This question was previously from the Haselbauer - Dickheiser Test For Exceptional Intelligence, a test that was previously available on the website www.highiqosociety.com, but was removed several years ago.

Consider two breeding strategies of the fictional Furble. Dominator Furbles can fight for a breeding territory, and if they win, will be able to rear 10 offspring. An alternative is to share territory with another Furble which will allow each to rear 5 offspring. Sharers who attempt to share with dominators will be forced out of the territory, although they will be able to find a new territory. Assume sharers become extra cautious after encountering a dominator and so will always find another territory to share the next time around, but due to lost time will only be able to produce 3 offspring. Dominators are always able to force sharers out of the territory and rear 10 young. Dominators who meet dominators will win 50% of the time. When they lose, they are not able to reproduce that season due to sustained injuries. Individual Furbles cannot switch strategies.

With a total population of 2000 dominator and sharer Furbles altogether, how many would you expect to dominate?

I would assume no Furble finds an empty location. They either find someone to share or fight for their territory. Otherwise the solution is trivially 2000 dominators.

  • $\begingroup$ Is there only enough territory for 1/2 the Furbles initially? $\endgroup$
    – JonTheMon
    Commented Nov 8, 2014 at 16:46
  • $\begingroup$ @JonTheMon, the number of territories is not given. $\endgroup$
    – Kenshin
    Commented Nov 8, 2014 at 16:52
  • $\begingroup$ do you mean 2000 of each? or there are 2000 furbles? $\endgroup$
    – d'alar'cop
    Commented Nov 8, 2014 at 17:03
  • $\begingroup$ @d'alar'cop, I meant altogether. I have edited. $\endgroup$
    – Kenshin
    Commented Nov 8, 2014 at 17:05
  • $\begingroup$ @Mew I've noticed that many of your posts have unsourced material - would it be possible for you to go back through them and attribute them? Our plagiarism policy is rather important in these cases. Thank you! $\endgroup$
    – user20
    Commented Nov 9, 2014 at 7:26

5 Answers 5


I assume that this question is asking what a stable ratio of dominators to sharers is. I'm also going to assume that their strategy comes from their parent (otherwise there's definitely not enough information to solve the puzzle), and that there at least some of each.

Let's look at what happens if we have a 50-50 ratio, so 1000 furble using each strategy. Since their numbers are equal, each dominator has a 50-50 chance of encountering a dominator or a sharer, and the same is true for the sharers. So with this generation, we have 250 territories with 2 dominators, 500 territories with a dominator and a sharer, and 250 territories with 2 sharers. With 2 dominators, one will win and rear 10 young while the other has no children. With a dominator and a sharer, the dominator rears 10 young while the sharer rears 3 young. With 2 sharers, they each raise 5 young.

So starting with 1000 of each, we get:

  • 250*10 + 500*10 = 7500 new dominators
  • 500*3 + 250*10 = 4000 new sharers

Whether or not we include the parents in the final count, the ratio of dominators to sharers has changed, so a 50-50 ratio is not stable.

Rather than guess and check at proper ratios, we'll set up some formulas. If there are $x$ dominators and $y$ sharers, then there will be

  • $\frac{x^2}{2(x+y)}$ dominator pairs
  • $\frac{2xy}{2(x+y)}$ dominator-sharer pairs
  • $\frac{y^2}{2(x+y)}$ sharer pairs

Thus in the next generation there will be

  • $X=\frac{10x^2+20xy}{2(x+y)}$ dominators
  • $Y=\frac{6xy+10y^2}{2(x+y)}$ sharers

Because we want a stable ratio, we want $\frac{x}{y}=\frac{X}{Y}=\frac{10x^2+20xy}{10y^2+6xy}$. Attempting to solve, we get:

  • $5xy^2+3x^2y=5x^2y+10xy^2$
  • $0=2x^2y+5xy^2$
  • $2x+5y=0$
  • $2x=-5y$

Well, that didn't go so well, but it fits the formulas. If we can somehow have negative numbers of furbles, then we can get a stable ratio. For 2000 total, this means we would want to have -1333.33 sharing furbles and 3333.33 dominating furbles.

That means that there are only two (trivial) stable solutions - either no sharing furbles or no dominating furbles. If there are ever a (positive) mix of the two, then with each successive generation the dominating furbles become more and more common. As such, I'd expect that if you have 2000 furbles, most of them will be dominators, so most of the initial 1000 pairings have at least one dominator. Thus, I would expect around 1000 furbles to dominate.

Of course, I could be overthinking this whole thing - if they're assuming an initial 50-50 ratio, then 750 furbles dominated (the 500 paired with a sharer + the 250 winners of the dominator pairs).

  • $\begingroup$ +1 Rob, nice solution (apart from negative furbles lol). I liked your approach and it helped me to prepare my solution. I have prepared an alternative solution that doesn't assume dominators will breed dominators, but rather assumes that for any child they have a probability, p, of being a dominator. The result is p = 0.5. $\endgroup$
    – Kenshin
    Commented Nov 9, 2014 at 14:27

I didn't know the answer when I posted this question either, but this is the answer I have come up with after reading the attempt by Rob Watts.

As Rob pointed out, if $x$ is the number of dominators, and $y$ is the number of shares, then for a particular round of breeding, the number of dominator-dominator pairs, the number of dominator-sharer pairs and the number of sharer-sharer pairs is given by the equations below:

  • $\frac{x^2}{2(x+y)}$ dominator pairs
  • $\frac{2xy}{2(x+y)}$ dominator-sharer pairs
  • $\frac{y^2}{2(x+y)}$ sharer pairs

The total number of children in a single breeding season, $C$ is thus given by,

$C=\frac{10x^2+20xy}{2(x+y)} + \frac{6xy+10y^2}{2(x+y)}$

My assumption is different to Rob's in that instead of assuming that dominators will always breed dominators, and sharers will always breed sharers, I will assume that the probability of being a dominator is $p$ and the probability of being a sharer is $1-p$.

Therefore, if we let $m$ be the total population, the total number of children in a breeding season is given by,

$C=\frac{10m^2p^2+20m^2p(1-p)}{2m} + \frac{6m^2p(1-p)+10m^2(1-p)^2}{2m}$

Which simplifies to,

$C = m[-3p^2 + 3p + 5]$

That is, the number of children in a season depends on current population, $m$, and the probability a child will be a dominator $p$.

$C$ is then maximized when the derivative, $C' = 0$, that is,

$-6p + 3 = 0$

And therefore,

$p = 0.5$

Therefore to maximize the number of children produced, the probability of being a dominator should be 50%, and therefore there are 1000 dominators and 1000 sharers.

Non mathematical solution:

The maximum number of children will occur when a dominator clashes with a sharer. This gives 10 children to the dominator and 3 children to the sharer, a total of 13 children.

When a dominator clashes with a dominator, a total of 10 children result, and the same is true for when a sharer clashes with a sharer.

Therefore you want to maximize the number of dominator-sharer clashes which will occur when you have an equal number of dominators and sharers.

  • $\begingroup$ Hi Mew, so the solution assumes that the goal is to maximise the number of offspring? if so, the question doesn't really state that goal. nice job regardless :) $\endgroup$
    – d'alar'cop
    Commented Nov 14, 2014 at 0:42
  • $\begingroup$ @d'alar'cop, I don't know that was just my assumption that the species will want to maximize offspring. I don't know that this is the correct answer, just the best answer so far. I should probably put a note that this isn't the offical answer just my attempt. $\endgroup$
    – Kenshin
    Commented Nov 14, 2014 at 0:59
  • $\begingroup$ I think it's probably the answer they expect... but imho it's flawed because of the assumption that the question doesn't explicitly contain. What do you think the answer is without that assumption? $\endgroup$
    – d'alar'cop
    Commented Nov 14, 2014 at 1:00
  • $\begingroup$ @d'alar'cop', my assumption isn't random though, because if there are two groups of furbles, one group that inherets p = 0.5 and another the inherits p = 0.7, the group that inherits p = 0.5 will have many more children and will soon outnumber the group that inherits p = 0.7. Over a long period of time, the group p=0.7 will thus die out, and so it safe to assume that the conditions that maximize offspring will be the most likely conditions when viewing a random group of furbles. $\endgroup$
    – Kenshin
    Commented Nov 14, 2014 at 1:15
  • 1
    $\begingroup$ @d'alar'cop', yes so of course it assumes that the value of p is initially any randomly distributed number. So you would expect some furbles to have p = 0.5, and the number of such furbles will increase at a greater rate over time than furbles with other p values. $\endgroup$
    – Kenshin
    Commented Nov 14, 2014 at 1:34

If there are infinitely many territories then they would all be best to share i.e. it is in everybody's best interests that everyone "shares" - this avoids fights so there's no risk of missing out on breeding (nobody can "lose"). And everyone will get to rear 5 offspring (it was never said that the goal is to maximise offspring). Their offspring will have no trouble moving off to share some other breeding grounds after this (since there is no limits to the number of territories imposed.)

  • $\begingroup$ @Mew true, so the question is about this generation? or in general, what percentage will be dominators? I'm pretty sure the strategy they adopt may end up being dependent on the availability of the territory they are competing over. $\endgroup$
    – d'alar'cop
    Commented Nov 8, 2014 at 17:09
  • $\begingroup$ d'alar'cop, you should assume that no Furble finds an empty location. They either share, or fight for a spot. $\endgroup$
    – Kenshin
    Commented Nov 8, 2014 at 17:25
  • $\begingroup$ @d'alar'cop And another change in profile pic! Who's that handsome bloke? $\endgroup$ Commented Aug 6, 2015 at 20:43

Given that a furble has to fight or share with someone, it seems that the maximum amount of furbles from the previous generation would come from equal dominators and sharers. Each dominator dominates a sharer, then the sharers get together and have offspring.

So, the 2000 furbles would end up with 1000 + 10000 dominators and 1000 + 3000 sharers.

  • $\begingroup$ this answer assumes that the Furbles can coordinate who they interact with. However given that some Frubles try to be more cautious about running into Dominators, I believe that it is an otherwise fairly random process as to whether a dominator runs into another dominator or not. I don't know the answer to the question myself however, so you could be correct. $\endgroup$
    – Kenshin
    Commented Nov 8, 2014 at 18:15
  • $\begingroup$ Why do you assume that " the maximum amount of furbles from the previous generation" was the case? For me much more reasonable assumption is that population is stable. $\endgroup$
    – klm123
    Commented Nov 9, 2014 at 9:33

I think the key part to all of this is that Individual Furbles can't change their strategies. That means only several Dominator Furbles together can become Sharer Furbles. This would seem to suggest that if two Dominator Furbles have been kicked-out of their turf and are too weak to reproduce that season, they can only find refuge with another Dominator Furble which has been defeaten.

For the sake of this argument I'm going to say that Sharer Furbles have arisen in Dominator Furble culture out of a necessity for survival, rather than that both co-existed since the very beginning. It would make more sense to me that the aggressive survival instinct came first, and the passive characteristics came later.

In this scenario I'd say the entire problem can begin with two Dominator Furbles. One defeats the other, resulting in 10 offspring produced. The 10 offspring in addition to the two original furbles makes 12.

It's made clearer in the figures below:

Generation: 0,  1,  2,   3,  4,

Dominator: 2, 12, 66, 396, 2376

Sharer: 0, 0, 6, 36, 216

Total: 2,   12, 72, 432, 2592

As you can see from the TOTAL row, it doesn't take long for the population to surpass 2000. It's somewhere in the late part of the 3rd generation.

Of course, I could run through every possibility regarding whether a Dominator Furble decides to fight another Dominator Furble or take a Sharer Furble's territory, but to make things simple I'm going to say, on average, the population growth is going to be constant. Because of this I'm going to say all these figures fit under a geometric progression. An interesting property about geometric progression is that consecutive terms a, b, and c satifty the equation:

b^2 = ac

And to make the terms of the question less pedantic I'm going to assume that Dominator Fubles only fight each other once, and so for each season there isn't just one Dominator Furble who stands, triumphant, over the rest, finally at peace to reproduce. If that's the case the question becomes trivial and also stupidly worded.

However, if you do apply the rules which geometric progression imposes then you end up with a scenario where no matter what, Dominator Furbles will turf Shares out of their territory (as seen in the figures above).

So, what you end up with after all of that is:

Dominator + Sharer = 2000

Dominator/Sharer = 11

Work through the simple substitution and you get:

Total = 2000

Sharer = 166.666666

Dominator = 1833.3333333

So, if anyone is still aware this thread exists, tell me what you think. I'd say the answer is 1833.


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