How to get rid of possible ties?

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#1

So for context, the players vote for one of three maps. The issue is I have no way of getting the most voted map while breaking up ties.

For instance, sometimes Map1 has 3 votes, Map2 has 1 vote and Map3 also has 3 votes. This is the script I have so far.

for i,v in pairs(VotingOptions) do
		if v[1].Votes == v[2].Votes then
			v[2].Votes += 1
		elseif v[2].Votes == v[3].Votes then
			v[3].Votes += 1
		elseif v[1].Votes == v[3].Votes then
			v[1].Votes += 1
		end
	end
end

What it does is it iterates through the votes and breaks up ties. I also don’t know how to get the most voted one of them all. Also will it even break up all ties?

Any help is appreciated.

#2

math.random() will guide you to your answer.

Also, you clearly do not understand some basics. It’s best to understand atleast the basics before going into making a game. Scripting Basics - Beginner guide to lua and Roblox

#3

Where did he said math.random? He does understands some basics, like for loops, and if statements. You can just compare the 3 values together, determine what is the tie, and get rid of it.

1 Like
#4 
local Votes = {
	{
		["Votes"] = 3,
		["Name"] = "A"
	},
	{
		["Votes"] = 2,
		["Name"] = "B"
	},
	{
		["Votes"] = 3,
		["Name"] = "C"
	},
}


local Highest = {}
for i, v in pairs(Votes) do
	if Highest[1] == nil then Highest[1] = v continue end
	if v.Votes > Highest[1].Votes then
		Highest[1] = v 
		continue
	elseif v.Votes == Highest[1].Votes and v ~= Highest[1] then
		Highest[2] = v
		continue
	end
end
if Highest[2] == nil then
	Highest = Highest[1]
else
	if Highest[1].Votes == Highest[2].Votes then
		Highest = Highest[math.random(1,2)]
	end
end
print(Highest)
1 Like
#5

Why are you even looping if you’re still going to check all of the indexes one by one? Also the solution above most likely works but here’s a less complicated one:

table.sort(VotingOptions, function(a, b) return a.Votes > b.Votes end)  -- Sort by votes
local winner = votes[1]  -- Take a single most voted one (even if it's tied)
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