How to Read the First 2026 Straw Poll for Choosing the Next UN Secretary-General. A Mathematical Model
By Javier Surasky
Drafting date: 08/01/2026
Originally written in Spanish. This is a ChatGPT translation reviewed by the author.
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Introduction
The first informal round to select the person who will succeed António Guterres at the helm of the United Nations produced a seemingly clear result: at first glance, Grynspan appears to be leading the race, with Rodrigues-Birkett as her main alternative. That is a reasonable conclusion, but an incomplete one, because in selecting the person the Security Council will recommend to the General Assembly for the post of Secretary-General, it is not enough to count how many favorable votes each candidacy receives; what is key is knowing who casts the negative votes.
For the Security Council to formally recommend a candidacy, at least nine favorable votes are required, with no negative vote from any of its five permanent members: China, France, Russia, the United Kingdom, and the United States.
In the first rounds of 2026, as is customary, identical ballots were used to record the positions of both permanent and non-permanent Council members. As a result, we still cannot know how many of the discourage votes received by each candidacy came from permanent members.
That lack of information creates a specific kind of uncertainty, since a candidacy that looks numerically viable may already be facing a possible veto that remains hidden from view.
In this analysis, I seek to show how that uncertainty can be measured without attributing votes to specific States, without inventing probabilities of change between rounds, and without confusing a combinatorial estimate with a political prediction.
What Previous First Rounds Teach Us
Historical precedents show that leading the first round is a strong sign of a candidacy’s viability, but not a guarantee.
In the three best-documented competitive processes since 1996, the candidate who led the first round ultimately became Secretary-General. Kofi Annan led in 1996 with twelve votes of support; Ban Ki-moon did the same in 2006; António Guterres repeated the pattern in 2016. All three maintained their advantage until the Council’s formal recommendation.
The clearest historical exception came in 1981. Prince Sadruddin Aga Khan received nine votes of support, one more than Javier Pérez de Cuéllar, but faced Soviet opposition. Pérez de Cuéllar, despite receiving fewer favorable votes, faced no opposition from any permanent member and was ultimately selected.
That case allows us to draw a first conclusion: relative position matters, but the absence of P5 opposition may be more decisive than leading the raw count.
There is also another relevant set of precedents: in 1996, Amara Essy received eleven votes of support and finished second behind Annan; in 2006, Shashi Tharoor received ten and finished behind Ban Ki-moon; and in 2016, Danilo Türk reached eleven but was surpassed by Guterres. None of the three became Secretary-General, despite having reached or exceeded the threshold of 10 votes of confidence, when another candidacy was ahead in the straw polls.
Thus, reaching ten or eleven votes of support shows that a candidacy is competitive, but if that number does not come together with first place, it is not enough to conclude that the person will be able to overcome discourage votes or avoid a possible block by a permanent member.
This is particularly relevant for interpreting the 2026 process. Grynspan not only received ten votes of support; she also led the round, while Rodrigues-Birkett reached nine and came in second. History favors the person in the lead, although the 1981 precedent requires close attention to the origin of the negative votes.
How Should the Results Be Read?
Each result contains three figures:
- The first corresponds to encourage votes (E), that is, votes of support.
- The second corresponds to discourage votes (D), that is, negative votes.
- The third corresponds to no opinion votes (N), that is, members that did not express a position. This may reflect either their view that they have not yet analyzed the candidacy sufficiently or a neutral stance toward it.
In every case, E + D + N = 15, the number of members of the Security Council.
The first straw poll to choose the person the Security Council will recommend to the General Assembly to lead the Secretariat produced the following result:
| Candidacy | E | D | N | Position |
|---|---|---|---|---|
| Rebeca Grynspan | 10 | 1 | 4 | 1 |
| Carolyn Rodrigues-Birkett | 9 | 2 | 4 | 2 |
| Rafael Mariano Grossi | 7 | 2 | 6 | 3 |
| Michelle Bachelet | 6 | 5 | 4 | 4 |
| Macky Sall | 6 | 7 | 2 | 5 |
| María Fernanda Espinosa | 5 | 2 | 8 | 6 |
| Olara Otunnu | 2 | 5 | 8 | 7 |
Source: 1 for 8 Billion
Thus, Grynspan’s 10-1-4 result means that she received ten votes of support, one discourage vote, and four no opinion votes, while Rodrigues-Birkett’s 9-2-4 means that she received nine votes of support, two discourage votes, and four no opinion votes, and so on.
The first step in the analysis is to determine whether a candidacy reaches the threshold of nine positive votes required to be selected by the Council. That already allows us to calculate a “deficit”:
If the result is zero, the candidacy already reaches or exceeds nine votes of support. If it is greater than zero, it indicates how many additional votes are needed.
Applied to 2026:
- Grynspan: (9 − 10 = −1), so the deficit is 0.
- Rodrigues-Birkett: (9 − 9 = 0).
- Grossi: (9 − 7 = 2).
- Bachelet: (9 − 6 = 3).
- Sall: (9 − 6 = 3).
- Espinosa: (9 − 5 = 4).
- Otunnu: (9 − 2 = 7).
The deficit does not estimate the probability that those votes will materialize. It only shows the arithmetic distance from the threshold.
How Many Neutral Votes Would Each Candidacy Have to Convert?
When a candidacy has not yet reached nine votes of support, it is possible to calculate what share of its no opinion votes would have to become support:
This measure also does not predict that neutral votes will change. It only shows whether the available neutral votes are arithmetically sufficient to cover the deficit.
Let us look at two examples:
Grossi has a deficit of two votes and six neutrals, that is:
He would need to convert at least 33.3% of his neutral votes into support.
She would have to convert half of her neutral votes into support.
Sall is in a different situation. He has a deficit of three votes, but only two neutrals:
The result is 150%, which shows that converting his neutral votes into support would not be enough: he necessarily has to reverse at least one discourage vote.
This calculation makes it possible to distinguish between candidacies that can reach nine votes of support by converting only neutral votes and those that also need to change negative positions.
A Key Unknown: Where the Discourage Votes Come From
Support numbers show one part of the picture, but as I said earlier, the possible origin of the negative votes matters, and matters a great deal: the five permanent members retain their veto power in the decision on the person to be selected for recommendation to the General Assembly.
In a round without differentiated ballots, we know the total number of discourage votes (D), but we do not know how many were cast by permanent members. We therefore call the unknown number of P5 negative votes (X).
The question then becomes the following:
Given that a candidacy received (D) discourage votes, what is the probability that exactly (x) of them were cast by permanent members?
To answer this, we can use a hypergeometric distribution. By definition, this involves choosing several elements from a group without replacement in order to calculate the probability of obtaining a certain number of elements of a particular type. It calculates probabilities by counting how many possible combinations meet a condition within a set when the elements are selected without replacement.
In this case, the operation does not describe how States actually voted. Rather, it measures the uncertainty that arises because we know the total number of negative votes, but not who cast them.
The formula we apply is:
The term 5x counts the ways of selecting x permanent members.
The term 10D − x counts the ways of choosing the remaining negative votes among the ten elected members.
The denominator 15D represents all possible ways of distributing the discourage votes among the fifteen Council members.
The underlying assumption is a criterion of indifference: in the absence of information about who cast the votes, all possible subsets of members are treated as equally probable.
This does not mean that votes are politically random, since States vote according to interests, alliances, and strategies. But it does make clear that the formula does not model those preferences. It only quantifies the uncertainty produced by not knowing who cast each vote.
Probability That No Discourage Vote Comes From a P5
The most favorable scenario for a candidacy is that all its negative votes come from non-permanent members.
The probability that none of the (D) discourage votes comes from a P5 is:
The numerator counts the ways of placing all negative votes among the ten elected members. The denominator represents all the ways of distributing them among the fifteen members.
To explain this operation more simply, consider the following. Starting with 10 non-permanent members and two votes:
- The first member, let us call it “A,” can be paired with 9 others; that is, member “A” can be paired with any other non-permanent member, but not with itself.
- The second, let us call it “B,” can be paired with 8 new ones; we discard the A→B combination, which already appeared in the previous step, because A→B = B→A.
- The third, let us call it “C,” can be paired with 7; we discard the A→C and B→C combinations, which already appeared in previous steps, and so on. This is equivalent to saying 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45.
The same applies to the denominator, except this time we calculate it among the 15 Council members and not only among the 10 non-permanent members = 105.
For Grynspan, who received a single discourage vote, we have:
The combinatorial possibility that her only negative vote came from a non-permanent member is 66.7%. Its complement, 33.3%, is therefore the probability that it was cast by a permanent member.
For Rodrigues-Birkett, who received two discourage votes:
This means there is a 42.9% combinatorial probability that neither of her two negative votes came from a P5 and, by complement, a probability of the same kind that at least one of those votes came from a P5 of 57.1%.
Grossi and Espinosa also received two negative votes. They therefore share the same P5 uncertainty distribution as Rodrigues-Birkett. The difference between them is not in the combinatorial risk of being blocked, but in the number of votes of support obtained.
With two discourage votes, three scenarios are possible:
- Neither comes from a P5.
- Only one comes from a P5, and the other from a non-permanent member.
- Both come from P5 members.
The probability that only one comes from a P5 is:
The probability is 47.6%.
Applying the same operation, the probability that both negative votes come from permanent members is:
That is, 9.50%.
Since the three scenarios—no P5, only one P5, or two P5—must add up to 100%, it follows that the possibility that the negative votes did not come from any P5 is 42.9%; that is:
This breakdown shows that Rodrigues-Birkett already reaches nine votes of support, but faces greater uncertainty than Grynspan over the possible existence of a P5 negative vote.
Why We Use 0.2 to Distribute the Identity of a Block
Suppose we know, or assume, that there is exactly one P5 negative vote, but we do not know which of the five permanent members cast it.
In the absence of additional information, the identity of the vote is distributed symmetrically, so that:
Each permanent member receives the same conditional weight as the others: 20%.
That 0.2 does not represent the probability that a P5 will change its position, nor does it measure the probability that the block will disappear. It simply distributes equally the unknown identity of a single permanent-member discourage vote.
In Grynspan’s case, the probability that her only negative vote came from some P5 is 33.3%. Conditional on that scenario, each P5 receives a weight of 20%, and the unconditional probability assigned to each specific permanent member is:
That is, 6.67%, the same figure we would obtain by dividing 1 by 15.
Under complete symmetry, each of the fifteen members has a probability of 1/15 of having cast that single negative vote.
But if there is more than one negative vote from permanent members, it is no longer appropriate to assign 0.2 to each pair, because the total number of combinations changes: it is no longer 1 out of 5, but 2 specific members out of 5, which we represent as:
In other words, we now have 10 possible pairs, so each combination receives a value of:
With this in mind, we can establish a general rule:
where S is the specific subset of permanent members.
What Does the Calculation Show for Each Candidacy?
Rebeca Grynspan
Grynspan is the only candidacy that exceeds the nine-vote threshold. Her 10-1-4 result combines the highest number of votes of support with the lowest number of discourage votes.
The combinatorial probability that her only negative vote came from an elected member is 66.7%, while the probability that it came from a P5 is 33.3%.
Her advantage is real, but it still does not allow us to say that she has no permanent-member opposition.
Carolyn Rodrigues-Birkett
Rodrigues-Birkett already reaches nine votes of support. However, her two discourage votes create a less favorable scenario than Grynspan’s.
The probability that neither is P5 is 42.9%. The probability that exactly one is P5 is 47.6%. The probability that both come from permanent members is 9.5%.
Her candidacy is competitive, but it faces considerably greater uncertainty over a possible block.
Rafael Mariano Grossi
Grossi received seven votes of support, two discourage votes, and six no opinion votes.
He needs to convert two neutrals to reach nine:
His combinatorial P5 risk is identical to Rodrigues-Birkett’s, because both received two discourage votes. The difference is that Grossi has not yet reached the minimum threshold.
María Fernanda Espinosa
Espinosa obtained five votes of support, two discourage votes, and eight no opinion votes.
She needs four additional votes of support:
Her neutral votes are arithmetically sufficient, but she would have to convert half of them. Her combinatorial P5 risk is the same as that of the other candidacies with two discourage votes.
Michelle Bachelet
Bachelet received six votes of support, five discourage votes, and four no opinion votes.
She needs three additional votes, meaning she must convert three of her four neutrals:
In addition, the combinatorial probability that at least one of her five discourage votes came from a P5 is much higher, reaching 91.6%. The combination of a support deficit and high opposition is her main weakness.
Macky Sall
Sall obtained six votes of support, seven discourage votes, and two no opinion votes.
He needs three additional votes, but has only two neutrals available. Even if he converted both, he would reach eight. He must also reverse at least one negative vote.
The combinatorial probability that at least one of his seven discourage votes came from a P5 rises to 98.1%, the highest among all candidacies.
Olara Otunnu
Otunnu received two votes of support, five discourage votes, and eight no opinion votes.
He needs seven additional votes:
78 = 87.5%
He would have to convert seven of his eight neutrals. At the same time, the combinatorial probability that at least one of his five discourage votes came from a P5 is the same as Bachelet’s: 91.6%.
His initial position is very weak.
What Will Change When Colored Ballots Appear?
Combinatorial uncertainty exists because we still do not know where the negative votes came from. But when the Council uses differentiated ballots to identify votes by permanent and non-permanent members separately, we will know how many discourage votes correspond to the P5 group. At that point, (X) will no longer be an unknown variable, and the hypergeometric distribution will no longer be necessary.
If a candidacy receives zero P5 discourage votes, the absence-of-block component required for the candidate’s election is satisfied and takes the value 1; it does not alter the possibilities produced by the other values. But if there is at least one veto, the condition is not met and the value of the component is 0, which brings the entire operation to 0 as a sign that the candidacy cannot move forward.
However, a zero value does not necessarily mean that the candidacy has been eliminated. It is still necessary to wait for the formal vote, since any State may change its position in a later round or even in the formal vote that follows the straw polls. The values must therefore be recalculated independently using the data that emerge from each round.
What the Model Can and Cannot Say
The calculation makes it possible to establish that Grynspan has the strongest initial position. She is the only candidate with more than nine votes of support and only one discourage vote. Rodrigues-Birkett is the main alternative because she already reaches nine votes, although her two discourage votes increase uncertainty around a possible P5 block.
Grossi has the strongest position among those who have not yet reached nine, because he needs to convert only two of his six neutrals. Espinosa still has room to grow, but she needs to turn half of her no opinion votes into support.
Bachelet, Sall, and Otunnu start from more adverse positions because of the combination of deficits and discourage votes.
The model cannot state that Grynspan has a given probability of being elected. Nor can it identify which States voted against her, or whether they are P5 members, until differentiated ballots are used. Nor can it anticipate whether those States will change their position. The hypergeometric distribution does not model the diplomatic process underlying the Security Council’s selection of a candidacy; it models the lack of information.
That distinction must not be lost, at the risk of misreading the scope of the proposed model: the percentages of 33.3%, 57.1%, or 91.6% do not represent the actual will of the P5. They only show how much the combinatorial possibility that at least one permanent-member negative vote exists increases as the total number of discourage votes grows.
Conclusions
The first 2026 straw poll confirms that the selection of the Secretary-General cannot be interpreted as a competition based exclusively on the accumulation of support, since the Security Council’s selection of a candidacy combines two different and concurrent logics:
- The majority logic, which requires at least nine votes;
- The logic of hierarchical differentiation, which allows any permanent member to block a recommendation.
Grynspan leads because she gathers more support than any other candidacy and faces fewer discourage votes, which in turn reduces the possibility that the discourage vote came from a P5, but does not eliminate it entirely.
Rodrigues-Birkett reaches nine votes, but her two discourage votes increase uncertainty and, with it, the possibility that at least one discourage vote came from a P5.
Grossi and Espinosa do not reach the threshold imposed by the majority logic, although they have enough room to make up that shortfall among the members that expressed no opinion. For them, the risk that one of the discourage votes came from a P5 is the same as the one facing Rodrigues-Birkett, since all three candidacies received two discourage votes. The difference among these three candidacies therefore lies mainly in whether they satisfy the majority logic, not in hierarchical differentiation.
The other candidacies face greater obstacles to achieving both conditions, since they do not reach the nine-vote majority and face comparatively much higher risks than the other candidacies that at least one of their discourage votes came from a permanent member.
The first straw poll tells us who is leading the race. But only once we have the information provided by colored ballots will we be able to come closer to knowing which candidate is in a position to satisfy the double logic required to be appointed Secretary-General of the United Nations.