# [FOM] Throwing Darts, Time, and the Infinite

Jeremy Gwiazda jeremygwiazda at gmail.com
Tue Apr 24 12:16:42 EDT 2012

“It's true that conditionalizing on the location of the first dart
gives probability 1 to this event…”

I take it that this is the role time plays. Assume ZFC and CH. Fix a
well-ordering. On Monday, a real is selected. On Friday, a real will
be selected. On Wednesday, you can bet at even odds whether Monday’s
or Friday’s real is further into the well-ordering. Is it really
coherent to take Monday’s? I am no expert on the metaphysics of time,
but I take the over (Friday’s). I think that this indicates that the
issue is, at the least, a ponens/tollens issue, as opposed to Freiling
being “totally right.” (Freiling’s: if the reals don’t know the order,
then not CH vs. my: if CH, then the reals know the order).

Best,
Jeremy

P.S. If we don’t care about the failure of countable additivity, we
can devise an Axiom of Symmetry for the countable case, which Freiling
did not do, but which assigns a finite set to each natural number (use
less than). Each natural number has a 0 chance of selection. Then,
with very slight changes, we can run Freiling's argument and 'prove'
that there is not a well-ordering of the natural numbers. I take this
as further evidence that certain structures do 'know' the order of the
darts/selections.

> I would just add - on the particular point at question here (that is,
> Freiling's claim that "the real number line does not really know which
> dart was thrown first or second" vs. Gwiazda's claim that the order
> that the darts are thrown in really does matter), it seems that
> Freiling is totally right.  The joint distribution of two dart throws
> gives a space in which the event of the first dart hitting an earlier
> number in the ordering than the second is unmeasurable, and so is the
> converse.  It's true that conditionalizing on the location of the
> first dart gives probability 1 to this event, but conditionalizing on
> the location of the second dart gives probability 0 to this event.
>
> If the probabilities are subjective probabilities, then there is no
> privilege to one of the conditionalizations over the other.  If the
> probabilities are objective chances, then whether one of them is
> privileged depends on the metaphysics of time.  Given the most
> plausible view for relativity (on which there is no distinguished time
> axis), neither one will be privileged, and so there's no reason to
> give up on the unmeasurability of this event, and thus no reason to
> think it will almost certainly go one way rather than the other.
>
> Kenny Easwaran
>
> On Wed, Apr 18, 2012 at 4:37 AM, Thomas Forster
> <T.Forster at dpmms.cam.ac.uk> wrote:
>>
>> This is an old chestnut. ?I am taking the liberty of inflicting
>> on listmembers the following email from my colleague Imre leader, with his
>> permission. (slightly edited).
>>
>> Dear Thomas,
>>
>> I was thinking about that thing you told me, that was supposedly
>> against CH: that one bijects $\Re$ with the set of countable ordinals
>> and then throws one dart and then another at the board.
>>
>> My reply then was (correctly) that this was just silly, as it confuses
>> what conditional probability means (one cannot condition on an event of
>> zero probability, as in the phrase given that I throw $\alpha$'), and
>> also it forgets that not all sets are measurable.
>>
>> That reply was entirely correct, but I have had two further thoughts. The
>> first one is: the fact that the event second dart beats first' is in
>> fact one of the absolutely most standard examples of a non-measurable set.
>> Indeed (assuming CH) one takes the subset of $[0,1] \times [0,1]$ given by
>> those points $\tuple{x,y}$ for which $x<y$ in the ordering induced from
>> $\omega_1$. Then each row is countable but each column is cocountable!
>>
>> My second thought is more important. It is that, even ignoring the
>> fact that the paradox' is rubbish because of conditional probability
>> and nonmeasurable sets, even then, it is {\bf not} against CH. What I
>> mean is, I will hereby run the {\bf exact} same paradox without any CH
>> assumption. Ready? Here goes \ldots
>>
>> Let $\kappa$ be the least cardinality of a set of positive measure. Let
>> such a set be $A$, and let us well-order $A$ in such a way that all
>> initial segments are smaller than kappa. [he means $\kappa$-like]. Note
>> that all initial segments have measure zero, by definition of $\kappa$.
>>
>> OK, our experiment is: throw a dart at the set $A$. Twice. [Note: as $A$
>> has positive measure, we can do this by throwing a dart at $[0,1]$ and only
>> counting it when it lands in $A$.] Then all the paradox still applies.
>>
>> Conclusion: there is no way, not even intuitively or anything, that this
>> paradox' has anything to do with CH.
>>
>> Imre
>>
>>
>> On Tue, 17 Apr 2012, Jeremy Gwiazda wrote:
>>
>>> Hello,
>>>
>>> Chris Freiling?s Axioms of Symmetry have, I believe, been discussed on
>>>
>>> FOM at least twice. In ?Axioms of Symmetry: Throwing Darts at the Real
>>> Number Line?, Freiling considers two darts thrown at [0, 1]. He
>>>
>>> writes, ?the real number line does not really know which dart was
>>> thrown first or second?, which leads to one of his axioms of symmetry.
>>>
>>> In a recently published paper, I suggest that a well-ordering of [0,
>>> 1] does know the order of the darts under certain assumptions. Fix a
>>> well-ordering of [0, 1]. Let r1 be the real hit by the first dart.
>>> Then assuming ZFC and CH, there are only countably many reals less
>>> than r1 in the well-ordering. Thus with probability 1 the second dart
>>> hits a real greater than r1 in the well-ordering. (Put again slightly
>>> differently: working in ZFC, Freiling demonstrates that assuming that
>>> the reals can?t tell the order of the darts proves not CH; I argue

>>>
>>> that assuming CH means that a well-ordering of [0, 1] can tell the
>>> order of the darts.) I go on to create a puzzle using special
>>> relativity. In case it is of interest, the paper is here:
>>>
>>>
>>>
>>> An earlier, countable version of a similar puzzle is available here:
>>>