Average multiplier as a function of win rate

[Panoply]

The internet went down at my work today, so I decided to see if I couldn't come up with the equation that describes the relationship between average multiplier and the probability of winning a round. I've been working on a tool for players to calculate upkeep and equipment, and I wanted to be able to determine average multiplier if given a certain win rate. In my previous encounters with this problem, I've either programmed a simulation, or I've used simple cases, such as a win rate of 50%.

Today, however, I decided to generalize it. Worse yet, I would do it by hand, without the aid of WaltF4's previous posts, which when I look at now, would have given me some solid hints on how to proceed. The algebra was a bitch by hand, but it was all worth it, when all the terms canceled out to yield this gem:

y = x^4 + x^3 + x^2 + x + 1

y is your average multiplier
x is your win rate, eg. if you win 50% of the time, x = 0.5; at 87.22%, x = 0.8722, etc.

So at 50% win rate, your average multiplier (not including valour) would be 1.9375. This is well established and simple to do by hand, since x = 1 - x.

I ran a simulation over 100,000 rounds at each percentile, and the data matches up perfectly with the above equation, so I can say it is correct with confidence.

Most of you are likely indifferent to this information, and some of you will hopefully have a passing curiosity, but I thought I'd share it anyway. Personally, I'm pretty proud, as probability often makes my head spin.

TL:DR Latest Recommendation: Only quit on x1.

[Christo]

y = x^4 + x^3 + x^2 + x + 1

wat

[Casimir]

visitors can't see pics , please register or login

[Kaelaen]

wat

I believe that is what could be considered as 'l33t speak.'  I understand it to have been popular with internet users a couple of years ago, though it seems to have fallen with style as people have grown older.

[Christo]

I believe that is what could be considered as 'l33t speak.'  I understand it to have been popular with internet users a couple of years ago, though it seems to have fallen with style as people have grown older.

I just did it for fun.

Obviously the post was well written, and is informative. You can see the effort put into it.

Hence my +1 on the OP's post.

Better, now?

[Belhade]

I was told there would be no math.   

[Paul]

Average multiplier depends on the average amount of rounds played in succession but goes rather quickly(after 4 rounds) towards to aforementioned 1.9375 (50% win chance). So just playing 1, 2 or 3 rounds and then quitting is bad for the average multiplier but I guess that's common sense.

So more interesting is a formular that contains the average amount of rounds(n) played in succession as a parameter instead of just assuming the case n->∞.

[Panoply]

EDIT AGAIN: Apparently my initial intuition was correct, and this post actually does have the correct math.

This post contains flawed reasoning, but I've kept it here for posterity.

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So more interesting is a formular that contains the average amount of rounds(n) played in succession as a parameter instead of just assuming the case n->∞.

Sure, now that I don't have to do the algebra by hand, this is pretty simple to do. Correct me if I'm wrong though. I have this nagging suspicion that I've missed something... but here it is as far as I can tell.

n - number of rounds played in one sitting on average
x - win rate
y - average multiplier

When:
n = 1, y = 1
n = 2, y = x + 1
n = 3, y = x^2 + x +1
n = 4, y = x^3 + x^2 + x + 1
n = 5, y = x^4 + x^3 + x^2 + x + 1
n > 5, y = x^4 + x^3 + x^2 + x + 1

As long as you play at least 5 rounds a sitting, and only quit on a x1 (since multipliers don't carry over sittings), the general equation of, y = x^4 + x^3 + x^2 + x + 1, should apply.

I'm not sure where you're getting the numbers you gave in your example. For x = 0.5 (50% win chance),

1 round, y = 1
2 rounds, y = 0.5*1 + 0.5*2 = 1.5
3 rounds, y = 0.5*1 + 0.5*0.5*2 + 0.5*0.5*3 = 1.75
4 rounds, y = 0.5*1 + 0.5*0.5*2 + 0.5*0.5*0.5*3 + 0.5*0.5*0.5*4 = 1.875
5 rounds, y = 1.9375

TL:DR, play at least 5 rounds per sitting, and only exit on a x1.

[Paul]

I posted wrong average earlier. But yours seem wrong too for small n. If you write down round results in a tree diagramm it looks like this:

1)                                   1
2)             1                                       2
3)     1                2                    1                 3
4)  1      2      1        3           1        2       1         4
5)1 2    1 3   1 2      1 4       1 2      1 3    1 2      1 5
.
.
n)

a=average multiplier
n=1: a = 1
n=2: a = ((1+1)/2 + (1+2)/2)/2 = (1+1 + 1+2)/(2*2) = 1.25
n=3: a = (1+1+1 + 1+1+2 + 1+2+1 + 1+2+3)/(3*4) = 17/12≈1.42
n=4: a = (1+1+1+1 + 1+1+1+2 + 1+1+2+1 + 1+1+2+3 + 1+2+1+1 + 1+2+1+2 + 1+2+3+1 + 1+2+3+4 )/(4*8) = 49/32≈1.53

To explain the easy case of n=2 in words for better understanding. Let's assume we play only 2 round for each session. With a win chance of 50% there are 2 possible cases with the same likeliness. I can win the first round and get x2 multi in the second. That way I have an average multiplier of 1.5 for this session. The second case is that I lose the first round and play both rounds with x1. So I have an average multi of 1. Because both cases happen with the same likeliness my overall average multiplier is then (1.5+1)/2 = 1.25. If I had a win chance of 60% my average would be (1.5*0.6 + 1*0.4)=1.3.

[Vibe]

Feels like high school again, yawn

[Byrdi]

This is great fun, I wish I could patitiepate in this "math duel".
Anyway nice job there.

[Torp]

Average multiplier depends on the average amount of rounds played in succession but goes rather quickly(after 4 rounds) towards to aforementioned 1.9375 (50% win chance). So just playing 1, 2 or 3 rounds and then quitting is bad for the average multiplier but I guess that's common sense.

So more interesting is a formular that contains the average amount of rounds(n) played in succession as a parameter instead of just assuming the case n->∞.

Many people only quit at x1, and if you have x1 after 3 rounds, it doesnt matter if you quit or stay - you'll still start the next played round with x1.

The fact that you can choose servers freely and quit whenever you want also allows people who want to, to get ahead of the statistics.

[Paul]

It's actually true. If you have the discipline to only quit at x1 then it is the same as n->∞ and we get an average multi of 1.9375 (or 2 if multi wasn't capped at x5).

[Panoply]

This post isn't actually that useful since it only really describes the situation if you're just starting out c-rpg, or if you like to quit randomly regardless of whether or not you have a multiplier. That said, I'm keeping it here for posterity.

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Ah, you're right. I had a feeling I forgot something:

I have this nagging suspicion that I've missed something... but here it is as far as I can tell.

My calculations basically told you your average multiplier at the nth round, but didn't average out with previous rounds. It also told you what average multipliers would be if the maximum multiplier were n.

But all is not lost! Since my formula tells me the average multiplier at a particular round, it follows that I can simply average these values to give the average multiplier based on n rounds played successively.

For example, I wrote:

1 round, y = 1
2 rounds, y = 0.5*1 + 0.5*2 = 1.5
3 rounds, y = 0.5*1 + 0.5*0.5*2 + 0.5*0.5*3 = 1.75
4 rounds, y = 0.5*1 + 0.5*0.5*2 + 0.5*0.5*0.5*3 + 0.5*0.5*0.5*4 = 1.875

So if you take the averages:
(1 + 1.5)/2 = 1.25
(1 + 1.5 + 1.75)/3 = 1.41667
(1 + 1.5 + 1.75 + 1.875)/4 =  1.53125
etc.

I'm not well versed in mathematical convention, but this is the general equation I see:

y = [(n-4)*x^4 + (n-3)*x^3 + (n-2)*x^2 + (n-1)*x + n] / n

I just don't know how to express that when a coefficient is negative, it is zero. In programming terms it's usually something like this:

y = [max (0, n-4)*x^4 + max (0, n-3)*x^3 + max (0, n-2)*x^2 + max (0, n-1)*x + n] / n

Here are a few examples:

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For n = 3, x = 0.5
y = (x^2 + 2*x + 3) / 3 = (0.5^2 + 2*0.5 + 3) / 3 = 1.41667

For n = 6, x = 0.6
y = (2x^4 + 3x^3 + 4x^2 + 5x + 6) / 6 = (2*0.6^4 + 3*0.6^3 + 4*0.6^2 + 5*0.6 + 6) / 6 = (0.2592 + 0.648 + 1.44 + 3 + 6) / 6 = 1.8912

For n = 1000, x = 0.5
y = (996x^4 + 997x^3 + 998x^2 + 999x + 1000) / 1000 = (62.25 + 124.625 +  249.5 + 499.5 + 1000) / 1000 = 1.935875
Getting damn close to the n->∞ of 1.9375

TL:DR, The more games you play in one sitting, the higher your average multiplier will be. Only quit when you're at x1

Let me know if I've made any mistakes.

[Siiem]