# Probability in Fighting Tournament in which Contestants consume each other

A tournament has 2 teams of 24 people each of weight 1. Each team sends one person into the pit where they fight until one consumes the other taking on all their weight.

The probability that one person consumes the other $$= \frac{their\ weight}{total\ weight\ in\ pit}$$ i.e. for first fight each person has probability $$\frac{1}{2}$$ of winning.

After 41 fights Team A has 4 members left of weights 16, 6, 5, 1 and Team B has 3 members of weights 10, 6, 4.

Team B has sent forth their member of weight 6.

What is the optimal strategy for Team A and what is their overall probability of winning this tournament after 41 fights.

• Do we always know who team B send in before we choose? And how does team B choose? Is it random with an equal chance between their remaining members, or do they use an optimal strategy? – Kruga Dec 13 '18 at 9:35
• puzzling.stackexchange.com/q/43771 [spoiler] – deep thought Dec 14 '18 at 0:22

## 2 Answers

Claim

I claim that no matter what A does (or indeed, what anyone does), team A currently has a $$(16+6+5+1)/(48) = 28/48 = 7/12$$ chance of winning. Indeed, this probability of winning is only dependent on the relative ratios of the total values of the piles on the two sides.

Proof

We proceed by induction on the number of total piles, which always decreases by 1 in every instance. The base case, where there are two piles, is obvious by definition of the problem: both sides must send out their entire weight and win in accordance to their ratios.

Now, suppose we have two teams, with total masses $$A$$ and $$B$$ respectively. They send out weights of masses $$x$$ and $$y$$ respectively. Now, let us compute the probability that team A wins. A loses this fight with a $$\frac{y}{x+y}$$ probability, for which then they will have $$A-x$$ mass compared to $$B+x$$ for B's side, and a probability of winning of $$\frac{A-x}{A+B}$$. A wins this fight with a $$\frac{x}{x+y}$$ probability, for which then they will have $$A+y$$ mass compared to $$B-y$$ for B's side, and a probability of winning of $$\frac{A+y}{A+B}$$. Taking weighted averages, it follows that A's probability of winning in the overall scenario is $$\frac{y}{x+y}\cdot\frac{A-x}{A+B}+\frac{x}{x+y}\cdot\frac{A+y}{A+B} = \frac{A(x+y)+x\cdot y-y\cdot x}{(A+B)(x+y)} = \frac{A(x+y)}{(A+B)(x+y)} = \frac{A}{A+B}$$, which is what we desired. This completes the induction.

• Does your answer hold, in the case that A has 10 units of mass 1, and B has 1 unit of mass 10, that they would then have an equivalent chance of winning? My intuition says that this wouldn't be the case – AHKieran Dec 13 '18 at 10:36
• @AHKieran I've wrote some code to test it and yes, I believe this claim is true (not that the proof wasn't solid to begin with). So now we know the answer to that 100 duck-sized horses vs. 1 horse-sized duck dilemma. My only doubt is whether B sending the weight 6 member first changes the result at all. – NudgeNudge Dec 13 '18 at 10:45

Retroactive Answer:

The optimal strategy would be to save the 16 for the latter fights and send in 6 against 6.

My primary thoughts are:

To send forth the person with weight 16 as they have the highest chance of winning this fight, and would then have an even higher chance of winning the following fights.

It might be an option to:

Hold back the person with weight 16 to save them for the fight against 10, in case they get defeated by the 6. However, this would mean a 6v6 which is a 1 in 2 chance of creating a 12 to fight as well.

Lets put some numbers to this:

If they just send 16 to the rest of the fights then the chance they win all three is either $$\frac{16}{22} \cdot \frac{22}{26} \cdot \frac{26}{36} = \frac{9152}{20592} = 0.\dot{4}$$
if they send out 4 second and 10 third, or
$$\frac{16}{22} \cdot \frac{22}{32} \cdot \frac{32}{36} = \frac{11264}{25344} = 0.\dot{4}$$
if they send out 10 second and 4 third. As you can see, these are the same probability.

If instead, they save the 16 to fight the 10, and use the 6 to fight the 6 and then the 4, there is a 50% probability that the 16 will have to fight a 10 and a 12 and a 4 (if their 6 loses and creates a 12 for the opposition), and a 50% probability that the 6 wins, becomes a 12, and then depending on whether they send the 4 or the 10 out next, a similar problem occurs.

Let's determine if the ordering of the last part matters:
$$\frac{16}{26} \cdot \frac{26}{30} = \frac{416}{780} = 0.5\dot{3}$$
$$\frac{16}{20} \cdot \frac{20}{30} = \frac{320}{600} = 0.5\dot{3}$$
So if A send 16 against both 10 and 4 then it doesn't matter what order it consumes them. (Which agrees with the first calculation). However, there if the opposition sends 4 out before 10, then there is the option to pit the 12 against the 4 and save the 16 for the 10, in case 16 loses to the 4 and they're left with a 20 and a 10 to fight with just a 12.
The chances that they win with pitting 12 against 4 then 16 against 10 are then:
$$\frac{12}{16} \cdot \frac{16}{26} = \frac{192}{416} = 0.\dot{4}6153\dot{8}$$

I realise now that I'm probably in way too over my head so lets round this off: $$\frac{1}{2} \cdot \frac{192}{416} + \frac{1}{2} \cdot \frac{320}{600} = 0.4\dot{9}7435\dot{8}$$
This number is the average of the chance to win depending on latter decisions if 16 is saved and they win the 6v6 battle.
Now, if they lose the 6v6 battle, 16 will have to fight 12 and 10 and win both to even have a remote possibility of winning overall:
$$\frac{16}{28} \cdot \frac{28}{38} = \frac{448}{1064} = 0.42105...$$
Next we combine the two probabilities to create an overall one of winning in this strategy:
$$\frac{1}{2} \cdot 0.4\dot{9}7435\dot{8} + \frac{1}{2} \cdot \frac{448}{1064} = 0.459244...$$

Finally:

$$0.459 > 0.\dot{4}$$
Therefore the optimal strategy would be to save the 16 for the latter fights and send in 6 against 6.

• I think you're missing some things here. For example, A could lose the first fight, with 16 say, and still win overall so the probability of winning while putting 16 out first will be higher than 0.44444... – hexomino Dec 13 '18 at 11:49
• @hexomino yeah I think I realised the actual number of possibilities when I was already partway through but just soldiered on to get a rough idea. My idea was that if 16 lost any, then the chance of A's remaining numbers being able to defeat B's new big boi would be very slim so not affect the answer much? – AHKieran Dec 14 '18 at 8:36