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Wynncraft, the Minecraft MMORPG. Play it now on your Minecraft client at (IP): play.wynncraft.com. No mods required! Click here for more info...

Guide Horse's Breeding Prices

Discussion in 'Wynncraft' started by Yuno F Gasai, Sep 17, 2015.

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should this get sticked

  1. yea!!!

    610 vote(s)
    81.8%

  2. nigh

    136 vote(s)
    18.2%
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  1. Windfall_

    Windfall_ Proud Owner of the Worst Sage HERO

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    Wait Wynn has horses... THAT YOU CAN RIDE :o :o

    Who invented this debunkers. Were horses imported from Fruma. There are no horses within the Fruman walls however...
     
    Windfall_, May 27, 2020
    #301
  2. BqwaOLD

    BqwaOLD Famous Adventurer

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    Seeing as though this thread hasn't been locked and forgotten and has been up since 2015, I think it deserves it.
     
    BqwaOLD, May 27, 2020
    #302
    Yuno F Gasai likes this.
  3. Yuno F Gasai

    Yuno F Gasai Forum God, FW

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    yea I (and 496 others) think it should get stickied
     
    Yuno F Gasai, May 27, 2020
    #303
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  4. StormKing3

    StormKing3 Famous Adventurer

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    So theoretically you can breed black horses and sell on them on the market for profit???

    I'm trying that
     
    StormKing3, May 27, 2020
    #304
  5. Melkor

    Melkor The dark enemy of the world HERO

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    Theoretically you can breed and sell any horse but brown and make a profit. It's just not likely.
     
    Melkor, May 27, 2020
    #305
  6. StormKing3

    StormKing3 Famous Adventurer

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    I'm just going off a black horse costing 2le and 16eb on average to breed, but then you can sell it for more on the market basically.

    But yeah you're right that in theory you could just grab 8 browns and get a white horse xD
     
    StormKing3, May 27, 2020
    #306
  7. Yuno F Gasai

    Yuno F Gasai Forum God, FW

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    It's worth it up to chestnut.
    White market value is no where near the breeding price
     
    Yuno F Gasai, May 27, 2020
    #307
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  8. Yuno F Gasai

    Yuno F Gasai Forum God, FW

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    And thanks to all 500 peeps that voted "yea!!!" on the poll.
    hopefully more people will vote
     
    Yuno F Gasai, May 28, 2020
    #308
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  9. OrangeEater9000

    OrangeEater9000 Travelled Adventurer

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    I'm a bit late to the party for this, but I noticed that every time someone said that they successfully bred a horse at a cheaper cost than the expected value, it was chalked up to luck. For the record, your math is correct, but the interesting part is that most people will not need that many emeralds to breed a horse. It sounds contradictory considering the expected value of combinations required is, indeed, 5 to get 1 successful breed, but I'll breed brown horses into black ones for this example (tier 1 to tier 2).

    Given a probability of success of 20%, what is the probability of success in one combination? Easy, that's just 20%. Nothing special here.

    What are the chances that we get at least one success after two combinations? That is a little trickier. Here is just one way to think of it. There are only two possible outcomes: either both combinations are failures or there is at least one success. Since the sum of all probabilities must always equal 100%, if we find the probability of both combinations being failures and subtract that from 1, we will get the probability of at least one success. Since the probability of success is 20%, the probability of failure is 80%. We all know from grade school that if you want to find the probability of 2 events you have to multiply the two probabilities together, and since the probability of failure is still 80% in the second combination, we just do 0.8 * 0.8 = 0.64. That means the probability of failing both combinations is 64%. Thus, 100%-64%=36% meaning there is a 36% chance of getting at least one success after two combinations. If many many people tried exactly two combinations each, 36% of those people would get at least one success. You can verify this by drawing a tree diagram and finding that there are 9 possible scenarios for success and 25 scenarios in total. 9/25=36%

    What are the chances that we get at least one success after n operations given a probability of success of p? We will do what we did in the last situation. What is the probability of failure in one operation? That is simply 1-p. What is the probability of failure after n operations? That is simply (1-p)^n (we multiply the probability of failure by itself for every operation). Finally, to get the probability of success after n operations, we subtract the probability of failure after n operations from 1 (100%). Thus, the probability of success after n operations is 1-(1-p)^n. For example, what is the probability I will roll a 6 on a die at least once after rolling 10 times? We know that the probability of rolling a 6 every roll is 1/6, so we just do 1-(1-1/6)^10≈0.8385. Thus, after rolling a die 10 times, the odds that I will have rolled a 6 once is 83.85%. As you plug in higher and higher numbers of n, you will see that the cumulative probability asymptotically approaches 100%. Indeed, if you rolled a die infinitely many times, then the act of never rolling a 6 is said to be almost never, but that is an interesting but unrelated topic.

    Lets apply this formula to the situation at hand. How many combinations must a large group of people each make such that after finishing all those combinations, half of them will have had a successful combination? Since we want half of these people to be successful, we should equate 1-(1-p)^n to 0.5 (50%). Given p=20%=0.2, we want to find n. We solve the equation 0.5=1-(1-0.2)^n. Using a little algebra we find that n≈3.106, meaning after having done 3.106 combinations, half of the people will have had a successful combination. Of course, 3.106 is not a whole number so this doesn't really make sense. If we round up to 4, then we should have a greater than 50% of the people getting a successful combination. Indeed, if we compute 1-(1-0.2)^4 we find the probability to be ≈59.04%. Recall that the expected average number of combinations was 5! However, most people, about 59.04% of all people to be exact, will be done in only 4 combinations, and thus only need 8 horses to combine. Both of these are correct, as the expected value of 5 is best thought of as the average, whereas 3.106 is best thought of as the median. It makes sense conceptually, as there may be some people who will fail after 20+ combinations, which skews the average higher than the median value.

    I hope you found this helpful and if you have any questions feel free to ask. I hope the math was easy to understand. I did my best to explain step by step.
     
    Last edited: Jul 9, 2020
    OrangeEater9000, Jul 9, 2020
    #309
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  10. Ackro

    Ackro The true Elite Quadbrid! HERO

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    I like you.
     
    Ackro, Jul 9, 2020
    #310
  11. Andyroo_88

    Andyroo_88 Travelled Adventurer VIP

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    U know u can make a black horse with 6 brown horses right cuz every 2 brown horses u use makes a new brown horse, same goes for theother tier.
    However, don't forget that there is the 30% chance of a downgrade, so you need to include that in the caluclations too (which I tried and did and it took a long, long time) but otherwise it was a nice simple guide!
     
    Last edited: Jul 17, 2020
    Andyroo_88, Jul 17, 2020
    #311
  12. Yuno F Gasai

    Yuno F Gasai Forum God, FW

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    It's already included in the calculations
     
    Yuno F Gasai, Jul 18, 2020
    #312
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  13. thatswhatido

    thatswhatido Supreme Wynncraft Player thatswhatido CHAMPION

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    Reopened because the thread owner requested.
     
    thatswhatido, Dec 29, 2020
    #313
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  14. Yuno F Gasai

    Yuno F Gasai Forum God, FW

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    ILY <3
     
    Yuno F Gasai, Dec 29, 2020
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  15. Castti

    Castti Kookie HERO

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    Nowadays with horse prices crashing this thread is needed again
     
    Castti, Dec 29, 2020
    #315
  16. Yuno F Gasai

    Yuno F Gasai Forum God, FW

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    tbh its always needed (altho like 90% of teh poll answerers arent active anymore)
     
    Yuno F Gasai, Dec 29, 2020
    #316
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  17. ocu

    ocu penguin gang CHAMPION

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    it returns
     
    ocu, Dec 29, 2020
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  18. IceResistance

    IceResistance Titans Valor [ANO] Founder CHAMPION

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    This is a lot of math for me to just breed two chestnuts and get a brown :P
     
    IceResistance, Dec 30, 2020
    #318
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  19. Yuno F Gasai

    Yuno F Gasai Forum God, FW

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    thats how we roll here
     
    Yuno F Gasai, Dec 30, 2020
    #319
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  20. HeyZeer0

    HeyZeer0 Wynncraft Developer Staff Member Admin Developer Game Developer Featured Wynncraftian CHAMPION

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    I wrote a simple "algorithm" that perfectly simulates a horse breeding process.

    My results are the following over 1 million simulated trades:
    So the average required amount of LEs required to breed a white horse is around 75.
    You have about 8 to 27 per million trades chance to get a white horse buying only 8 horses.
    You are also able to be damn unlucky (1 in a million) and need to buy 2110 horses (~12 stacks).

    I really prefer to accept the whole average as a good amount, because otherwise, the data is insanely positively biased. For example, if you consider only the top 20 trades that mostly occurred, the average is 72 horses, but the top 20 only happened 87.613 times, which is about 8,7% of the total attempts.

    There's a lot of stuff that you have to consider while calculating it, including the not-so-random java default random algorithm that's used over wynn. This also takes into consideration trades that the horse tier went down.

    The approach I had is brute-forcing.
    We are not asking what's the chance of getting a white horse, instead, we question how much trades do we need until getting a white horse. So basically my "algo" simulates singular breeding until white is reached. Then makes that 1 million more times and creates some averages based on the values obtained.
    You can get very precise values by brute-forcing a simple task like that, simply because we have a million data available, not 100 or 1 thousand.

    Have fun with my data :saltedhappy:

    My java code:
    Made to be easily understandable, meaning it has a bit higher complexity that increases the calculation time but it's still fast (took about 3s to calculate the 1 million attempts).

    Code:
        public static int calculateBreedAttempts(int maxTier) {
        int boughtHorses = 0;

        // the java random instance
        Random r = new Random();

        // map -> K = tier | V = amount of horses
        HashMap<Integer, Integer> horseTiers = new HashMap<>();

        // loop until one horse of the max tier was created
        int currentTier = 0;
        while (horseTiers.getOrDefault(maxTier, 0) != 1) {
            // if current tier == 0, means we need to buy until we reach 2 horses
            if (currentTier == 0) {
                boughtHorses += (2 - horseTiers.getOrDefault(0, 0));
                horseTiers.put(0, 2);

                currentTier++;
                continue;
            }

            // if the horses of the current tier is below 2 that means we have to breed!
            if (horseTiers.getOrDefault(currentTier, 0) < 2) {
                // we need two horses of the previous tier to breed if not met, reduce the tier
                if (horseTiers.get(currentTier-1) != 2) {
                    currentTier--;
                    continue;
                }

                // always 1 horse will need to be removed from the previous tier
                horseTiers.put(currentTier-1, horseTiers.get(currentTier-1) - 1);

                double chance = r.nextDouble() * 100;

                // breed was a success (add 1 horse the current tier and clear all from the last one)
                if (chance <= 20) {
                    horseTiers.put(currentTier, horseTiers.getOrDefault(currentTier, 0) + 1);
                    horseTiers.put(currentTier-1, 0);
                    continue;
                }

                // same tier horse or current tier is 1 (because there's not a tier -1 horse lmao)
                if (chance <= (20 + 50) || currentTier == 1) continue;

                // horse tier went down
                horseTiers.put(currentTier-1, 0);
                horseTiers.put(currentTier-2, horseTiers.get(currentTier-2) + 1);
                continue;
            }

            // required breeds for the next tier were met, let's increase the tier
            currentTier++;
        }

        return boughtHorses;
    }
     
    Last edited: Dec 31, 2020
    HeyZeer0, Dec 30, 2020
    #320
    btdmaster, Meow_Pig, violetlord and 13 others like this.
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