Honor in Summoner's Rift and Twisted Treeline - probabilities

SonicAF·10/18/2018, 4:53:28 PM·2 votes·2,540 views

Well, hello there.

I've had a discussion with Voldymort@EUNE about probability to receive honor in specific gamemode(5v5 vs 3v3) and I promised to have my take on it. He also stated that he might make it into a post through the magic of copy-paste, when he is relatively sure about the numbers. But I am overconfident so, obviously, I am stealing his idea and writing this post myself.

Here we go!


First of all, players are incentivized to honor other people due to spilling some honor upon themselves in the process. While, obviously, not every player honors every game, be it out of spite or genuine righteous fury, we will ignore this aspect for now and count the probability for the case of "everyone honors someone".

The second assumption is that all players have equal chance to receive honor, which is, obviously, not the case - out of 4 people you are more likely to honor someone outstanding, but let's assume that noone is toxic or awesome. We have a game of 5 average Joes.

The third is a valuable disclamer that makes our life simple - receiving a honor point, obviously, doesn't interact with your chance to receive other honor points, which simplifies our calculations even further.

This is basically what we are calculating: https://www.youtube.com/watch?v=s5DWg4o6h8Q In this case there were 3 missiles, each having a chance of 20% to hit a specific target independently 20%*20%*20% = 0.8% chance for it to happen. Easy.


Summoner's Rift:

We have four free honor points. Assuming that each player has equal chance to receive each specific point, we are dealing with 25% chance per point and they roll independently. Here is an easy way to calculate the odds of honors hitting you:

  1. 25%(obviously)
  2. (the previous was a hit)25%*25% = 6.25%
  3. (two previous were hits)6.25%*25% = 1.5625%
  4. (three previous were hits)1.5625%*25% = 0.390625%

This is quite a lottery. https://www.youtube.com/watch?v=ymPpIzaanhY

Twisted Treeline

We have two free honor points. Assuming that each player has equal chance to receive each specific point, we are dealing with 50% chance per point and they roll independently. Here is an easy way to calculate the odds of honors hitting you:

  1. 50%(obviously)
  2. (the previous was a hit)50%*50% = 25%

https://imgur.com/2oNu26e


If we take into account that people don't honor in, let's assume, 10% games, we have our chances reduced by 10%(we are only interested in this disappointment if it would otherwise roll successfully) on each step, which means:

SR would be 22.5% 22.5%*22.5% = 5.0625% 22.5%*22.5%*22.5% = 1.1390625% 22.5%*22.5%*22.5%*22.5% = 0.256289062%

TT would be: 45% 45%*45% = 20,25%

How and where did I fuck that up? https://www.youtube.com/watch?v=k7HMQuoISNg

15 Comments

Voldymort10/18/2018, 4:55:59 PM4 votes

[sg-shisa]

so, obviously, I am stealing his idea and writing this post myself.

http://www.quickmeme.com/img/22/227aebe3abfb2d8e0ea323801e44b3ea5015ca1739bf8c2f2430a4db16346899.jpg

that said, you make a lot of assumptions in that math of yours

https://i.kym-cdn.com/photos/images/newsfeed/001/038/936/710.jpg

AeroWaffle10/18/2018, 5:21:21 PM4 votes

Provided chances are equal to receiving honor, your math is correct if you assume the goal is to receive all possible honor in a game. But "all possible honor" are two different results between the two game modes.

TT is 2 possible honor SR is 4 possible honor

So instead lets calculate the chance of receiving at least 2 honor to keep things equal.

TT remains the 25% that you calculated.

but SR is a bit more tricky (edited out previous one as I noticed the error, recalculating math).

It would be 1 - (probability of zero honors) - (probability of one honor).

The probability of zero is (3/4 x 3/4 x 3/4 x 3/4) = ~32% The probability of getting one is (1/4 x 3/4 x 3/4 x 3/4) x (4 variations) = ~42%

so 1 - 32% - 42% = 26% to at least get 2 honors in SR.

And this isn't even taking into account that SR has a chance of getting 3 and 4 honors, but TT has 0% chance.

Edit 2:

Should you take into account the chance of people honoring no one, the gap widens between the two modes.

(assuming 10% chance to not honor)

90% chance to honor split between two people = 45% chance per honor point. TT: (.45 x .45) = ~20% chance.

90% chance to honor split between four people = 22.5% chance to get honored per honor point. (77.5% to not get honored)

SR: 1 - (77.5% ^ 4) - ((22.5% x (77.5% ^ 3)) x 4) = ~22% chance

Increasing the chance that someone doesn't honor at all increases the gap further.

Kei14310/18/2018, 5:58:26 PM2 votes

Have you seen these stats from March 21-23 2018? These were from Kim Voll in a GDC presentation.

https://imgur.com/gallery/mDRHzer

Also ..

https://i.imgur.com/XO6j8Xe.jpg

There are lots of factors that skew how you get honors. But mathematically, for 1 honor, you need to take in account for the full scenario, as having 1 player honor you and the other 3 NOT honor you is mutually exclusive, meaning this is what you need to take account of:

P(A1 n B0 n C0 n D0) u
P(A0 n B1 n C0 n D0) u P(A0 n B0 n C1 n D0) u P(A0 n B0 n C0 n D1)

Or in simpler terms

4P(1 n P(0)^3)

Where ABCD are other players and 0= they don't honor you and 1= they honor you.

If you want to add accuracy to the model, you need to take in account the probability of people honoring other players and the probability of them not honoring at all. Which brings your model from 4 scenarios to 32 scenarios:

P(1 n 0 n 0 n 0) u P(1 n 0 n 0 n -1) u P(1 n 0 n -1 n 0) u P(1 n -1 n 0 n 0) u P(1 n -1 n -1 n 0) u P(1 n -1 n 0 n -1) u P(1 n 0 n -1 n -1) u P(1 n -1 n -1 n -1)

x4

Where -1 = P( not honoring ) 0 = P( honoring BUT not you ) 1 = P( honoring you )

EDIT:

High school math question, based on what I showed, what is the true average probability that someone honors you?

Subdue10/18/2018, 6:15:16 PM2 votes

Your math is wrong.

It's blatantly obvious that it's wrong because your numbers don't add up to 100%.

If you make the HUGE assumption that honor is given at random (it's not), then the probability of receiving honor is as follows:

0 Honor: 31.64% 1 Honor: 42.19% 2 Honor: 21.09% 3 Honor: 4.69% 4 Honor: 0.39%

Probability of 0 Honor = .75^4 = 31.64%

What's probably more interested is the 1 honor scenario. What you actually need is the probability of 1 player giving honor and 3 players not giving honor times the number of players, which is 4.

(.25 x .75^3)*4 = 42.19%.

I'm not going to go through all of the scenarios for you. You can do that on your own.

Edit:

From the probabilities you can calculate expected values.

On SR:

0 Honor: 31.64% x 1 = 0.00 1 Honor: 42.19% x 2 = 0.42 2 Honor: 21.09% x 3 = 0.42 3 Honor: 4.69% x 4 = 0.14 4 Honor: 0.39% x 5 = 0.02

Total = 1.00

Expected Value on SR = 1 Honor / Game.

On TT: 0 Honor = 25% x 0 = 0.00 1 Honor = 50% x 1 = 0.50 2 Honor = 25% x 2= 0.50

Total = 1.00

Expected Value on TT = 1 Honor / Game.

usul120210/18/2018, 5:28:29 PM1 votes

But riot accounts for stuff like this. Just like how they confirmed that you couldn't cheat honor by spamming oddessy for 1 minute game times just to honor people...

Kei14310/18/2018, 10:36:47 PM1 votes

Going to start a separate reply to give you more visibility.

From the formulas of the other reply, using 60% chance of people honoring others and 25% of them honoring you.

P(get 1 honor in SR): 36.848 % P(get 1 honor in TT): 25.5%

P(get 2 honors in SR): 9.754% P(get 2 honors in TT): 2.25%

If we were to use Riot's numbers of P(get 1 Honor), then it means the chances of someone honoring you is 16.55% (instead of your assumed number of 25%).

In that scenario.

P(1 honor in SR): 29.0% P(1 honor in TT): 17.88%

For one to get 2 honors in SR with Riot's stats, it means that they will have to get honored 21.275% of the time. Meaning:

P(2 honor in SR): 7.1% P(2 honor in TT): 1.629%




Numerically, it makes sense that it is harder to get 2 honors within the game. Afterall, you'd have to make a stronger impressions to get 2 honors.

In your assumption of having 25% chance to get honored, SR has 44.5% relative better chance to get 1 honor, but with Riot's numbers, SR has a 62.25% relative improved chance to get 1 honor over TT.




Fun fact, you need to have a 26.125% chance to get honored to get Riot's average stats of getting 3 honors, and 33.75% chance to get honored to get 4 honors based on Riot's stats.

Then again, it's probably wrong to use 60% chance to honor others when you are venturing into the space of getting 3 honors and 4 honors. But based on the info we have right now, that's all we can do.