their frequency, averaged.
EX: you get 2, 3, and 4 wins. 2 wins = 120 value; 3 wins = 145 value; 4 wins = 165 value. that's an average of ~143, not 145. (These numbers are completely made up)
On average: 0, 1, 2 wins are achieved most often, 3, 4, 5 are achieved less often, and 6+ are achieved the least often. You would need to know the average yield of each reward level and the weight of each level's frequency to be achieved. Frequency will change over time: as the playing field evens out over time, there will be less 6+ wins and more 3/3 wins.
Finally there's another thing to consider: in play mode, a new player can achieve a deck which is very powerful based on online recommendations (find the flavor of the day cookie-cutter deck) and have a better chance at winning. In arena, that player would need to know everything that goes into making a good arena deck every time he plays. This coupled with play mode's focus on new packs (which are better for new players) would make play mode better for new players and, respectively, make arena mode potentially better for experienced players.
It's not the apex. If you have a probability density function (i.e. probability of winning N games), the probability weighted average is not necessarily the same as the apex of that curve. This is only the case in specific circumstances, in particular symmetric PDFs. Based on your usage of "probability bell curve" it's clear this is what you're assuming, but there's no guarantee that the PDF for number of games won looks like a bell curve.
In fact, it intrinsically does *not* in this case - both the 3-loss exit condition and 9-win condition introduce asymmetric skew. For example, what is your probability of winning 1 or 2 games? 13 games? If it were a bell curve, it would be nonzero.
Here's what the PDF actually looks like:
This is most certainly not a bell curve, and the average (nearly 3) is certainly not the apex of the distribution.
Note that this is not a Monte-Carlo simulation - the "one million games" thing was brought up by someone else. This is an analytical (i.e. exact) solution to the problem using simple combinatorics.
Well, it's not the integration of this chart, strictly speaking. For one, the chart isn't continuous; but if we integrate discretely (or more simply, cumulative sum) you still don't get a bell curve. It would be a monotonically increasing function that starts at 0 at x=0 and goes up to 1 at x=9:
That said, I do see what you're saying. You're answering the question, "given M trials (decks played to completion), what is the probability of getting an average of 2.95 wins?" That should indeed look like a normal distribution (bell curve) thanks to the Central Limit Theorem, unless I'm mistaken.
I found this page searching Google for Arena win rate math. Theck, if you're still there (or if anyone knows how to contact you)... I have some questions.
This chart assumes each match is a coinflip (50% win chance for each participant). That's a fair assumption for the overall population because the Arena tries to match players by win rate - as I understand it if you went 0-2, they will try to give you an opponent who has also gone 0-2 also, so even though they're both worse players, they should be worse by the same amount.
What would be more useful is actually charts modelling win rates for a characters with a known skill levels. This is practical because then we can look at the charts and say, hm, if you're at the 30th percentile of skill, you'll win on average X games, which isn't worth it, but if you're at the 80th percentile of skill, you'll actually manage to break even with Y games most of the time. Let me give you an example of what I mean, why it's different to the 50% win rate assumption you use for the entire population...
Say we assign players a skill rating between 0 and 1, and when two players play, the player with the higher skill rating wins. Say skill rating is distributed normally in the population with mean 0.5 and standard deviation 0.165, truncating at 0 and 1.
The first player you fight is drawn from a population with mean 0.5 and standard deviation 0.165
If you lost the first match, you'll be matched with a weaker opponent. Presumably we can find out exactly how much weaker with some maths, but I don't know it. I just plugged the numbers into Excel (10,000 samples, two players match, look at the distribution of skill of the losers coming out of game 1 and going into game 2). A person who has lost one game is drawn from a population with mean 0.405 and standard deviation of 0.135 - this is a significantly weaker choice of opponent compared to your first game. A person who has lost two games in a row is drawn from a population with mean 0.327 and standard deviation 0.115.
The maths would look something like this: "you" are a person with a skill distribution of mean 0.6 and standard deviation 0.05 (you're a 60th percentile player, and the standard deviation is small because it's only the variance in the decks you build and hand draws). The first match you fight is against a person drawn from a population with mean 0.5 and standard deviation 0.165. If you lose it, you then go on to fight a weaker person with mean 0.405 and standard deviation 0.135. If you won, however, you get given a person who also went 1-0 - this person is drawn from a population with mean 0.595 and standard deviation 0.135.
And so on. This has some stabilizing effect on the win rate forcing it to tend to a mean of some sorts, reducing the probability of outliers. The shape of the graph would be very different from the chart you show with a flat 50% win rate. The sum of every person's individual chart though, would sum up to the entire population chart which you showed.
I vaguely remember some methods for adding / subtracting / using conditional formulas on PDFs but it's all forgotten to me now...
it's simple...pack is 100, arena is 150. As long as you make 50 gold per arena run, you got your packs worth of gold value. Anything over 50 is profit compared to getting a pack.
At 9 wins you get 150g + 1 pack + 1 pack/Golden card + 1 random + 1 random
150g is worth one arena ticket or 1.5 packs
1 pack is worth at least 40 dust
1 Golden card is worth at least 50 dust
At 9 wins you are guaranteed at least 140 dust, but have a very high chance of getting more dust.
Gold on the other hand cannot be converted from anything, gold can be changed into stuff but stuff cannot be changed into gold.
This limits the least amount of gold to 150 for 9 wins.
What we need to determine is the pool of stuff the 2 random slots can be.
We know they include 1 pack, 1 random golden expert card, but we do not know the amounts of gold or arcane dust possible for maximums and minimums per level of rewards.
This determines the worth of an arena run.
You are all too smart to get over a very stupid invalid assumption: This is NOT a closed system.
1. New decks are entering arena all the time.
2. Bad decks/players are getting wiped out below 3 wins.
3. Long streakers are getting wiped out some of the time.
4. Even longer streakers are getting fewer and more far between.
It is a very fluid, very dynamic system, and therefore cannot assume to be closed regardless of your guys' statistical analysis. One can, however, capitalize on the population fluctuations to some degree, if they have all hours of the day available to invest in Hearthstone.
Wow, you must've digged down deep to find this thread! =)
Anyway, does anyone know what current 12 wins arena reward system looks like? How many wins do you need for a guaranteed card or pack? I've got 12 wins only once and was awarded 1 pack, 1 normal rare card, 90 gold, 150 gold, 235 gold. If I remember correctly, you get
You can get a card if you get only 3 chests, maybe even less. So I guess 1st chest is a guaranteed card pack. 2nd and 3rd can be X gold or Y dust or a card. 4th and 5th can be dust, gold, card or a card pack? Also not sure if a number of wins affects the probability of getting a golden card instead of normal.
Whoa, I forgot I made this lol. I suppose it's relevant again since the gold-per-win was upped (since I made the thread) and arena wins were increased to 12 possible, making a lot of the previous math invalid/outdated (although the general picture is basically the same).
I'm not sure if this has been posted yet.. But you don't break even if you get 50 gold from arena. Don't forget that you would get 10g per 3 wins if you invested that time into constructed, unless you hit your gold cap on a daily basis. So if you've earned 50 gold in a 5/3 run, you've actually lost about 17.7 gold.
Come join me @ twitch.tv/Creeepling ! Only fun and unusual decks!
According to this site http://www.arenamastery.com/ you have a chance to get a second card pack only at 12 wins (about 30% chance) and a golden card only at 8-12 wins (~50% chance), while you can get a normal card in any of 0-12 outcomes (10-30%, depending on # of wins).
Old thread, with old posts and outdated info. This was made with the old "max 9 win" arena system.
The thread been necroed (I see no reason for it), but I'm sure some people won't realize that and will read the first few posts nonetheless (I did, and then I found something "off" about the numbers).
Since this thread got necroed and information provided is way outdated, I'm closing it here.
If you have something to add, just use one of the current ones.
Please report toxic behaviour and unwanted threads, so the moderators can deal with them.