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Random mathematics question

dylancatlow
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5/10/2016 9:01:43 PM
Posted: 6 months ago
In the card game Hearthstone there's a game mode called Arena where you battle against random players using decks drafted from a random selection of cards. You keep playing until you either win 12 times or lose 3 times. Despite Blizzard's best efforts there's still an advantage to going first such that the person who goes first wins 55 percent of the time. When this fact came to my attention I realized that the win rates of good players are lower than they should be. First turn advantage effectively means that 5 percent of the time the person going first "steals" a win from the player going second. Since the person who goes first is decided by a coin flip this basically means that in 5 percent of games the skill level of the players is totally irrelevant. This means that for a good player -- someone who wins, say, 70 percent of the time -- that 5 percent is "locked in" at a 50 percent win rate instead of their usual 70, which means that their win rate is ~1 percent lower than it should be (they're winning 50 percent of that 5 percent instead of 70, the difference being 1 percent overall: .20 x5=1).

I also realized that oscillating back and forth between "high" and "low" win rates hurts skilled players as well. The chance to win 4 games in a row, for instance, is higher when your chance to win each individual game is 70 percent rather than a mix between 65 and 75 (24.01% vs 23.65%).

I need help solving the following:

1) The mean number of Arena wins for someone whose odds of winning each game is 71 percent.

2) The mean number of Arena wins for someone whose odds of winning each game has a 50 percent chance of being 65 and a 50 percent chance of being 75 percent.
dylancatlow
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5/10/2016 9:09:57 PM
Posted: 6 months ago
And yes, I realize that I oversimplified some stuff and totally ignored the fact that someone's win rate varies depending on what their current record is and hence the skill levels of the people they're facing.
RuvDraba
Posts: 6,033
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5/10/2016 9:34:20 PM
Posted: 6 months ago
Dylan, I haven't played the game or reviewed your analysis. Just to answer the math...

At 5/10/2016 9:01:43 PM, dylancatlow wrote:
1) The mean number of Arena wins for someone whose odds of winning each game is 71 percent.
The expected number of wins is given by the sum:
Sum (n=1.. infinity) (p ^ n) * n

where p is the probability of a win. (0.71 in this case), and n represents the number of wins in a row.

So the sum looks like:

0.71 * 1 + 0.71^2 *2 + 0.71^3 * 3 +...

There's bound to be an algebraic way to convert that expansion to a finite formula, but I haven't thought it through or looked it up. But it's the sort of thing you can also do in a spreadsheet. Crunching the numbers out to 50 terms though, you get an expected win rate of about 8.44 -- and it's stable to at least three decimal places by this time.

2) The mean number of Arena wins for someone whose odds of winning each game has a 50 percent chance of being 65 and a 50 percent chance of being 75 percent.

That gives you a weighted success rate of 0.5*0.65 + 0.5*0.75 = 0.7 (i.e. halfway between the two chances.)

Crunching expected wins out to 50 terms again for comparison, we get an expected 7.78 wins.

Hope that may help. :)
dylancatlow
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5/10/2016 9:49:55 PM
Posted: 6 months ago
At 5/10/2016 9:34:20 PM, RuvDraba wrote:
Dylan, I haven't played the game or reviewed your analysis. Just to answer the math...

At 5/10/2016 9:01:43 PM, dylancatlow wrote:
1) The mean number of Arena wins for someone whose odds of winning each game is 71 percent.
The expected number of wins is given by the sum:
Sum (n=1.. infinity) (p ^ n) * n

where p is the probability of a win. (0.71 in this case), and n represents the number of wins in a row.

So the sum looks like:

0.71 * 1 + 0.71^2 *2 + 0.71^3 * 3 +...



There's bound to be an algebraic way to convert that expansion to a finite formula, but I haven't thought it through or looked it up. But it's the sort of thing you can also do in a spreadsheet. Crunching the numbers out to 50 terms though, you get an expected win rate of about 8.44 -- and it's stable to at least three decimal places by this time.

2) The mean number of Arena wins for someone whose odds of winning each game has a 50 percent chance of being 65 and a 50 percent chance of being 75 percent.

That gives you a weighted success rate of 0.5*0.65 + 0.5*0.75 = 0.7 (i.e. halfway between the two chances.)

Crunching expected wins out to 50 terms again for comparison, we get an expected 7.78 wins.

Hope that may help. :)

It doesn't seem like this equation takes account of the fact that an Arena run ends as soon as someone loses 3 times though. I think it would have to be quite a bit more complex.
dylancatlow
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5/10/2016 9:51:19 PM
Posted: 6 months ago
At 5/10/2016 9:34:20 PM, RuvDraba wrote:
Dylan, I haven't played the game or reviewed your analysis. Just to answer the math...

At 5/10/2016 9:01:43 PM, dylancatlow wrote:
1) The mean number of Arena wins for someone whose odds of winning each game is 71 percent.
The expected number of wins is given by the sum:
Sum (n=1.. infinity) (p ^ n) * n

where p is the probability of a win. (0.71 in this case), and n represents the number of wins in a row.

So the sum looks like:

0.71 * 1 + 0.71^2 *2 + 0.71^3 * 3 +...



There's bound to be an algebraic way to convert that expansion to a finite formula, but I haven't thought it through or looked it up. But it's the sort of thing you can also do in a spreadsheet. Crunching the numbers out to 50 terms though, you get an expected win rate of about 8.44 -- and it's stable to at least three decimal places by this time.

2) The mean number of Arena wins for someone whose odds of winning each game has a 50 percent chance of being 65 and a 50 percent chance of being 75 percent.

That gives you a weighted success rate of 0.5*0.65 + 0.5*0.75 = 0.7 (i.e. halfway between the two chances.)

Crunching expected wins out to 50 terms again for comparison, we get an expected 7.78 wins.

Hope that may help. :)

I also don't think that averaging together the 65 percent win rate and the 75 percent win rate is valid.
RuvDraba
Posts: 6,033
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5/10/2016 11:38:40 PM
Posted: 6 months ago
At 5/10/2016 9:49:55 PM, dylancatlow wrote:
At 5/10/2016 9:34:20 PM, RuvDraba wrote:
Dylan, I haven't played the game or reviewed your analysis. Just to answer the math...

At 5/10/2016 9:01:43 PM, dylancatlow wrote:
1) The mean number of Arena wins for someone whose odds of winning each game is 71 percent.
The expected number of wins is given by the sum:
Sum (n=1.. infinity) (p ^ n) * n

It doesn't seem like this equation takes account of the fact that an Arena run ends as soon as someone loses 3 times though. I think it would have to be quite a bit more complex.

Ah. I'm beginning to see the issue. i didn't understand how the game works.

So it's a series of opponents? Any player is ejected after three losses, and a champ is ejected after twelve wins?

If so, it seems that what happens to other players irrelevant to the length of your game. All that predicts your game length is where your wins and losses fall. If I understand the rules, your game-length can never exceed fifteen matches -- at which time you'll have either accumulated twelve wins, or three losses.

Is that correct?
RuvDraba
Posts: 6,033
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5/10/2016 11:40:19 PM
Posted: 6 months ago
At 5/10/2016 11:38:40 PM, RuvDraba wrote:
At 5/10/2016 9:49:55 PM, dylancatlow wrote:
At 5/10/2016 9:34:20 PM, RuvDraba wrote:
Dylan, I haven't played the game or reviewed your analysis. Just to answer the math...

At 5/10/2016 9:01:43 PM, dylancatlow wrote:
1) The mean number of Arena wins for someone whose odds of winning each game is 71 percent.
The expected number of wins is given by the sum:
Sum (n=1.. infinity) (p ^ n) * n

It doesn't seem like this equation takes account of the fact that an Arena run ends as soon as someone loses 3 times though. I think it would have to be quite a bit more complex.

Ah. I'm beginning to see the issue. i didn't understand how the game works.

So it's a series of opponents? Any player is ejected after three losses, and a champ is ejected after twelve wins?

If so, it seems that what happens to other players irrelevant to the length of your game. All that predicts your game length is where your wins and losses fall. If I understand the rules, your game-length can never exceed fifteen matches -- at which time you'll have either accumulated twelve wins, or three losses.

Is that correct?

Er.. I mean 14 matches. Because 13 gives you 11 wins + 2 losses, and the fourteenth will see you ejected no matter what.
dylancatlow
Posts: 12,242
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5/10/2016 11:46:28 PM
Posted: 6 months ago
At 5/10/2016 11:38:40 PM, RuvDraba wrote:
At 5/10/2016 9:49:55 PM, dylancatlow wrote:
At 5/10/2016 9:34:20 PM, RuvDraba wrote:
Dylan, I haven't played the game or reviewed your analysis. Just to answer the math...

At 5/10/2016 9:01:43 PM, dylancatlow wrote:
1) The mean number of Arena wins for someone whose odds of winning each game is 71 percent.
The expected number of wins is given by the sum:
Sum (n=1.. infinity) (p ^ n) * n

It doesn't seem like this equation takes account of the fact that an Arena run ends as soon as someone loses 3 times though. I think it would have to be quite a bit more complex.

Ah. I'm beginning to see the issue. i didn't understand how the game works.

So it's a series of opponents? Any player is ejected after three losses, and a champ is ejected after twelve wins?

If so, it seems that what happens to other players irrelevant to the length of your game. All that predicts your game length is where your wins and losses fall. If I understand the rules, your game-length can never exceed fifteen matches -- at which time you'll have either accumulated twelve wins, or three losses.

Is that correct?

Yes.

The matchmaking system tries to pair you up with opponents with similar win records, so as you accumulate more wins your win rate should fall, which complicates matters greatly. But I'm just going to ignore this.
dylancatlow
Posts: 12,242
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5/10/2016 11:48:06 PM
Posted: 6 months ago
At 5/10/2016 11:40:19 PM, RuvDraba wrote:
At 5/10/2016 11:38:40 PM, RuvDraba wrote:
At 5/10/2016 9:49:55 PM, dylancatlow wrote:
At 5/10/2016 9:34:20 PM, RuvDraba wrote:
Dylan, I haven't played the game or reviewed your analysis. Just to answer the math...

At 5/10/2016 9:01:43 PM, dylancatlow wrote:
1) The mean number of Arena wins for someone whose odds of winning each game is 71 percent.
The expected number of wins is given by the sum:
Sum (n=1.. infinity) (p ^ n) * n

It doesn't seem like this equation takes account of the fact that an Arena run ends as soon as someone loses 3 times though. I think it would have to be quite a bit more complex.

Ah. I'm beginning to see the issue. i didn't understand how the game works.

So it's a series of opponents? Any player is ejected after three losses, and a champ is ejected after twelve wins?

If so, it seems that what happens to other players irrelevant to the length of your game. All that predicts your game length is where your wins and losses fall. If I understand the rules, your game-length can never exceed fifteen matches -- at which time you'll have either accumulated twelve wins, or three losses.

Is that correct?

Er.. I mean 14 matches. Because 13 gives you 11 wins + 2 losses, and the fourteenth will see you ejected no matter what.

Yeah, 14 is the most, oops.
user13579
Posts: 822
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5/11/2016 1:05:45 AM
Posted: 6 months ago
At 5/10/2016 9:01:43 PM, dylancatlow wrote:
In the card game Hearthstone there's a game mode called Arena where you battle against random players using decks drafted from a random selection of cards. You keep playing until you either win 12 times or lose 3 times. Despite Blizzard's best efforts there's still an advantage to going first such that the person who goes first wins 55 percent of the time. When this fact came to my attention I realized that the win rates of good players are lower than they should be. First turn advantage effectively means that 5 percent of the time the person going first "steals" a win from the player going second. Since the person who goes first is decided by a coin flip this basically means that in 5 percent of games the skill level of the players is totally irrelevant. This means that for a good player -- someone who wins, say, 70 percent of the time -- that 5 percent is "locked in" at a 50 percent win rate instead of their usual 70, which means that their win rate is ~1 percent lower than it should be (they're winning 50 percent of that 5 percent instead of 70, the difference being 1 percent overall: .20 x5=1).

I'm guessing that somebody who wins 70% overall wins about (70+x)% of the time when going first and (70-x)% of the time when going second. If they have an equal number as each side it's 70% overall. I'm too lazy to figure out what x is, but x=5 maybe
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Mhykiel
Posts: 5,987
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5/11/2016 3:52:20 AM
Posted: 6 months ago
At 5/10/2016 11:38:40 PM, RuvDraba wrote:
At 5/10/2016 9:49:55 PM, dylancatlow wrote:
At 5/10/2016 9:34:20 PM, RuvDraba wrote:
Dylan, I haven't played the game or reviewed your analysis. Just to answer the math...

At 5/10/2016 9:01:43 PM, dylancatlow wrote:
1) The mean number of Arena wins for someone whose odds of winning each game is 71 percent.
The expected number of wins is given by the sum:
Sum (n=1.. infinity) (p ^ n) * n

It doesn't seem like this equation takes account of the fact that an Arena run ends as soon as someone loses 3 times though. I think it would have to be quite a bit more complex.

Ah. I'm beginning to see the issue. i didn't understand how the game works.

So it's a series of opponents? Any player is ejected after three losses, and a champ is ejected after twelve wins?

If so, it seems that what happens to other players irrelevant to the length of your game. All that predicts your game length is where your wins and losses fall. If I understand the rules, your game-length can never exceed fifteen matches -- at which time you'll have either accumulated twelve wins, or three losses.

Is that correct?

14. Is most games 12 wins 2 losses or 11 wins 3 losses.
Mhykiel
Posts: 5,987
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5/11/2016 4:06:10 AM
Posted: 6 months ago
At 5/10/2016 11:40:19 PM, RuvDraba wrote:
At 5/10/2016 11:38:40 PM, RuvDraba wrote:
At 5/10/2016 9:49:55 PM, dylancatlow wrote:
At 5/10/2016 9:34:20 PM, RuvDraba wrote:
Dylan, I haven't played the game or reviewed your analysis. Just to answer the math...

At 5/10/2016 9:01:43 PM, dylancatlow wrote:
1) The mean number of Arena wins for someone whose odds of winning each game is 71 percent.
The expected number of wins is given by the sum:
Sum (n=1.. infinity) (p ^ n) * n

It doesn't seem like this equation takes account of the fact that an Arena run ends as soon as someone loses 3 times though. I think it would have to be quite a bit more complex.

Ah. I'm beginning to see the issue. i didn't understand how the game works.

So it's a series of opponents? Any player is ejected after three losses, and a champ is ejected after twelve wins?

If so, it seems that what happens to other players irrelevant to the length of your game. All that predicts your game length is where your wins and losses fall. If I understand the rules, your game-length can never exceed fifteen matches -- at which time you'll have either accumulated twelve wins, or three losses.

Is that correct?

Er.. I mean 14 matches. Because 13 gives you 11 wins + 2 losses, and the fourteenth will see you ejected no matter what.

Ah oops I should have kept reading.

I play the game. Each opponent picks 30 cards from 30 choices of 3 fairly equal value cards. The goal is to pick cards that combo together or allow for a tactic that inflicts more damage against your opponent,

The 5 percent is that when 2 equally skilled players equipped with equally constructed decks face off. The person going first wins 5 percent of the time.

I don't know where the OP get's that number.

It has been said that 95 percent of the matches played are blind chance. Like a game of war or gold fish. The illusion of mastering the game when in reality you can go up and down the ranks with even random flips of the finger.

I'm not sure if I accept that but I know blizzard is leveraging game theory and mechanics to make it addictive to the common population. And it certainly can feel like a night at the slot machine or black jack table.

Each card has a cost to play attached with it called mana. The second player will get a 'coin card' that gives one free mana for one round. This usually evens out the power available to each opponent.

But the person who went first will have a monster ready to attack secound round. That 1 round of inflicting 2 more damage than the 2 turn player is what I feel accounts for any advantage going first has.
Mhykiel
Posts: 5,987
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5/11/2016 4:07:59 AM
Posted: 6 months ago
At 5/10/2016 11:48:06 PM, dylancatlow wrote:
At 5/10/2016 11:40:19 PM, RuvDraba wrote:
At 5/10/2016 11:38:40 PM, RuvDraba wrote:
At 5/10/2016 9:49:55 PM, dylancatlow wrote:
At 5/10/2016 9:34:20 PM, RuvDraba wrote:
Dylan, I haven't played the game or reviewed your analysis. Just to answer the math...

At 5/10/2016 9:01:43 PM, dylancatlow wrote:
1) The mean number of Arena wins for someone whose odds of winning each game is 71 percent.
The expected number of wins is given by the sum:
Sum (n=1.. infinity) (p ^ n) * n

It doesn't seem like this equation takes account of the fact that an Arena run ends as soon as someone loses 3 times though. I think it would have to be quite a bit more complex.

Ah. I'm beginning to see the issue. i didn't understand how the game works.

So it's a series of opponents? Any player is ejected after three losses, and a champ is ejected after twelve wins?

If so, it seems that what happens to other players irrelevant to the length of your game. All that predicts your game length is where your wins and losses fall. If I understand the rules, your game-length can never exceed fifteen matches -- at which time you'll have either accumulated twelve wins, or three losses.

Is that correct?

Er.. I mean 14 matches. Because 13 gives you 11 wins + 2 losses, and the fourteenth will see you ejected no matter what.

Yeah, 14 is the most, oops.

Oh yeah and the opponent pairing system is by wins. Blizzard has said the matchmaking is so that you win and. Lose 50 percent of the time.
RuvDraba
Posts: 6,033
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5/11/2016 5:03:06 AM
Posted: 6 months ago
At 5/10/2016 11:48:06 PM, dylancatlow wrote:
At 5/10/2016 11:40:19 PM, RuvDraba wrote:
At 5/10/2016 11:38:40 PM, RuvDraba wrote:
It seems that what happens to other players irrelevant to the length of your game. All that predicts your game length is where your wins and losses fall. If I understand the rules, your game-length can never exceed fifteen matches -- at which time you'll have either accumulated twelve wins, or three losses.

Is that correct?

Er.. I mean 14 matches. Because 13 gives you 11 wins + 2 losses, and the fourteenth will see you ejected no matter what.

Yeah, 14 is the most, oops.

(Sorry for delay -- I had to chip away at this between bouts of work")

Assuming games are always played to completion (i.e. no forfeits), let's define
E(n) as the expected number of wins if a game is of length exactly n, for n= 3 .. 14

Since games of length up to 11 bouts always terminate with three losses, we know that:
E(3) = 0,
E(4) = 1,
E(5) = 2,
E(6) = 3,
..
E(11) = 8


We'll need to treat E(12 ... 14) separately, so let's come back to them.

Now let's look at the probability of games of length 3..11. These games always terminate with a third loss on the last game. So if w is your chance of win per game, define G(n, 3) as the probability that a game of length n sees exactly 3 losses, with your final loss on the last game. So you have to find all the ways to fit exactly two losses into a game of length n-1, and then get a loss on the final game.

If my calculations are right, that gives us:
G(3,3) = 2C2 * (1-w)^3 * w^0
G(4,3) = 3C2 * (1-w)^3 * w^1
G(5,3) = 4C2 * (1-w)^3 * w^2
..
G(n,3) = (n-1)C2 * (1-w)^3 * w^(n-3)


And Expected wins for up to 11 games in a run is:

Sum (n = 3 .. 11) E(n) * G(n,3)

There are some algebraic substitutions we could do to simplify all this, but since we won't be reusing it all that much, let's just crunch it on a spreadsheet. I get"
G(3,3) = 0.024389
G(4,3) = 0.051949
G(5,3) = 0.073767
G(6,3) = 0.087291
G(7,3) = 0.092965
G(8,3) = 0.092407
G(9,3) = 0.087479
G(10,3) = 0.079856
G(11,3) = 0.070872


And so far the expected number of wins for just the games of length 3 .. 11 with w = 0.71 2.946084.

To that we have to add the calculations for G(12,3), G(12,0), G(13,3), G(13,1), G(14,3), G(14,2), where the final result for on G(n,k) is a win if k <= 2, but a loss if k=3. And then we have to work out E(n) for n = 12..14 based on those numbers.

Anyone want to finish it off?
RainbowDash52
Posts: 294
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5/11/2016 2:32:58 PM
Posted: 6 months ago
Answer: 6.68113751508077

Calculations done here: http://rextester.com...

You can edit the winPercent variable value from .71 to something else to find the mean number of wins for other win percents too.
dylancatlow
Posts: 12,242
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5/11/2016 4:22:59 PM
Posted: 6 months ago
At 5/11/2016 2:32:58 PM, RainbowDash52 wrote:
Answer: 6.68113751508077

Calculations done here: http://rextester.com...

You can edit the winPercent variable value from .71 to something else to find the mean number of wins for other win percents too.

Thanks, this is exactly what I needed! Any idea on how to solve the second question?
Burzmali
Posts: 1,310
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5/11/2016 4:27:04 PM
Posted: 6 months ago
At 5/11/2016 5:03:06 AM, RuvDraba wrote:
At 5/10/2016 11:48:06 PM, dylancatlow wrote:
At 5/10/2016 11:40:19 PM, RuvDraba wrote:
At 5/10/2016 11:38:40 PM, RuvDraba wrote:
It seems that what happens to other players irrelevant to the length of your game. All that predicts your game length is where your wins and losses fall. If I understand the rules, your game-length can never exceed fifteen matches -- at which time you'll have either accumulated twelve wins, or three losses.

Is that correct?

Er.. I mean 14 matches. Because 13 gives you 11 wins + 2 losses, and the fourteenth will see you ejected no matter what.

Yeah, 14 is the most, oops.

(Sorry for delay -- I had to chip away at this between bouts of work")

Assuming games are always played to completion (i.e. no forfeits), let's define
E(n) as the expected number of wins if a game is of length exactly n, for n= 3 .. 14

Since games of length up to 11 bouts always terminate with three losses, we know that:
E(3) = 0,
E(4) = 1,
E(5) = 2,
E(6) = 3,
..
E(11) = 8


We'll need to treat E(12 ... 14) separately, so let's come back to them.

Now let's look at the probability of games of length 3..11. These games always terminate with a third loss on the last game. So if w is your chance of win per game, define G(n, 3) as the probability that a game of length n sees exactly 3 losses, with your final loss on the last game. So you have to find all the ways to fit exactly two losses into a game of length n-1, and then get a loss on the final game.

If my calculations are right, that gives us:
G(3,3) = 2C2 * (1-w)^3 * w^0
G(4,3) = 3C2 * (1-w)^3 * w^1
G(5,3) = 4C2 * (1-w)^3 * w^2
..
G(n,3) = (n-1)C2 * (1-w)^3 * w^(n-3)


And Expected wins for up to 11 games in a run is:

Sum (n = 3 .. 11) E(n) * G(n,3)

There are some algebraic substitutions we could do to simplify all this, but since we won't be reusing it all that much, let's just crunch it on a spreadsheet. I get"
G(3,3) = 0.024389
G(4,3) = 0.051949
G(5,3) = 0.073767
G(6,3) = 0.087291
G(7,3) = 0.092965
G(8,3) = 0.092407
G(9,3) = 0.087479
G(10,3) = 0.079856
G(11,3) = 0.070872


And so far the expected number of wins for just the games of length 3 .. 11 with w = 0.71 2.946084.

To that we have to add the calculations for G(12,3), G(12,0), G(13,3), G(13,1), G(14,3), G(14,2), where the final result for on G(n,k) is a win if k <= 2, but a loss if k=3. And then we have to work out E(n) for n = 12..14 based on those numbers.

Anyone want to finish it off?

To clarify, Ruv, the Arena system works like this:
I just created (drafted) my deck and my record is 0-0. I hit the matchmaking button and presumably get matched to someone who is also 0-0. We play a single game. Let's say I win, so my record is now 1-0. I hit the matchmaking button again and should be matched to someone else who is 1-0. We play one game, etc. I repeat until I have 12 wins or 3 losses. Most people don't drop out before that point because you get better and better rewards based on your record when you are finally done, and it costs either real money or in-game currency (which can be earned) to start an Arena run.
Burzmali
Posts: 1,310
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5/11/2016 4:31:39 PM
Posted: 6 months ago
Not to drag this too far sideways, but does anyone have an opinion of the recent change to split to Standard and Wild formats? I personally am not great at deck-building, and I don't care to spend more than my initial $20 investment in the game. So I'm struggling a bit to compete with the new set being out and only having like 5 packs worth of the new cards that they gave me.
RainbowDash52
Posts: 294
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5/11/2016 4:38:22 PM
Posted: 6 months ago
At 5/11/2016 4:22:59 PM, dylancatlow wrote:
At 5/11/2016 2:32:58 PM, RainbowDash52 wrote:
Answer: 6.68113751508077

Calculations done here: http://rextester.com...

You can edit the winPercent variable value from .71 to something else to find the mean number of wins for other win percents too.

Thanks, this is exactly what I needed! Any idea on how to solve the second question?

For the second question, wouldn't a 50% chance of 75% win rate, and a 50% chance of a 65% win rate be identical to a 70% win rate?
dylancatlow
Posts: 12,242
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5/11/2016 5:04:09 PM
Posted: 6 months ago
At 5/11/2016 4:38:22 PM, RainbowDash52 wrote:
At 5/11/2016 4:22:59 PM, dylancatlow wrote:
At 5/11/2016 2:32:58 PM, RainbowDash52 wrote:
Answer: 6.68113751508077

Calculations done here: http://rextester.com...

You can edit the winPercent variable value from .71 to something else to find the mean number of wins for other win percents too.

Thanks, this is exactly what I needed! Any idea on how to solve the second question?

For the second question, wouldn't a 50% chance of 75% win rate, and a 50% chance of a 65% win rate be identical to a 70% win rate?

No, you'd much rather have a consistent win rate. For example, someone whose chances of winning are 70 across all games has a 1.3 percent probability of getting 12 wins in a row, whereas someone whose chances of winning are 75 percent for the first six games and 65 for the latter six games only has a .8 percent chance of getting 12 wins in a row. The consistent person has a 2.7 percent chance of getting zero wins, whereas the inconsistent person has a 2.76 percent chance.
user13579
Posts: 822
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5/11/2016 5:11:30 PM
Posted: 6 months ago
At 5/11/2016 5:04:09 PM, dylancatlow wrote:
At 5/11/2016 4:38:22 PM, RainbowDash52 wrote:
At 5/11/2016 4:22:59 PM, dylancatlow wrote:
At 5/11/2016 2:32:58 PM, RainbowDash52 wrote:
Answer: 6.68113751508077

Calculations done here: http://rextester.com...

You can edit the winPercent variable value from .71 to something else to find the mean number of wins for other win percents too.

Thanks, this is exactly what I needed! Any idea on how to solve the second question?

For the second question, wouldn't a 50% chance of 75% win rate, and a 50% chance of a 65% win rate be identical to a 70% win rate?

No, you'd much rather have a consistent win rate. For example, someone whose chances of winning are 70 across all games has a 1.3 percent probability of getting 12 wins in a row, whereas someone whose chances of winning are 75 percent for the first six games and 65 for the latter six games only has a .8 percent chance of getting 12 wins in a row. The consistent person has a 2.7 percent chance of getting zero wins, whereas the inconsistent person has a 2.76 percent chance.

That doesn't make sense, because the second person has an average win rate of 70 ((65+75)/2 assuming an equal number of games as each side)! Those two people are basically the same person. If I play 1000 games at 65% and 1000 games at 75%, my average win rate is 70%.
Science in a nutshell:
"Facts are neither true nor false. They simply are."
"All scientific knowledge is provisional. Even facts are provisional."
"We can be absolutely certain that we have a moon, we can be absolutely certain that water is made out of H2O, and we can be absolutely certain that the Earth is a sphere!"
"Scientific knowledge is a body of statements of varying degrees of certainty -- some most unsure, some nearly sure, none absolutely certain."
dylancatlow
Posts: 12,242
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5/11/2016 5:12:44 PM
Posted: 6 months ago
At 5/11/2016 4:31:39 PM, Burzmali wrote:
Not to drag this too far sideways, but does anyone have an opinion of the recent change to split to Standard and Wild formats? I personally am not great at deck-building, and I don't care to spend more than my initial $20 investment in the game. So I'm struggling a bit to compete with the new set being out and only having like 5 packs worth of the new cards that they gave me.

First of all, you should be able to get 13 packs for free if you win 9 games in standard.

I think the change was absolutely necessary. The game would have remained in roughly the same state till the end of time if they didn't take action, which would have been a disaster. If you don't have enough of the new cards you can always do wild, although I'm not sure if it's balanced.
dylancatlow
Posts: 12,242
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5/11/2016 5:13:50 PM
Posted: 6 months ago
At 5/11/2016 5:11:30 PM, user13579 wrote:
At 5/11/2016 5:04:09 PM, dylancatlow wrote:
At 5/11/2016 4:38:22 PM, RainbowDash52 wrote:
At 5/11/2016 4:22:59 PM, dylancatlow wrote:
At 5/11/2016 2:32:58 PM, RainbowDash52 wrote:
Answer: 6.68113751508077

Calculations done here: http://rextester.com...

You can edit the winPercent variable value from .71 to something else to find the mean number of wins for other win percents too.

Thanks, this is exactly what I needed! Any idea on how to solve the second question?

For the second question, wouldn't a 50% chance of 75% win rate, and a 50% chance of a 65% win rate be identical to a 70% win rate?

No, you'd much rather have a consistent win rate. For example, someone whose chances of winning are 70 across all games has a 1.3 percent probability of getting 12 wins in a row, whereas someone whose chances of winning are 75 percent for the first six games and 65 for the latter six games only has a .8 percent chance of getting 12 wins in a row. The consistent person has a 2.7 percent chance of getting zero wins, whereas the inconsistent person has a 2.76 percent chance.

That doesn't make sense, because the second person has an average win rate of 70 ((65+75)/2 assuming an equal number of games as each side)! Those two people are basically the same person. If I play 1000 games at 65% and 1000 games at 75%, my average win rate is 70%.

Your average win rate is 70, but your average number of wins PER RUN is less.
user13579
Posts: 822
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5/11/2016 5:16:18 PM
Posted: 6 months ago
At 5/11/2016 5:13:50 PM, dylancatlow wrote:
At 5/11/2016 5:11:30 PM, user13579 wrote:
At 5/11/2016 5:04:09 PM, dylancatlow wrote:
At 5/11/2016 4:38:22 PM, RainbowDash52 wrote:
At 5/11/2016 4:22:59 PM, dylancatlow wrote:
At 5/11/2016 2:32:58 PM, RainbowDash52 wrote:
Answer: 6.68113751508077

Calculations done here: http://rextester.com...

You can edit the winPercent variable value from .71 to something else to find the mean number of wins for other win percents too.

Thanks, this is exactly what I needed! Any idea on how to solve the second question?

For the second question, wouldn't a 50% chance of 75% win rate, and a 50% chance of a 65% win rate be identical to a 70% win rate?

No, you'd much rather have a consistent win rate. For example, someone whose chances of winning are 70 across all games has a 1.3 percent probability of getting 12 wins in a row, whereas someone whose chances of winning are 75 percent for the first six games and 65 for the latter six games only has a .8 percent chance of getting 12 wins in a row. The consistent person has a 2.7 percent chance of getting zero wins, whereas the inconsistent person has a 2.76 percent chance.

That doesn't make sense, because the second person has an average win rate of 70 ((65+75)/2 assuming an equal number of games as each side)! Those two people are basically the same person. If I play 1000 games at 65% and 1000 games at 75%, my average win rate is 70%.

Your average win rate is 70, but your average number of wins PER RUN is less.

Wait a minute. You're saying somebody's win rate is 70%, even when you already know what side they're playing? Never mind in that case. (Although, how can that be? You already said the first player has a big advantage, so that doesn't make sense...)
Science in a nutshell:
"Facts are neither true nor false. They simply are."
"All scientific knowledge is provisional. Even facts are provisional."
"We can be absolutely certain that we have a moon, we can be absolutely certain that water is made out of H2O, and we can be absolutely certain that the Earth is a sphere!"
"Scientific knowledge is a body of statements of varying degrees of certainty -- some most unsure, some nearly sure, none absolutely certain."
dylancatlow
Posts: 12,242
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5/11/2016 5:17:41 PM
Posted: 6 months ago
At 5/11/2016 5:16:18 PM, user13579 wrote:
At 5/11/2016 5:13:50 PM, dylancatlow wrote:
At 5/11/2016 5:11:30 PM, user13579 wrote:
At 5/11/2016 5:04:09 PM, dylancatlow wrote:
At 5/11/2016 4:38:22 PM, RainbowDash52 wrote:
At 5/11/2016 4:22:59 PM, dylancatlow wrote:
At 5/11/2016 2:32:58 PM, RainbowDash52 wrote:
Answer: 6.68113751508077

Calculations done here: http://rextester.com...

You can edit the winPercent variable value from .71 to something else to find the mean number of wins for other win percents too.

Thanks, this is exactly what I needed! Any idea on how to solve the second question?

For the second question, wouldn't a 50% chance of 75% win rate, and a 50% chance of a 65% win rate be identical to a 70% win rate?

No, you'd much rather have a consistent win rate. For example, someone whose chances of winning are 70 across all games has a 1.3 percent probability of getting 12 wins in a row, whereas someone whose chances of winning are 75 percent for the first six games and 65 for the latter six games only has a .8 percent chance of getting 12 wins in a row. The consistent person has a 2.7 percent chance of getting zero wins, whereas the inconsistent person has a 2.76 percent chance.

That doesn't make sense, because the second person has an average win rate of 70 ((65+75)/2 assuming an equal number of games as each side)! Those two people are basically the same person. If I play 1000 games at 65% and 1000 games at 75%, my average win rate is 70%.

Your average win rate is 70, but your average number of wins PER RUN is less.

Wait a minute. You're saying somebody's win rate is 70%, even when you already know what side they're playing? Never mind in that case. (Although, how can that be? You already said the first player has a big advantage, so that doesn't make sense...)

Not sure what you're saying
user13579
Posts: 822
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5/11/2016 5:22:09 PM
Posted: 6 months ago
At 5/11/2016 5:17:41 PM, dylancatlow wrote:
At 5/11/2016 5:16:18 PM, user13579 wrote:
At 5/11/2016 5:13:50 PM, dylancatlow wrote:
At 5/11/2016 5:11:30 PM, user13579 wrote:
At 5/11/2016 5:04:09 PM, dylancatlow wrote:
At 5/11/2016 4:38:22 PM, RainbowDash52 wrote:
At 5/11/2016 4:22:59 PM, dylancatlow wrote:
At 5/11/2016 2:32:58 PM, RainbowDash52 wrote:
Answer: 6.68113751508077

Calculations done here: http://rextester.com...

You can edit the winPercent variable value from .71 to something else to find the mean number of wins for other win percents too.

Thanks, this is exactly what I needed! Any idea on how to solve the second question?

For the second question, wouldn't a 50% chance of 75% win rate, and a 50% chance of a 65% win rate be identical to a 70% win rate?

No, you'd much rather have a consistent win rate. For example, someone whose chances of winning are 70 across all games has a 1.3 percent probability of getting 12 wins in a row, whereas someone whose chances of winning are 75 percent for the first six games and 65 for the latter six games only has a .8 percent chance of getting 12 wins in a row. The consistent person has a 2.7 percent chance of getting zero wins, whereas the inconsistent person has a 2.76 percent chance.

That doesn't make sense, because the second person has an average win rate of 70 ((65+75)/2 assuming an equal number of games as each side)! Those two people are basically the same person. If I play 1000 games at 65% and 1000 games at 75%, my average win rate is 70%.

Your average win rate is 70, but your average number of wins PER RUN is less.

Wait a minute. You're saying somebody's win rate is 70%, even when you already know what side they're playing? Never mind in that case. (Although, how can that be? You already said the first player has a big advantage, so that doesn't make sense...)

Not sure what you're saying

How can anybody have a "consistent win rate" if there's a big first turn advantage? If somebody has a win rate of 70, isn't that very likely to mean that player wins e.g. 75% of the time when going 1st and 65% of the time when going 2nd? In order to be a"consistent" win rate, it looks like it would have to be 70% when going 1st and 70% when going 2nd!
Science in a nutshell:
"Facts are neither true nor false. They simply are."
"All scientific knowledge is provisional. Even facts are provisional."
"We can be absolutely certain that we have a moon, we can be absolutely certain that water is made out of H2O, and we can be absolutely certain that the Earth is a sphere!"
"Scientific knowledge is a body of statements of varying degrees of certainty -- some most unsure, some nearly sure, none absolutely certain."
dylancatlow
Posts: 12,242
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5/11/2016 5:24:25 PM
Posted: 6 months ago
At 5/11/2016 5:22:09 PM, user13579 wrote:
At 5/11/2016 5:17:41 PM, dylancatlow wrote:
At 5/11/2016 5:16:18 PM, user13579 wrote:
At 5/11/2016 5:13:50 PM, dylancatlow wrote:
At 5/11/2016 5:11:30 PM, user13579 wrote:
At 5/11/2016 5:04:09 PM, dylancatlow wrote:
At 5/11/2016 4:38:22 PM, RainbowDash52 wrote:
At 5/11/2016 4:22:59 PM, dylancatlow wrote:
At 5/11/2016 2:32:58 PM, RainbowDash52 wrote:
Answer: 6.68113751508077

Calculations done here: http://rextester.com...

You can edit the winPercent variable value from .71 to something else to find the mean number of wins for other win percents too.

Thanks, this is exactly what I needed! Any idea on how to solve the second question?

For the second question, wouldn't a 50% chance of 75% win rate, and a 50% chance of a 65% win rate be identical to a 70% win rate?

No, you'd much rather have a consistent win rate. For example, someone whose chances of winning are 70 across all games has a 1.3 percent probability of getting 12 wins in a row, whereas someone whose chances of winning are 75 percent for the first six games and 65 for the latter six games only has a .8 percent chance of getting 12 wins in a row. The consistent person has a 2.7 percent chance of getting zero wins, whereas the inconsistent person has a 2.76 percent chance.

That doesn't make sense, because the second person has an average win rate of 70 ((65+75)/2 assuming an equal number of games as each side)! Those two people are basically the same person. If I play 1000 games at 65% and 1000 games at 75%, my average win rate is 70%.

Your average win rate is 70, but your average number of wins PER RUN is less.

Wait a minute. You're saying somebody's win rate is 70%, even when you already know what side they're playing? Never mind in that case. (Although, how can that be? You already said the first player has a big advantage, so that doesn't make sense...)

Not sure what you're saying

How can anybody have a "consistent win rate" if there's a big first turn advantage? If somebody has a win rate of 70, isn't that very likely to mean that player wins e.g. 75% of the time when going 1st and 65% of the time when going 2nd? In order to be a"consistent" win rate, it looks like it would have to be 70% when going 1st and 70% when going 2nd!

Oh, the consistent player is hypothetical. I'm trying to figure out what the effects of first turn advantage are on win rates of skilled players. I know that it's bad, but I'm trying to calculate HOW bad.
user13579
Posts: 822
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5/11/2016 5:31:12 PM
Posted: 6 months ago
At 5/11/2016 5:24:25 PM, dylancatlow wrote:
At 5/11/2016 5:22:09 PM, user13579 wrote:
At 5/11/2016 5:17:41 PM, dylancatlow wrote:
At 5/11/2016 5:16:18 PM, user13579 wrote:
At 5/11/2016 5:13:50 PM, dylancatlow wrote:
At 5/11/2016 5:11:30 PM, user13579 wrote:
At 5/11/2016 5:04:09 PM, dylancatlow wrote:
At 5/11/2016 4:38:22 PM, RainbowDash52 wrote:
At 5/11/2016 4:22:59 PM, dylancatlow wrote:
At 5/11/2016 2:32:58 PM, RainbowDash52 wrote:
Answer: 6.68113751508077

Calculations done here: http://rextester.com...

You can edit the winPercent variable value from .71 to something else to find the mean number of wins for other win percents too.

Thanks, this is exactly what I needed! Any idea on how to solve the second question?

For the second question, wouldn't a 50% chance of 75% win rate, and a 50% chance of a 65% win rate be identical to a 70% win rate?

No, you'd much rather have a consistent win rate. For example, someone whose chances of winning are 70 across all games has a 1.3 percent probability of getting 12 wins in a row, whereas someone whose chances of winning are 75 percent for the first six games and 65 for the latter six games only has a .8 percent chance of getting 12 wins in a row. The consistent person has a 2.7 percent chance of getting zero wins, whereas the inconsistent person has a 2.76 percent chance.

That doesn't make sense, because the second person has an average win rate of 70 ((65+75)/2 assuming an equal number of games as each side)! Those two people are basically the same person. If I play 1000 games at 65% and 1000 games at 75%, my average win rate is 70%.

Your average win rate is 70, but your average number of wins PER RUN is less.

Wait a minute. You're saying somebody's win rate is 70%, even when you already know what side they're playing? Never mind in that case. (Although, how can that be? You already said the first player has a big advantage, so that doesn't make sense...)

Not sure what you're saying

How can anybody have a "consistent win rate" if there's a big first turn advantage? If somebody has a win rate of 70, isn't that very likely to mean that player wins e.g. 75% of the time when going 1st and 65% of the time when going 2nd? In order to be a"consistent" win rate, it looks like it would have to be 70% when going 1st and 70% when going 2nd!

Oh, the consistent player is hypothetical. I'm trying to figure out what the effects of first turn advantage are on win rates of skilled players. I know that it's bad, but I'm trying to calculate HOW bad.

Can some decks reduce the first turn advantage, even at the expense of lower overall win rate? By making the win rate more consistent, you could get more wins per run even with a lower overall win rate. So if everybody did that, it would get closer and closer to there being no first turn advantage! Heh.
Science in a nutshell:
"Facts are neither true nor false. They simply are."
"All scientific knowledge is provisional. Even facts are provisional."
"We can be absolutely certain that we have a moon, we can be absolutely certain that water is made out of H2O, and we can be absolutely certain that the Earth is a sphere!"
"Scientific knowledge is a body of statements of varying degrees of certainty -- some most unsure, some nearly sure, none absolutely certain."
RuvDraba
Posts: 6,033
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5/11/2016 5:43:18 PM
Posted: 6 months ago
At 5/11/2016 4:22:59 PM, dylancatlow wrote:
At 5/11/2016 2:32:58 PM, RainbowDash52 wrote:
Answer: 6.68113751508077

Calculations done here: http://rextester.com...

You can edit the winPercent variable value from .71 to something else to find the mean number of wins for other win percents too.

Thanks, this is exactly what I needed! Any idea on how to solve the second question?

Okay -- less stats for me to distract myself with! :D

Good luck, DC. :)