Monte Carlo analysis of King’s Gauntlet

How much cheese do you need to succeed in the King’s Gauntlet?  Well, that depends…

Log summaries have given us a pretty good idea of how many potions are dropped per hunt on each tier.  From that you might be tempted to just multiply the cheese used by the drop rate and the potion conversion rate per tier to get an estimated # of eclipse caught.  But this has some flaws.  Since drops are rare and independent, the potion drops follow a Bernoulli distribution for each tier which looks different from a normal ‘bell-shaped’ distribution.  Also, a change of single pot drop, especially at high tiers, can have a huge effect on Eclipse Mouse catching.  Furthermore, wouldn’t it be great to see the whole range of possibilities for your run, rather than just a single average number of Eclipse expected?

A Monte Carlo simulation uses the estimated probabilities of events occurring and then does a simulated run thousands of times.  Applying this technique to the King’s Gauntlet of Mousehunt gives a set of simulations from which we can get an excellent estimate of the true distribution of possible runs, given a starting cheese amount and which tier you start converting with SuperBrie+.

Using MATLAB, I simulated 10,000 King’s Gauntlet run for each possible SB+ Tier conversion level, for multiple starting cheese amounts.  The results for 100 Brie as starting cheese are below.  The Y-axis indicates the cumulative proportion of runs that have equal or exceed the value of the X-axis.  For example, using SB+ conversion from Tier 2 pots and above, 54% of the time you will catch at least 1 Eclipse mouse with 100 starting Brie.  You will earn less than 1.9 million points half the time, and more than 1.9 million half the time.  And overall, you will lose 2.1 million or more in gold about half the time, assuming a SB+ cost of 5000 (+10% tax).

If you start with 1,000 Brie, here’s what your run distributions look like :

So now that we can see the range of possibilities for each strategy of SB+ conversion, we can answer the questions that are on every mousehunter’s mind.  Which tier should I switch to SB+ conversion on if I want to, on average, break even on profits in my hunting?  Converting Tier 5 pots and above with SB+ will let you break even across the entire run, on average.  Maximum average profit of 160,000 gold per 100 starting Brie is achieved when converting pots with SB+ only on tier 7 and 8, though starting on tier 6 is almost as good and will net you more eclipse. Starting with SB+ from the very beginning will be VERY expensive. You will lose about 2,000,000 per 100 starting Brie.  However, you will probably get an Eclipse even with only 100 starting cheese!

What about catching that Eclipse mouse?  How many cheese do you need to use to have a 50% chance of catching it?  What about a 90% chance?  If you SB+ convert from tier 5 onwards, you need to start with 265 Brie for a 50% chance and 880 for a 90% chance.

Of course, this data is only as good as our potion drop estimates, which are based on YOUR data.  Please consider saving copies of your logs and summarizing them with Nathan Yang’s tool found here.  You can submit your data to a massive mousehunt database by clicking here.

The estimates used for this simulation were:

If you are interested in the MATLAB code used to generate this data it can be found below :


If you have further questions re: mousehunt please leave a comment.

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6 Responses to Monte Carlo analysis of King’s Gauntlet

  1. Mike says:

    Thanks for this outstanding analysis!

    One question if you have any time/inclination.. is there any possibility that hunting in the KG is stragically justifiable from a points per hunt perspective? I was looking at this problem just based on my own hunting results but never finished the analysis to get an answer. I’m very curious what even a rough estimate of points/hunt would be in KG; I think its unlikely to be competitive with Balack’s Cove or Seasonal Garden+Tower, but I don’t think its impossible!

  2. andrewhires says:

    On average, if you start with 1000 Brie, and use SuperBrie to convert T2 pots and above (which will give you max points / hunt overall) you will do :

    5635 hunts
    Lose 21M gold (superbrie – proft)
    Gain 22M points
    for 3900 points per hunt overall.

    With T5 convert
    3070 hunts
    break even on gold
    Gain 7M points
    for 2280 points per hunt.

    So its not terrible, but Nerg, Elub, Balack’s Cove, Seasonal Garden and Zugzwang’s Tower are all significantly more efficient places to get points.

  3. culteras says:

    Suppose I have only 100 SB+ to spend, which tier of pots to spend the 100SB+ will net the most eclipse and how to prove this statistically?

    My hypothesis is that tier 2 pots will produce more cheese using SB+ and will be most efficient to use the 100 SB+, but it may be that using SB+ at higher tiers will net more eclipse as pot drops are rarer then.

    • ian w says:

      proving this statistically is easy its T7 pots as SB+ increases the number of T7 chesses this will statistically result in more T8 potions and as every tier of the gauntlet gives only a bit more than a 10% yield of potions given you will get diminishing returns. how ever this will change if you want to do this for a selected number of starting cheese however I think the problem of the least time taken to get the most eclipses with only 100 SB+ for conversion is over specified but could be solved using linear programming functions.

  4. ian w says:

    lol, I did the same kind of thing trying out an ant colony optimisation algorithm last year unfortunately I can’t find the data it generated but it came to the same conclusion but I used 0.125 for all the potion drop rates to make it easer for me

    I might try Nathan’s spread sheet and excel to ‘try’ Monte-Carlo as iv herd its possible and never know when its going to be useful
    Fortran for life :^)

  5. Daan Eeckhout says:

    If i start whit a 1000 brie run and starting to convert whit sb on T5 what whil the number of eclipse that i catch be (±)

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