2 Definitions and notation

Let C be the set of all $\kappa $ candidates contesting an IRV election. For a candidate c, denote that candidate’s vote total $v_c$ . $\beta $ will represent the ordering of candidates on a ranked ballot, where a voter may make $L \leq \kappa $ choices, with the positions indexed by i. When considering a list $\lambda $ , the ordered sub-list from index a up to index b (inclusive, indexing from 1) will be denoted $\lambda _{a:b}$ . We will represent the order in which candidates are dropped as a list of dropped candidates $S = [S_1, S_2, \ldots , S_{\kappa -1}]$ . For brevity let $S_{-1} \equiv S_{\kappa - 1}$ . We will also consider the list of dropped candidates with the winner w concatenated to the end, denoted by A, so that $A = [S_1, S_2, \ldots , S_{\kappa -1}, w]$ . To represent the number of voters who rank candidate c in any of the ballot positions between 1 and r, given that they assigned every higher ballot position to some candidate in the sub-list $S_{1:n}$ , we will use $\mu _{c}^{r}|S_{1:n}$ .

I use “ballot” to mean a voter’s cast ranking $\beta $ , “ballot length” to mean the number L of candidates who can be ranked, and “round” for each act of comparing votes (e.g., $S_1$ is dropped in the “first round”). There are two mutually exclusive ways that a ballot could be pivotal:

• • Direct pivotality: because a candidate is ranked on a ballot, they win the election.

• • Indirect pivotalty: because candidate A is ranked on a ballot, candidate B wins the election.

When I state a pivotal probability, the implicit comparison is to abstention, as explained in Section 2 of the Supplementary Material. Section 4 of the Supplementary Material discusses the history of indirect pivotality.

3 Example

Imagine a voter in Alaska’s 2022 U.S. House contest expects ballots to be cast in the following (unrealistic but illustrative) distribution.Footnote ² For exposition, suppose Bye always loses tie-breakers and Palin always wins them.

Default: If the voter abstains, Palin wins as follows:

Direct pivotality: If the voter ranks Peltola first (e.g., $\beta = $ [Peltola, Bye, Begich, Palin]), they cause Peltola to win:

Indirect pivotality: Imagine the voter casts $\beta $ = [Bye, Begich, Palin], leaving the last spot blank. By ranking Bye, they cause Peltola to win:

4 Direct pivotality

After d candidates have been dropped, candidate c has the following vote total:

(1) $$ \begin{align} v_{c} = \overset{c \ \text{first}}{\underset{\text{vacuous condition}}{\overbrace{\mu_{c}^1}\underbrace{|S_{1:0}}}} + \overbrace{\mu_{c}^{2}|S_{1:1}}^{\substack{c \ \text{second}, \\ \text{any dropped} \\ \text{candidate first}}} + \cdots + \underbrace{\mu_{c}^{d+1}|S_{1:d}}_{\substack{c \ \text{ranked in} \ d + 1, \\ \text{all candidates ranked} \\ 1 \ \text{to} \ d \ \text{dropped}}} \end{align} $$

(2) $$ \begin{align} v_{c} = \sum_{q = 0}^{d} \mu_{c}^{\ d+1} | S_{1:q}. \qquad\qquad\qquad\qquad\qquad \end{align} $$

Denote the probability that candidate c has k more votes than some candidate j by

(3) $$ \begin{align} \mathbb{P}(v_c - v_j = k). \end{align} $$

The probability that placing some candidate c in position i on the ballot will cause that candidate to win is the probability that, after every candidate has been eliminated except two, c is among those two remaining candidates, is ranked on the ballot above the other remaining candidate, and is either a) one vote short of winning and would not win the tie-breaker, or b) two votes short of winning and would win the tie-breaker. The probability that c and some other candidate have the same vote totals after some number of eliminations is:

(4) $$ \begin{align} \mathbb{P}(v_c - v_{S_{-1}} = 0). \end{align} $$

In order to reach a pivotal contest against candidate $S_{-1}$ , candidate c must exceed the vote total of every other candidate at the time at which they are dropped. But this is only pivotal if c also is not among the dropped candidates.

In Section 3 of the Supplementary Material, I state two independence assumptions, which I introduce only for ease of communication, so that we can multiply the probability of events without specifying their joint probabilities. By Assumption 2, the probability $p_d|S$ of a directly pivotal contest involving candidate c (for now just considering the case in which a tie can be broken), given a specific sequence S in which the other candidates are dropped, is:

(5) $$ \begin{align} \begin{aligned} p_d|S &= \mathbb{P}(v_c = v_{S_{-1}}) \times \bigg[\overset{\substack{\text{Probability }c \\ \text{not eliminated 1st}}}{\overbrace{\big(\mathbb{P}(\mu_{c}^{1}> \mu_{S_1}^1)\big)}} \underset{\substack{\text{Probability }c \\ \text{not eliminated 2nd}}}{\underbrace{\big(\mathbb{P}(\mu_{c}^2|S_1 > \mu_{S_2}^2|S_1)\big)}} \ldots \underset{\substack{\text{Probability }c\text{ not} \\ \text{eliminated 2nd-last}}}{\underbrace{\big(\mathbb{P}(\mu_c^{\kappa-2}|S_{1:\kappa-3} > \mu_{S_{\kappa-2}}^{\kappa-2}|S_{1:\kappa-3}\big)}} \bigg] \\ &= \mathbb{P}(\mu_c^{\kappa-1}|S_{1:-1} = \mu_{S_{-1}}^{\kappa-1}|S_{1:-1}) \times \prod_{h = 1}^{\kappa - 2} \mathbb{P} \big(\mu_c^h|S_{1:h-1} > \mu_{S_h}|S_{1:h-1}\big). \end{aligned} \end{align} $$

This equation is conditional on the candidates being dropped in the order of some list S. The probability of S occurring is the probability that candidate $S_1$ has fewer initial votes than any other candidate, that $S_2$ has the fewest votes once $S_1$ has been dropped, and so on. Recall that A is the list obtained by concatenating the winning candidate to the end of S. By Assumptions 1 and 2, the probability $p_d(S)$ that S is the list of candidates dropped, followed by c being in a directly pivotal contest, is:

(6) $$ \begin{align} \begin{aligned} p_d(S) &= \mathbb{P}\big(v_{S_1}^1 < v_{S_2}^{1}\big) \mathbb{P}\big(v_{S_1}^1 < v_{S_3}^{1}\big) \cdots \mathbb{P}\big(v_{S_{1}}^1 < v_{S_{-1}}^{1}\big)\mathbb{P}\big(v_{S_1}^1 < v_{c}^{1}\big) \times \\ &\quad \mathbb{P}\big(v_{S_2}^2 < v_{S_3}^{2}\big) \cdots \mathbb{P}\big(v_{S_{2}}^2 < v_{S_{-1}}^2\big)\mathbb{P}\big(v_{S_2}^2 < v_{c}^2\big) \times \\ &\quad \vdots \\ &\quad \mathbb{P}\big(v_{S_{\kappa-2}}^{\kappa-2} < v_{S_{-1}}^{\kappa-2}\big)\mathbb{P}\big(v_{S_{\kappa-2}}^{\kappa-2} < v_{c}^{\kappa-2}\big) \times \\ &\quad \mathbb{P}\big(v_{S_{-1}}^{\kappa-1} = v_{c}^{\kappa-1}\big) \times \frac{1}{2} \\ &= \bigg[\overset{\substack{\text{Probability the candidates} \\ \text{are dropped in order }A}}{\overbrace{\prod_{\ell = 1}^{\kappa - 2} \prod_{r=\ell+1}^{\kappa} \mathbb{P} \big(v_{A_\ell}^\ell < v_{A_r}^\ell \big)}} \bigg] \overset{\substack{\text{Probability that }c\text{ is} \\ \text{in a first-place tie} \\ \text{given the drop order }A}}{\overbrace{\mathbb{P} \big(v_{A_{\kappa-1}}^{\kappa-1} = v_c^{\kappa-1}\big)}} \ \overset{\substack{\text{Probability} \\ \text{after tying} \\ c \text{ loses}}}{\times \overbrace{\frac{1}{2}}} \\ \\ &= \bigg[\prod_{\ell = 1}^{\kappa - 2} \prod_{r=\ell+1}^{\kappa} \mathbb{P} \bigg(\sum_{q=0}^{\ell-1} \mu_{A_{\ell}}^{q+1} | A_{1:q} < \sum_{q=0}^{\ell-1} \mu_{A_{r}}^{q+1} | A_{1:q} \bigg) \bigg] \mathbb{P} \bigg(\sum_{q=0}^{\kappa-2} \mu_{A_{\kappa-1}}^{q+1} | A_{1:q} = \sum_{q=0}^{\kappa-2} \mu_{c}^{q+1} | A_{1:q}\bigg) \times \frac{1}{2}. \end{aligned} \end{align} $$

S is mutually exclusive with any other drop sequence, so the probability that ranking candidate c first will be directly pivotal is obtained by summing over Equation 6 for every possible drop sequence:

(7) $$ \begin{align} \begin{aligned} p_d = & \sum_{\substack{S \in \\ \text{Sym}(C \,{\backslash}\, c)}} \bigg\{ \bigg[\prod_{\ell = 1}^{\kappa - 2} \prod_{r=\ell+1}^{\kappa} \mathbb{P} \bigg(\sum_{q=0}^{\ell-1} \mu_{A_{\ell}}^{q+1} | A_{1:q} < \sum_{q=0}^{\ell-1} \mu_{A_{r}}^{q+1} | A_{1:q} \bigg) \bigg] \frac{1}{2} \cdot \mathbb{P} \bigg(\sum_{q=0}^{\kappa-2} \mu_{A_{\kappa-1}}^{q+1} | A_{1:q} = \sum_{q=0}^{\kappa-2} \mu_{c}^{q+1} | A_{1:q}\bigg) \bigg\} \end{aligned} \end{align} $$

where $\text {Sym}(C \,{\backslash}\, c)$ is the symmetric group on the set of other candidates.

To move beyond the first ballot position, impose a simple restriction: the ith ballot position can only be pivotal if all candidates ranked higher on the ballot have been dropped. Note also that a voter can be pivotal not just by breaking a first-place tie that the candidate would not have won, but also by creating a first-place tie that the candidate wins. The direct pivotal probability of the ballot $\beta $ is obtained from summing over Equation 7 for all positions on the ballot:

(8) $$ \begin{align} \begin{aligned} p_{\text{direct}}(\beta) = & \sum_{i=1}^{L} \bigg(\sum_{\substack{S \in \\ \text{Sym}(C \,{\backslash}\, i) \\ \beta_{1:i-1} \subset S}} \bigg\{ \bigg[\prod_{\ell = 1}^{\kappa - 2} \prod_{r=\ell+1}^{\kappa} \mathbb{P} \bigg(\sum_{q=0}^{\ell-1} \mu_{A_{\ell}}^{q+1} | A_{1:q} < \sum_{q=0}^{\ell-1} \mu_{A_{r}}^{q+1} | A_{1:q} \bigg) \bigg] \times \\ & \bigg[\frac{1}{2} \cdot \mathbb{P} \bigg(\sum_{q=0}^{\kappa-2} \mu_{A_{\kappa-1}}^{q+1} | A_{1:q} = \sum_{q=0}^{\kappa-2} \mu_{i}^{q+1} | A_{1:q}\bigg) + \\ & \frac{1}{2} \cdot \mathbb{P} \bigg(\sum_{q=0}^{\kappa-2} \mu_{A_{\kappa-1}}^{q+1} | A_{1:q} = 1 + \sum_{q=0}^{\kappa-2} \mu_{i}^{q+1} | A_{1:q}\bigg) \bigg] \bigg\}\bigg). \end{aligned} \end{align} $$

5 Indirect Pivotality

Suppose that candidates will be dropped according to $A \equiv [S_1, S_2, \ldots , S_{\kappa }]$ , but because a voter ranks some candidate c in position i on their ballot, instead candidates are dropped according to a different sequence $A'$ . In Section 4 of the Supplementary Material, I identify two conditions specifying the cases in which a switch from A to $A'$ represents an indirectly pivotal event. By the reasoning there, the probability of c being involved in a potentially pivotal tie with another candidate is the sum of the probability of c being in a tie or a near-tie with each remaining candidate t. So, where y is the index of c in A,

(9) $$ \begin{align} p_{\text{tie}}|A' = \sum_{t \in A_{y+1:\kappa}} \bigg[\frac{1}{2} \ \mathbb{P} \big(v_c^y = v_t^y\big) + \frac{1}{2} \ \mathbb{P} \big(v_c^y = v_t^y - 1\big) \bigg]. \end{align} $$

To obtain the probability of switching from A to $A'$ , it is necessary to know not just the probability that there was a tie to create or break, but also that the end of $A'$ will be the specific sequence that follows creating or breaking that tie involving c. That is the probability that every candidate d in $A^{\prime }_{y+1:\kappa }$ defeats every candidate prior to it. For brevity denote $G \equiv A^{\prime }_{y+1:\kappa }$ , that is, G is the alternate ending in the hypothetical pivotal event. Recall that the candidate that c is in a last-place tie with must be $A^{\prime }_{y}$ , and let $t \equiv A^{\prime }_y$ . Then, by Assumption 1:

(10) $$ \begin{align} p_{\text{tie}}(A'|A) = \overset{\substack{\text{Probability that }G\text{ is the} \\ \text{sequence after a tie with }c}}{\overbrace{\prod_{d = 1}^{|G|} \bigg[ \prod_{h = 1}^{d-1} \mathbb{P} \big( v_{G_{d}}> v_{G_{h}} \big) \bigg]}} \times \overset{\substack{\text{Probability that }c\text{ enters a tie they will win,} \\ \text{or avoids a tie they would not have won}}}{\overbrace{\bigg[\frac{1}{2} \ \mathbb{P} \big(v_c^y = v_t^y\big) + \frac{1}{2} \ \mathbb{P}\big(v_c^y = v_t^y - 1\big) \bigg]}}. \end{align} $$

This is the conditional probability of turning some A into a specific sequence $A'$ . But there is not just one valid $A'$ . Let $\mathbf {A}$ denote the set of all $A'$ that, for a given A, fulfill Conditions 1 and 2 from Section 4 of the Supplementary Material, and note that any two $A'$ represent mutually exclusive events. Then the probability of any indirectly pivotal event arising from the ranking of c in position i given that the drop sequence would otherwise have followed A is the sum of Equation 10 over all possible $A'$ :

(11) $$ \begin{align} p_{\text{tie}}|A = \sum_{A' \in \mathbf{A}} \bigg\{ \prod_{d = 1}^{|G|} \bigg[ \prod_{h = 1}^{d-1} \mathbb{P} \big(v_{G_d}> v_{G_{h}} \big) \bigg] \times \bigg[\frac{1}{2} \ \mathbb{P}\big(v_c^y = v_t^y\big) + \frac{1}{2} \ \mathbb{P}\big(v_c^y = v_t^y - 1\big) \bigg] \bigg\}. \end{align} $$

To obtain the probability of $A'$ arising because of a vote for c, it is now necessary to know the probability that A occurs. By Assumptions 1 and 2, the probability $p_{\neg \text {d}}(A \to A')$ that a vote for c changes the sequence from A to $A'$ is the product of Equation 11 with the probability of A:

(12) $$ \begin{align} \begin{aligned} p_{\neg \text{d}} & (A \to A') = \overset{\text{Probability of }A\text{ occurring}}{\overbrace{\prod_{\ell = 2}^{\kappa} \bigg[ \prod_{h = 1}^{\ell - 1} \mathbb{P} \big(v_{A_\ell}> v_{A_{h}} \big) \bigg]}} \times \\ & \underset{\text{Probability of any }A'\text{ arising from }A\text{, with }c\text{ tied for being dropped at some point}}{\underbrace{\sum_{A' \in \mathbf{A}} \bigg\{ \prod_{d = 1}^{|G|} \bigg[ \prod_{h = 1}^{d-1} \mathbb{P} \big(v_{G_d} > v_{G_{h}} \big) \bigg] \times \bigg[\frac{1}{2} \ \mathbb{P} \big(v_c^y = v_t^y\big) + \frac{1}{2} \ \mathbb{P} \big(v_c^y = v_t^y = -1\big) \bigg] \bigg\}}}. \end{aligned} \end{align} $$

What remains is to sum over the possible sequences A, and also conduct the calculation for every ballot location. The one important detail in the sum over different values of A is that the only sequences that should be included are those which involve dropping every candidate listed before i on the ballot, which we can obtain by summing Equation 12 over all possible A and all ballot locations:

(13) $$ \begin{align} \begin{aligned} p_{\text{indirect}}(\beta) & = \sum_{i = 1}^{L} \bigg( \sum_{\substack{A \in \text{Sym}(C) \\ \beta_{1:i} \subset A_{1:y}}} \bigg[ \prod_{\ell = 2}^{\kappa} \bigg[ \prod_{h = 1}^{\ell - 1} \mathbb{P} \big(v_{A_\ell}> v_{A_{h}} \big) \bigg] \\ & \times \sum_{A' \in \mathbf{A}} \bigg\{ \prod_{d = 1}^{|G|} \bigg[ \prod_{h = 1}^{d-1} \mathbb{P} \big(v_{G_d} > v_{G_{h}} \big) \bigg] \sum_{t \in F} \bigg[\frac{1}{2} \ \mathbb{P} \big(v_c^y = v_t^y\big) + \frac{1}{2} \ \mathbb{P} \big(v_c^y = v_t^y - 1\big) \bigg] \bigg\} \bigg). \end{aligned} \end{align} $$

Finally, substitute Equation 2, and account for making as well as breaking ties, to obtain the full expression for the indirect pivotal probability of $\beta $ :

(14) $$ \begin{align} \begin{aligned} p_{\text{indirect}}(\beta) = \sum_{i = 1}^{L} \bigg( & \sum_{\substack{A \in \text{Sym}(C) \\ \beta_{1:i} \subset A_{1:y}}} \bigg[ \prod_{\ell = 2}^{\kappa} \bigg[ \prod_{h = 1}^{\ell - 1} \mathbb{P} \bigg(\sum_{q=0}^{h-1} \mu_{A_{\ell}}^{q+1} \big \vert A_{1:q} > \sum_{q = 0}^{h-1} \mu_{A_h}^{q+1} \big \vert A_{1:q} \bigg) \bigg] \\& \times \sum_{A' \in \mathbf{A}} \bigg\{ \prod_{d = 1}^{|G|} \bigg[ \prod_{h = 1}^{d-1} \mathbb{P} \bigg(\sum_{q=0}^{y+d} \mu_{G_d}^{q+1} \big \vert A_{1:q} \ > \sum_{q=0}^{y+d} \mu_{G_h}^{q+1} \big \vert A_{1:q} \bigg) \bigg] \\& \times \bigg[\frac{1}{2} \ \mathbb{P} \bigg(\sum_{q=0}^{y} \mu_{c}^{q+1} \big \vert A_{1:q} \ = \sum_{q=0}^{y} \mu_{t}^{q+1} \big \vert A_{1:q}\bigg) + \frac{1}{2} \ \mathbb{P} \bigg(\sum_{q=0}^{y} \mu_{c}^{q+1} \big \vert A_{1:q} = \sum_{q=0}^{y} \mu_{t}^{q+1} \big \vert A_{1:q} - 1 \bigg) \bigg] \bigg\} \bigg] \bigg). \end{aligned} \end{align} $$

6 Full Pivotal Probability

The probability that a ballot $\beta $ is pivotal is Equation 8 (direct pivotal probability) plus Equation 14 (indirect pivotal probability). We can obtain the expected utility, or the probability of causing a pivotal event times the utility obtained from that change, by multiplying by the change in the voter’s utility as a result of the pivotal event. This yields the full expression for expected utility in IRV as a function of the ballots cast:

(15) $$ \begin{align} \begin{aligned} u(\beta) =& \overset{\substack{\text{Sum over} \\ \text{ballot}}}{\overbrace{\sum_{i=1}^{L}}} \bigg( \overset{\substack{\text{Sum over} \\ \text{possible drop} \\ \text{sequences}}}{\overbrace{\sum_{\substack{S \in \\ \text{Sym}(C \,{\backslash}\, i) \\ \beta_{1:i-1} \subset S}}}} \bigg\{\bigg[\overset{\text{Probability of that drop sequence occurring}}{\overbrace{\prod_{\ell = 1}^{\kappa - 2} \prod_{r=\ell+1}^{\kappa} \mathbb{P} \bigg(\sum_{q=0}^{\ell-1} \mu_{A_{\ell}}^{q+1} | A_{1:q} < \sum_{q=0}^{\ell-1} \mu_{A_{r}}^{q+1} | A_{1:q} \bigg)}} \bigg] \times \\ & \bigg[\overset{\substack{\text{Probability of breaking first-place tie} \\ \text{for a candidate who would not have won}}}{\overbrace{\frac{1}{2} \ \mathbb{P} \bigg(\sum_{q=0}^{\kappa-2} \mu_{A_{\kappa-1}}^{q+1} | A_{1:q} = \sum_{q=0}^{\kappa-2} \mu_{i}^{q+1} | A_{1:q}\bigg)}} + \overset{\substack{\text{Probability of creating first-place tie} \\ \text{for a candidate who then wins}}}{\overbrace{\frac{1}{2} \ \mathbb{P} \bigg(\sum_{q=0}^{\kappa-2} \mu_{A_{\kappa-1}}^{q+1} | A_{1:q} = 1 + \sum_{q=0}^{\kappa-2} \mu_{i}^{q+1} | A_{1:q}\bigg)}} \bigg] \bigg\} \times \\ & \overset{\substack{\text{Net utility from causing} \\ \text{that candidate to win}}}{\overbrace{\bigg[u(i) - u(S_{-1})\bigg]}} + \overset{\substack{\text{Sum over eligible} \\ \text{drop sequences}}}{\overbrace{\sum_{\substack{A \in \text{Sym}(C) \\ \beta_{1:i} \subset A_{1:y}}}}} \bigg[ \overset{\text{Probability of that drop sequence occurring}}{\overbrace{\prod_{\ell = 2}^{\kappa} \bigg[ \prod_{h = 1}^{\ell - 1} \mathbb{P} \bigg(\sum_{q=0}^{h-1} \mu_{A_{\ell}}^{q+1} \big \vert A_{1:q} > \sum_{q = 0}^{h-1} \mu_{A_h}^{q+1} \big \vert A_{1:q} \bigg)}} \bigg] \times \\ & \overset{\substack{\text{Sum over alternative} \\ \text{drop sequences}}}{\overbrace{\sum_{A' \in \mathbf{A}}}} \bigg\{\overset{\text{Conditional probability of alternative drop sequence}}{\overbrace{\prod_{d = 1}^{|G|} \bigg[ \prod_{h = 1}^{d-1} \mathbb{P} \bigg(\sum_{q=0}^{y+d} \mu_{G_d}^{q+1} \big \vert A_{1:q} \ > \sum_{q=0}^{y+d} \mu_{G_h}^{q+1} \big \vert A_{1:q} \bigg)}} \bigg] \times \\ & \bigg[\overset{\substack{\text{Probability of breaking first-place tie} \\ \text{for a candidate who would not have won}}}{\overbrace{\frac{1}{2} \ \mathbb{P} \bigg(\sum_{q=0}^{y} \mu_{c}^{q+1} \big \vert A_{1:q} \ = \sum_{q=0}^{y} \mu_{t}^{q+1} \big \vert A_{1:q}\bigg)}} + \overset{\substack{\text{Probability of creating first-place tie} \\ \text{for a candidate who then wins}}}{\overbrace{\frac{1}{2} \ \mathbb{P} \bigg(\sum_{q=0}^{y} \mu_{c}^{q+1} \big \vert A_{1:q} \ = \sum_{q=0}^{y} \mu_{t}^{q+1} \big \vert A_{1:q} - 1 \bigg)}} \bigg] \bigg\} \bigg] \times \\ & \overset{\substack{\text{Net utility from causing} \\ \text{that candidate to win}}}{\overbrace{\bigg[u(A^{\prime}_{-1}) - u(A_{-1})\bigg]}} \bigg) \end{aligned} \end{align} $$

where $\text {Sym}(C)$ is the symmetric group of the set C, L is the length of the ballot $\beta $ such that $L \leq \kappa $ , $\mu _{a}^{b}|S$ denotes the number of voters expected to rank candidate a in any of the ballot positions 1 through b conditional on assigning any higher ballot position to candidates in the set of previously dropped candidates S, y is the index in the list A of the candidate ranked at position i, t is the candidate in $A^{\prime }_y$ , and $\mathbf {A}$ is the set of all ordered lists $A'$ formed by pre-pending $A_{1:y}$ to G, for every G in the symmetric group on the set $C \,{\backslash}\, A_{1:y}$ .

Equation 15 accounts for partially blank ballots or contests where not all candidates can be ranked. If any i in $\beta $ is blank, the probability that candidate ties is vacuously zero, so add zero to the pivotal probability. When $L < \kappa $ , simply sum up to L. Section 5 of the Supplementary Material provides worked examples, §6 provides pseudocode, §7 discusses the caveat that multiple ballots may have the same expected utility, and §8 introduces one idea for modeling probabilities.

Section 9 of the Supplementary Material performs an initial, illustrative comparison between the pivotal probability in IRV versus SMDP of ballots cast by hypothetical electorates with two types of preferences: uniformly distributed preferences, or preferences that follow a power law. These highly stylized simulations produce numbers of very similar orders of magnitudes, suggesting that IRV and SMDP may provide similar strategic opportunities.