How people vote in contests: new findings from Immortal Songs 2

Abstract

Many important contests, such as job interviews and political debates, are presented in sequential order. Previous studies found serial position effects such that the later presenter has a higher probability of winning. However, no previous studies use both random assignments of contestants and a large number of contestants and judges. We use Immortal Songs 2, a popular TV program that satisfies both conditions, to confirm the findings of serial position effects. In addition, Immortal Songs 2 has round-by-round competition rules. The first round is a competition between the first and second contestants. Then, the winner of the first round is announced. The second round is a contest between the winner of the first round and the third contestant. As the rounds continue, in theory, the winner of the previous round should have a higher probability of winning; however, in fact, the second contestant’s probability of winning is always 0.5.

Appendix

1.1 Asymptotic win probabilities

We first state an assumption that is maintained throughout the analysis. Let $\left( {U_{i}^{1} , \ldots , U_{i}^{m} , \ldots ,U_{i}^{M} } \right)$ be the utility of person $i$ in the randomly collected audience of size $n$ from the m-th position singer’s song in the (m − 1)-th round. The voting rule is that i votes for the m-th position $\left( {y_{i}^{m} = 1} \right)$ if she is sufficiently satisfied with the singer’s performance and does not vote for it $\left( {y_{i}^{m} = 0} \right)$ otherwise. Assuming that i’s utility from the m-th position singer’s performance is not affected by all previous position singers’ songs and that i’s vote in the (m − 1)-th round is solely determined by his or her utility from the m-th position singer’s performance, we may regard $y_{i}^{m} = I\{ U_{i}^{m} > c^{m} )$, where $I\left\{ A \right\}$ is an indicator function that takes the value of one if condition $A$ holds and zero otherwise, and $c^{m}$ is the threshold above which $i$ feels that the m-th position singer’s performance was satisfactory. If we assume that i treats every position equally, then we can say that $c^{m} = c$ for all m. Let $\Pr \left[ {U_{i}^{m} \le \nu } \right] = F\left( \upsilon \right)$. Then, without loss of generality, $\Pr \left[ {y_{i}^{m} = 1} \right] = \Pr \left[ {U_{i}^{m} > c} \right] = 1 - F\left( c \right) = q$, that is, $y_{i}^{m}$ is a Bernoulli random variable with fixed success probability $q$.

Let $Y^{m}$ be the total number of votes that the m-th position receives in the (m − 1)-th round. The probability that the (m + 1)-th position wins in the m-th round is the probability that $Y^{m + 1}$ is greater than or equal to all previous votes, that is, $\Pr \left[ {Y^{m + 1} \ge \mathop {\hbox{max} }\nolimits_{s \le m} Y^{s} } \right]$. To derive this probability, we need the following lemmas.

Lemma 1

For every m, $\frac{1}{\sqrt n }\mathop \sum \nolimits_{i = 1}^{n} \left( {y_{i}^{m} - q} \right)\mathop \to \limits^{p} \sigma Z_{m} \sim N\left( {0, \sigma^{2} } \right), \;{\text{where}}\;\sigma \text{ := }\sqrt {q\left( {1 - q} \right)}$ as $n \to \infty$.

Proof of Lemma 1 It is obvious because $y_{i}^{m}$ is an independent and identically distributed (i.i.d.) Bernoulli random variable with fixed success probability $q$ for all i and m.

Lemma 2

Consider independent and identically distributed standard normal random variables $Z_{1} , Z_{2} , \ldots , Z_{m}$. Let us define $V^{\left( m \right)} \text{ := }\mathop {\hbox{max} }\nolimits_{s \le m} Z_{s}$. $\Pr \left[ {V^{\left( m \right)} \le x} \right] = [\Phi \left( x \right)]^{m}$ and $\frac{\text{d}}{{{\text{d}}x}}\Pr \left[ {V^{\left( m \right)} \le x} \right] = m\left[ {\Phi \left( x \right)} \right]^{m - 1} \phi \left( x \right)$, where $\Phi \left( \cdot \right)$ and $\phi \left( \cdot \right)$ are the distribution and density functions associated with the standard normal distribution.

Lemma 2 is easily proved by using $\Pr \left[ {V^{\left( m \right)} \le x} \right] = \Pr \left[ {Z_{1} \le x, Z_{2} \le x, \ldots , Z_{m} \le x} \right]$.

Lemma 3

Consider the i.i.d. standard normal random variables $Z_{1} , Z_{2} , \ldots , Z_{m + 1}$.

$$\Pr \left[ {Z_{m + 1} \ge \mathop {\hbox{max} }\limits_{s \le m} Z_{s} } \right] = \frac{1}{m + 1}.$$

Proof of Lemma 3 $\begin{aligned} & \Pr \left[ {Z_{m + 1} \ge \mathop {\hbox{max} }\limits_{s \le m} Z_{s} } \right] \\& = \int \Pr \left[ {Z_{m + 1} \ge x |\mathop {\hbox{max} }\limits_{s \le m} Z_{s} = x} \right]m[\Phi \left( x \right)]^{m - 1} \phi \left( x \right){\text{d}}x \\ & = 1 - m\smallint \left[ {\Phi \left( x \right)} \right]^{m} \phi \left( x \right){\text{d}}x = 1 - m\mathop \smallint \limits_{0}^{1} u^{m} {\text{d}}u = \frac{1}{m + 1}. \\ \end{aligned}$.

By using Lemmas 1, 2, and 3, we can see that the following holds.

Proposition

As ${\text{n}} \to \infty$,

$$\begin{aligned} & \Pr \left[ {Y^{m + 1} \ge \mathop {\hbox{max} }\limits_{s \le m} Y^{s} } \right] = \Pr \left[ {\mathop \sum \limits_{i = 1}^{n} \left( {y_{i}^{m + 1} - q} \right) \ge \mathop {\hbox{max} }\limits_{s \le m} \mathop \sum \limits_{i = 1}^{m} \left( {y_{i}^{s} - q} \right)} \right] \\ & \quad = \Pr \left[ {\frac{1}{{\sqrt {nr} }}\mathop \sum \limits_{i = 1}^{n} \left( {y_{i}^{m + 1} - q} \right) \ge \mathop {\hbox{max} }\limits_{s \le m} \frac{1}{{\sqrt {n\sigma } }}\mathop \sum \limits_{i = 1}^{n} \left( {y_{i}^{s} - q} \right)\mathop \to \limits^{p} \Pr \left[ {Z_{m + 1} \ge \mathop {\hbox{max} }\limits_{s \le m} Z_{s} } \right]} \right. \\ & \quad = \frac{1}{m + 1} \\ \end{aligned}$$

Thus, the probability that the (m + 1)-th position wins in the m-th round is approximately 1/(m + 1) when n is sufficiently large.

1.2 Exact round win probabilities

We can compute the exact winning probabilities with (potentially) varying $\Pr \left[ {y_{i}^{m} = 1} \right] = q^{m}$. For a given audience size n and number of contestants m, $Y^{m}$ follows a binomial distribution with the number of trials n and the probability of success $q^{m}$. Let us denote two functions: $b\left( {s:n,q} \right) = \left( {\begin{array}{*{20}c} n \\ s \\ \end{array} } \right)q^{s} \left( {1 - q} \right)^{n - s}$; $B\left( {s:n,q} \right) = \mathop \sum \nolimits_{r = 0}^{s} b\left( {r:n,q} \right)$, where $\left( {\begin{array}{*{20}c} n \\ s \\ \end{array} } \right) = \frac{n!}{{s!\left( {n - s} \right)!}}$, such that $\Pr \left[ {Y^{m} = s } \right] = b\left( {s:n,q^{m} } \right)$ and $\Pr \left[ {Y^{m} \le s } \right] = B\left( {s:n,q^{m} } \right)$. For notational compactness, we let $B\left( {s:n,q^{m} } \right) = 0$ for any $s < 0$ regardless of $q^{m}$. Because $\Pr \left[ {\mathop {\hbox{max} }\limits_{s \le m} Y^{s} = k} \right] = \Pr \left[ {Y^{1} = k,Y^{2} = k, \ldots ,Y^{m} = k} \right] + \Pr \left[ {Y^{1} = k,Y^{2} \le k - 1, \ldots ,Y^{m} \le k - 1} \right] + \Pr \left[ {Y^{1} \le k - 1,Y^{2} = k, \ldots ,Y^{m} \le k - 1} \right] + \cdots + \Pr \left[ {Y^{1} \le k - 1,Y^{2} \le k - 1, \ldots ,Y^{m} = k} \right]$, we can see $\Pr \left[ {\mathop {\hbox{max} }\nolimits_{s \le m} Y^{s} = k} \right] = \mathop \prod \nolimits_{r = 1}^{m} b\left( {k:n,q^{r} } \right) + \mathop \sum \nolimits_{l = 1}^{m} b\left( {k:n,q^{l} } \right)\mathop \prod \nolimits_{r = 1,r \ne l}^{n} B\left( {k - 1:n,q^{r} } \right) \equiv b_{max} \left( {k:n,\left( {q^{1} ,q^{2} , \ldots ,q^{m} } \right)} \right).$

The exact probability of the (m + 1)-th position wins at the m-th round, which we may denote as $\Xi _{m + 1} \left( {n,\left( {q^{1} ,q^{2} , \ldots ,q^{m + 1} } \right)} \right)$ is

$$\Xi _{m + 1} \left( {n,\left( {q^{1} ,q^{2} , \ldots \infty ,q^{m + 1} } \right)} \right) = \Pr \left[ {Y^{m + 1} > \mathop {\hbox{max} }\limits_{s \le m} Y^{s} } \right] + 0.5 \Pr \left[ {Y^{m + 1} = \mathop {\hbox{max} }\limits_{s \le m} Y^{s} } \right],$$

where the latter term deals with tie-break situations. With given $\left( {q^{1} ,q^{2} , \ldots ,q^{m} ,q^{m + 1} } \right)$, $\Pr \left[ {Y^{m + 1} > \mathop {\hbox{max} }\limits_{s \le m} Y^{s} } \right] = 1 - \Pr \left[ {Y^{m + 1} \le \mathop {\hbox{max} }\limits_{s \le m} Y^{s} } \right] = 1 - \mathop \sum \nolimits_{s = 0}^{n} B\left( {s:n,q^{m + 1} } \right) \cdot b_{max} \left( {s:n,\left( {q^{1} ,q^{2} , \ldots ,q^{m} } \right)} \right)$ and $\Pr \left[ {Y^{m + 1} = \mathop {\hbox{max} }\limits_{s \le m} Y^{s} } \right] = \mathop \sum \nolimits_{s = 0}^{n} b\left( {s:n,q^{m + 1} } \right) \cdot b_{max} \left( {s:n,\left( {q^{1} ,q^{2} , \ldots ,q^{m} } \right)} \right)$.

1.3 The conundrum of 0.5

With given n and $q^{1}$, we can find $q^{2}$ such that the second contestant wins the first round with probability 0.5 by solving $\Xi _{2} \left( {n,\left( {q^{1} ,q^{2} } \right)} \right) = 0.5$. With n, $q^{1}$, and $q^{2}$, we can solve $\Xi _{3} \left( {n,\left( {q^{1} ,q^{2} ,q^{3} } \right)} \right) = 0.5$ to find $q^{3}$ with which the third contestant wins the second round with probability 0.5 and so on. Table 7 shows solutions for n = 500, M = 9, and $q^{1} = 0.5$ and $q^{1} = 0.7654$.

Regarding Table 7, for example, if individuals in the audience vote “yes” for the first contestant with probability 0.5, then they vote “yes” for subsequence contestants with probabilities 0.5, 0.5016, 0.5141, and so on. Basically, for the 0.5 winning probabilities at each round to happen, individuals must increasingly vote “yes” for the second contestant as the rounds progress.

Table 8 shows that the number of votes calculated by the theoretical model and the actual data are similar.