3 contestants are left in the finale of a reality TV show. The public must then vote for these contestants (one vote per person). The contestant with the most votes wins. Most means more than every other contestant individually.
eg A = 40%, B = 35%, C = 25% means that A wins.
Contestant A belongs to a group (Group 1) which comprises 1/3 of the population.
Contestants B and C belong to the other 2/3 of the population.
Group 1 is known to vote tribally, thus they all vote for contestant A.
The remainder of the population votes randomly for the 3 contestants.
What is the probability that contestant A wins the competition?
Can you generalize the result for more than 3 contestants?
Update:
Steve: I’m a little unclear with your terminology. What does “Binomial[2n+2,2]” mean? Is this a value, a summation…?
Update 2:
OK tell me if I’m making any mistake here.
There are 3n balls, A already has n.
A = final no. of balls A gets
B = final no. of balls B gets
Expectations
E(A) = n + 2n/3
E(B) = 2n/3
Variances
Var(A) = n*(1/3)*(2/3) = 2n/9
Var(B) = 2n/9
E(B-A) = -n
Var(B-A) = 4n/9
For large n
B-A ~ N(-n, 4n/9)
So P(B-A>0) = P(Z > 1.5√n)
where Z ~ N(0,1)
As n –> ∞, this prob approaches 0.
Update 3:
This is the probability that B wins. Double it for C winning, but it still approaches zero. Thus shouldn’t P(A winning) approach 1 as n –> ∞?
Update 4:
Also Steve, shouldn’t a 3^(4n) appear somewhere in your denominator?
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Think of this as 3 bins, A B and C, and you are placing 2n balls in the bins, with n balls already in bin A. The only way A can’t win (assuming no one wins with a tie), is if, say, bin B has n+b balls, and bin C has c balls, with 2n-b-c ≤ n+b, or n-c ≤ 2b. I will assume n is even, to avoid floor/ceiling functions. Then the number of ways this happens, allowing b to range from 0 to n, is
(n/2 + 1)². Doubling (to count the symmetric case C has more balls), and discarding the second counting of b=0,c=n, gives
2 * (n/2 + 1)² – 1. So the probability A wins is
1 – [ 2 * (n/2 + 1)² – 1] / Binomial[2n+2,2].
Steve
EDIT – in the general case, a nice intuitive picture is to assume some continuity, then we are finding the measure of a (hyper)plane intersect a (hyper)cube. Of course, in the discrete case, we are counting lattice points, which is ridiculously difficult, even in relatively low dimensions.
I think the overall simplifying notion here is to look at when A *doesn’t* win. Then, we don’t care what the other (k-1) candidates have, it only matters what the candidate with the *most* votes has. This sufficiently linearizes the problem, and I think makes it one of linear programming and optimization. Basically, we have (k-1) non-negative integers, whose sum is bounded above. Assuming B has the most votes (and dealing with symmetry afterward), we have upper and lower bounds on B’s votes, and for each integral value in this range, we have an upper bound on the sum of the remaining (k-2) vote counts. This is something a computer can easily do.
EDIT 2 – Binomial[2n+2,2] is the binomial coefficient, as in
( 2n+2 )
(… 2 …)
EDIT 3 – there really is no need to write the binomial coefficient that way, we can instead write the probability as
n(3n+2) / [ 2 * (n+1) * (2n+1) ]. This is nice, because it shows the probability is steadily increasing, yet the limiting value is not 1, but 3/4. So Zanti’s first assertion is incorrect; it’s nice because it shows how counter-intuitive probability can be.
EDIT 4 – Sorry for so many edits! OK, here is a function that works with even and odd. So your hypotheses imply the population size is divisible by 3, so write it as 3n. Then the probability A wins is
( n(2n+1) – IntegerPart[n²/2] ) / ( (n+1) * (2n+1) ).
As you can see, not as pretty as the even case 🙂
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For a reasonably large number of voters, the probability A wins is surely very close to 1%. Since A already has 1/3 of the votes locked up, B and C have almost no chance to catch him. Also, the more votes being made, the closer A’s odds get to 1%.
With more than 3 contestants, A will still be the prohibitive favorite, but the odds improve for one of the other contestant to pull off the huge upset, for two reasons: (a) A starts off with fewer votes in his pocket, and (b) there are more contestants, hence a better chance one of them will get lucky.
IM not sure, but I think its
(1/3)+((1/3)*(2/3))