<p>Completeness: given any pair, I have a preference, I can make a choice.</p>

<p>If had to choose between marrying Rachel and Monica, I would go for Rachel. Good looks, fun, etc.</p>

<p>If had to choose between marrying Chandler and Rachel, I would go for Chandler. Corny sense of humor but aware of it, even temperament, etc.</p>

<p>If had to choose between marrying Monica and Chandler, I would go for Monica. Loyal, very diligent etc.</p>

<p>Wait, if I had to choose between Rachel, Monica and Chandler, who would I marry? (what is my ordering, who is my 1st choice)</p>

CC BY-SA 4.0
2019-10-30T20:27:16.860Z
To add on to this, I think the source of John's confusion may be the fact that complete non-transitive preferences, when they exist, imply that preferences are being conditioned on the set of attainable alternatives. This doesn't mean that preferences aren't complete, but it does mean that you won't be able to create a utility function.
H Rogers
CC BY-SA 4.0
2019-10-30T20:29:24.450Z
@HRogers What you write seems interesting, but I have never heard of preferences being conditioned on a set of alternatives so I am afraid I do not understand.
Giskard
CC BY-SA 4.0
2019-10-30T20:34:52.167Z
Take your example above: Who would you choose if you could have your pick between Rachel, Monica, or Chandler? Since your preferences are complete it must be the case that you can pick someone (or else declare indifference). Let's say you pick Rachel. This implies that you have a weak preference for Rachel over Chandler, which contradicts your earlier decision to marry Chandler over Rachel back when Monica wasn't an option. So in effect, having Monica as an option altered your preferences between Chandler and Rachel (as a side note this is equivalent to violating the independence axiom).
H Rogers
CC BY-SA 4.0
2019-10-30T20:36:17.477Z
@HRogers I get why this violates IIA, but I don't understand the general concept of conditioning. Is it like an ordering over a subset? Can you recommend reading material?
Giskard
CC BY-SA 4.0
2019-10-30T20:40:51.373Z
All I meant was that preferences change depending on what the set of alternatives are, in the same way that if you have two variables X and Y such that $(Y|X) \neq Y$, you could say that Y is being conditioned on X.
H Rogers
CC BY-SA 4.0
2019-10-30T20:42:46.173Z
@HRogers Seems to me like the ordering may change while preferences remain the same.
Giskard
CC BY-SA 4.0
2019-10-30T20:43:38.250Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/100490/discussion-between-h-rogers-and-giskard).
H Rogers
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2019-10-31T04:14:20.983Z
@Giskard See my answer to this question. Let me know your thoughts
EconJohn
CC BY-SA 4.0
2019-10-30T20:18:13.327Z
2019-10-30T20:18:13.327Z
Giskard
<p>Here's an example of an incomplete but transitive preference.</p>

<p>Consider three fruits, an apple (<span class="math-container">$A$</span>), a banana (<span class="math-container">$B$</span>), and a coconut (<span class="math-container">$C$</span>). I cannot choose between individual fruits, i.e. I don't have a preference over <span class="math-container">$A$</span>, <span class="math-container">$B$</span>, or <span class="math-container">$C$</span> --- not that I'm indifferent between them, I just can't compare them. However, I do prefer more variety to less, namely, I'd choose a bundle with two fruits, e.g. <span class="math-container">$\{A,B\}$</span> over a bundle with only one fruit, say <span class="math-container">$\{C\}$</span>. </p>

<p>It's easy to see that my preference over bundles of fruits (the powerset of <span class="math-container">$\{A,B,C\}$</span>) is <strong>incomplete</strong>, since it is not defined over singletons. However, my preference is still <strong>transitive</strong>, in that whenever bundle <span class="math-container">$X$</span> is preferred to <span class="math-container">$Y$</span> and <span class="math-container">$Y$</span> preferred to <span class="math-container">$Z$</span>, <span class="math-container">$X$</span> is preferred to <span class="math-container">$Z$</span>, for example, <span class="math-container">$\{A,B,C\}\succ\{B,C\}\succ\{C\}\;\Rightarrow\;\{A,B,C\}\succ\{C\}$</span>. </p>

CC BY-SA 4.0
2019-10-30T20:26:12.213Z
@Giskard: Oh... I read the text "transitivity requires completeness" and thought to provide a counterexample.
Herr K.
CC BY-SA 4.0
2019-10-30T20:35:57.583Z
This answer is important and correct based on a typo in the original question. Thank you
EconJohn
CC BY-SA 4.0
2019-10-30T20:22:14.543Z
2019-10-30T20:22:14.543Z
Herr K.
<p>It may be helpful to consider a phenomenon from politics called the <a href="https://en.wikipedia.org/wiki/Condorcet_paradox" rel="nofollow noreferrer">Condorcet paradox</a>. This is a situation that can happen in votes, in which the overall population would vote for A over B, would vote for B over C, and would vote for C over A. It is not a purely theoretical possiblity: <a href="https://blogs.lse.ac.uk/politicsandpolicy/brexit-condorcet/" rel="nofollow noreferrer">it is the present reality in the UK over the best resolution to the Brexit crisis</a>. Either link gives a good explanation of how the situation can arise, but for completeness the simplest example has three voters and three candidates with preferences as follows:</p>

<pre><code>1: A &gt; B &gt; C
2: B &gt; C &gt; A
3: C &gt; A &gt; B
</code></pre>

<p>When taken to a vote, 2/3 would prefer A to B, 2/3 would prefer B to C, and 2/3 would prefer C to A. Transitivity would require that A be prefered over C. Therefore the preference of the group (although not the individuals) is complete and non-transitive. </p>

CC BY-SA 4.0
2019-10-31T15:17:01.620Z
This is a nice post, but unfortunately there was a mix-up at deltapoll. That is, they got the data wrong, and this is not an example of a Condorcet cycle: https://twitter.com/deltapolluk/status/1067787089360637952?lang=en
Bayesian
CC BY-SA 4.0
2019-10-31T12:18:55.680Z
2019-10-31T12:18:55.680Z
Josiah
<p>From what I understand from @Giskard's answer is that the fact that we have comparablity of items over sets of 2, does not necessarily imply the fact that we have comparablity over sets of 3.</p>

<p>Consider the case where we are considering bundles <span class="math-container">$X=\{x_1,x_2\}$</span>,<span class="math-container">$Y=\{y_1,y_2\}$</span> and <span class="math-container">$Z=\{z_1,z_2\}$</span>. Suppose now we have for the pairwise comparison between bundles:
<span class="math-container">$$(1)\ \{X,Y\}\in A,\ \ \ Y\in C(A)$$</span>
<span class="math-container">$$(2) \ \{Y,Z\}\in B,\ \ \ Z\in C(B)$$</span>
<span class="math-container">$$(3) \ \{Z,X\}\in D,\ \ \ X\in C(D)$$</span></p>

<p>Where <span class="math-container">$C(\cdot)$</span> is the choice in environment <span class="math-container">$A,B,D$</span> ect.</p>

<p>We have shown that we in fact do have completeness of preferences since we can compare any two bundles. However, transitivity is not established because given the above results we cannot say for <span class="math-container">$ \{X,Y,Z\}\in Q$</span> there is a choice that follows since we have (3) as our source of irrationality fudging the transitivity of preferences.</p>

<p>Graphically we have the following situation:</p>

<p><a href="https://i.sstatic.net/hKRgx.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/hKRgx.png" alt=""></a></p>

<p><sub></sub></p>

CC BY-SA 4.0
2019-10-31T10:03:56.077Z
My understanding is that we are talking about binary relations, hence completeness will always be about whether a relation exists between two bundles. A preference relation is complete "over 3 bundles" if it is complete for all pairs, where pairs are selected from the three bundles. What you seem to be talking about is not completeness, but an order. A complete (and reflexive...) relation can order any 2 bundles, but without transitivity there may be no order over larger sets.
Giskard
CC BY-SA 4.0
2019-10-30T21:25:45.760Z
2019-11-01T14:12:19.260Z
2019-11-01T14:12:19.260Z
EconJohn
<p>I'm quite surprised nobody has picked the obvious one:</p>

<ul>
<li>I prefer rock over scissors.</li>
<li>I prefer scissors over paper.</li>
<li>I prefer paper over rock.</li>
</ul>

<p>Complete, definitely not transitive.</p>

CC BY-SA 4.0
2019-10-31T15:03:32.470Z
2019-10-31T15:03:32.470Z
R.. GitHub STOP HELPING ICE
<p>I've been doing my own reading on non-rational preference relations.</p>
<p>I'm currently under the impression that transitivity follows as a direct result of completeness of preferences. However my (much more advanced colleagues) have told me this is not necessarily the case.</p>
<p>I'm still to understand how this can be. What would such preferences look like?</p>

CC BY-SA 4.0
2019-10-30T20:27:14.200Z
As we have discovered after answering your question, the body and the title refer to different kinds of preference relations. (See answers for details.)
Giskard
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2019-10-30T20:31:23.570Z
@Giskard whoops i miswrote. Editing now
EconJohn
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2019-10-30T20:33:02.680Z
Please at least make a mention of the typo, so you do not exclude either answer.
Giskard
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2019-10-30T20:37:48.457Z
9 comments and 4 upvotes (in total Q&As), pretty good traffic for a question with 6 views!
Giskard
CC BY-SA 4.0
2019-10-30T20:39:54.583Z
@Giskard im glad to see that others have also thought about this obscure question!
EconJohn
CC BY-SA 4.0
2019-10-31T21:19:42.343Z
[Nontransitive dice](https://en.wikipedia.org/wiki/Nontransitive_dice) lead to a lot of interesting answers to this because they provide such nice concrete probabilities to crunch.
Cort Ammon
<microeconomics><mathematical-economics>
Is it possible to have a preference relation that is complete but not transitive?
EconJohn
# Is it possible to have a preference relation that is complete but not transitive?

(Asked by: username_0 on 2019-10-30)

I've been doing my own reading on non-rational preference relations.

I'm currently under the impression that transitivity follows as a direct result of completeness of preferences. However my (much more advanced colleagues) have told me this is not necessarily the case.

I'm still to understand how this can be. What would such preferences look like?

**Tags:** microeconomics, mathematical-economics

> **Comments:**
> **username_1**: As we have discovered after answering your question, the body and the title refer to different kinds of preference relations. (See answers for details.)  
> **username_0**: @username_1 whoops i miswrote. Editing now  
> **username_1**: Please at least make a mention of the typo, so you do not exclude either answer.  
> **username_1**: 9 comments and 4 upvotes (in total Q&As), pretty good traffic for a question with 6 views!  
> **username_0**: @username_1 im glad to see that others have also thought about this obscure question!  
> **username_2**: [Nontransitive dice](https://en.wikipedia.org/wiki/Nontransitive_dice) lead to a lot of interesting answers to this because they provide such nice concrete probabilities to crunch.  

---

## 5 answers

### Answer by username_7 on 2019-10-31. Score: 11 | Accepted answer

I'm quite surprised nobody has picked the obvious one:

* I prefer rock over scissors.
* I prefer scissors over paper.
* I prefer paper over rock.

Complete, definitely not transitive.


---

### Answer by username_1 on 2019-10-30. Score: 6

Completeness: given any pair, I have a preference, I can make a choice.

If had to choose between marrying Rachel and Monica, I would go for Rachel. Good looks, fun, etc.

If had to choose between marrying Chandler and Rachel, I would go for Chandler. Corny sense of humor but aware of it, even temperament, etc.

If had to choose between marrying Monica and Chandler, I would go for Monica. Loyal, very diligent etc.

Wait, if I had to choose between Rachel, Monica and Chandler, who would I marry? (what is my ordering, who is my 1st choice)

> **Comments:**
> **username_3**: To add on to this, I think the source of John's confusion may be the fact that complete non-transitive preferences, when they exist, imply that preferences are being conditioned on the set of attainable alternatives. This doesn't mean that preferences aren't complete, but it does mean that you won't be able to create a utility function.  
> **username_1**: @HRogers What you write seems interesting, but I have never heard of preferences being conditioned on a set of alternatives so I am afraid I do not understand.  
> **username_3**: Take your example above: Who would you choose if you could have your pick between Rachel, Monica, or Chandler? Since your preferences are complete it must be the case that you can pick someone (or else declare indifference). Let's say you pick Rachel. This implies that you have a weak preference for Rachel over Chandler, which contradicts your earlier decision to marry Chandler over Rachel back when Monica wasn't an option. So in effect, having Monica as an option altered your preferences between Chandler and Rachel (as a side note this is equivalent to violating the independence axiom).  
> **username_1**: @HRogers I get why this violates IIA, but I don't understand the general concept of conditioning. Is it like an ordering over a subset? Can you recommend reading material?  
> **username_3**: All I meant was that preferences change depending on what the set of alternatives are, in the same way that if you have two variables X and Y such that $(Y|X) \neq Y$, you could say that Y is being conditioned on X.  
> **username_1**: @HRogers Seems to me like the ordering may change while preferences remain the same.  
> **username_3**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/100490/discussion-between-h-rogers-and-giskard).  
> **username_0**: @username_1 See my answer to this question. Let me know your thoughts  

---

### Answer by username_4 on 2019-10-30. Score: 4

Here's an example of an incomplete but transitive preference.

Consider three fruits, an apple ($A$), a banana ($B$), and a coconut ($C$). I cannot choose between individual fruits, i.e. I don't have a preference over $A$, $B$, or $C$ --- not that I'm indifferent between them, I just can't compare them. However, I do prefer more variety to less, namely, I'd choose a bundle with two fruits, e.g. $\{A,B\}$ over a bundle with only one fruit, say $\{C\}$.

It's easy to see that my preference over bundles of fruits (the powerset of $\{A,B,C\}$) is **incomplete**, since it is not defined over singletons. However, my preference is still **transitive**, in that whenever bundle $X$ is preferred to $Y$ and $Y$ preferred to $Z$, $X$ is preferred to $Z$, for example, $\{A,B,C\}\succ\{B,C\}\succ\{C\}\;\Rightarrow\;\{A,B,C\}\succ\{C\}$.

> **Comments:**
> **username_4**: @username_1: Oh... I read the text "transitivity requires completeness" and thought to provide a counterexample.  
> **username_0**: This answer is important and correct based on a typo in the original question. Thank you  

---

### Answer by username_5 on 2019-10-31. Score: 1

It may be helpful to consider a phenomenon from politics called the Condorcet paradox. This is a situation that can happen in votes, in which the overall population would vote for A over B, would vote for B over C, and would vote for C over A. It is not a purely theoretical possiblity: it is the present reality in the UK over the best resolution to the Brexit crisis. Either link gives a good explanation of how the situation can arise, but for completeness the simplest example has three voters and three candidates with preferences as follows:
1: A > B > C
2: B > C > A
3: C > A > B

When taken to a vote, 2/3 would prefer A to B, 2/3 would prefer B to C, and 2/3 would prefer C to A. Transitivity would require that A be prefered over C. Therefore the preference of the group (although not the individuals) is complete and non-transitive.

> **Comments:**
> **username_6**: This is a nice post, but unfortunately there was a mix-up at deltapoll. That is, they got the data wrong, and this is not an example of a Condorcet cycle: https://twitter.com/deltapolluk/status/1067787089360637952?lang=en  

---

### Answer by username_0 on 2019-10-30. Score: 0

From what I understand from @username_1's answer is that the fact that we have comparablity of items over sets of 2, does not necessarily imply the fact that we have comparablity over sets of 3.

Consider the case where we are considering bundles $X=\{x\_1,x\_2\}$,$Y=\{y\_1,y\_2\}$ and $Z=\{z\_1,z\_2\}$. Suppose now we have for the pairwise comparison between bundles:
$$(1)\ \{X,Y\}\in A,\ \ \ Y\in C(A)$$
$$(2) \ \{Y,Z\}\in B,\ \ \ Z\in C(B)$$
$$(3) \ \{Z,X\}\in D,\ \ \ X\in C(D)$$

Where $C(\cdot)$ is the choice in environment $A,B,D$ ect.

We have shown that we in fact do have completeness of preferences since we can compare any two bundles. However, transitivity is not established because given the above results we cannot say for $ \{X,Y,Z\}\in Q$ there is a choice that follows since we have (3) as our source of irrationality fudging the transitivity of preferences.

Graphically we have the following situation:

[![](https://i.sstatic.net/hKRgx.png)](https://i.sstatic.net/hKRgx.png)

> **Comments:**
> **username_1**: My understanding is that we are talking about binary relations, hence completeness will always be about whether a relation exists between two bundles. A preference relation is complete "over 3 bundles" if it is complete for all pairs, where pairs are selected from the three bundles. What you seem to be talking about is not completeness, but an order. A complete (and reflexive...) relation can order any 2 bundles, but without transitivity there may be no order over larger sets.  

---

