On Santamaths (An xmaths conversation with a 5 year old)

OK, let's get straight to the point: it is a fallacy, or to put it another way, a gross oversimplification to think that Santa will deem you, in binary fashion, to have been either naughty or nice, discretely, categorically, one way or the other, and that this will in turn delineate whether you receive presents or not.  This is — plainly, to put it politely, and as I will show in what follows — utter horse nonsense.  There are gradations of naughtiness or niceness: one person can be naughtier than another, or nicer, and it is evident to boot that naughtiness is a function of niceness, and vice-versa.  To put it another way, an increase in naughtiness (say η) and a decrease in niceness (say η) are biconditional:

\[\rlap{-} \eta \;\; \Leftrightarrow \;\;\eta \]

To elaborate: Sometimes you will be naughty and sometimes you will be nice. And of course the more you are naughty the less time you have to be nice — and vice-versa: the more you are nice the less time you have to be naughty.  So, and in other words, the naughtier you are the less likely it is that you will be nice, and the nicer you are the less likely it is that you will be naughty.  Yes?  The truth is that two naughty people will be naughty to different degrees: one will be more or less naughty than the other.  (The same is obviously true for two nice people.)  And of course, one naughty act might not be as naughty as another, so there are gradations of naughtiness and, indeed, gradations of niceness.  It is nice to say please and thank you, for example, but it is nicer to slot a pound into a Noddy car ride in the shopping centre and leave it to be discovered by an unsuspecting — but soon to be overjoyed — toddler.  Similarly, it is naughty to take your sister's cuddly Pooh bear when she doesn't want you to, but it is naughtier to hide Pooh and tell your sister that he’s run away because he doesn’t love her her anymore.

Santa is not stupid; he knows all of this.  And he uses Santamaths to deal with the inherent variability in η and hence η.  Put simply, he uses relative frequencies to assign a weighted 'niceness index' (known as the η_i — pronounced /niː/) to each person (𝓍), and the presents each person receives are thus dependent on such.  In short:

\[\begin{array}{l}\forall x,\;0 \le {\eta _i} = \frac{\eta }{{\eta  + \rlap{-} \eta }} \le 1\\\\and\;as\;{\eta _i} \to 1,\;\forall x\left( {\rlap{-} \eta  \Rightarrow \eta } \right),\\and\;P\left( {x\;gets\;what\;s/he\;wants} \right) \to 1\\\\Similarly,\;as\;{\eta _i} \to 0,\;\forall x\left( {\eta  \Rightarrow \rlap{-} \eta } \right),\\and\;P\left( {x\;gets\;what\;s/he\;wants} \right) \to 0\end{array}\]
Santa then uses probability theory and combinatorics to calculate what and how many presents you get.  As such, this year, as you know, I have been keeping a tally of all the times you have done something naughty (η) and all of the times you have done something nice (η).  You get one scratch for a naughty act and one for a nice act, an act of kindness if you will, and dependent of course on the act.  Remember that time you fooled your sister —  your type 1 hypersensitive to tomatoes sister — into thinking that ketchup you had mixed in her ice-cream was actually raspberry juice, well that got you ten scratches in the naughty column, for example, whilst tidying up your Lego without being asked that one time you did it in the whole year got you five scratches in the nice column.  

Here's a little section of the tallies I've been keeping for this year:

Overall you had 740 naughty scratches this year and 623 nice scratches, giving η_i = 0.46:

\[\begin{array}{l}\begin{align}\frac{{623}}{{740 + 623}} &= \frac{{623}}{{1363}}\\\\ &= 0.46\end{align}\end{array}\]
Now, this doesn't mean that you have a 46% chance of getting your presents this year.  Oh no.  You wrote to Santa to ask for the following five presents: 

• An umbrella in the shape of an Octopus, 
• A Lego Batman car [I presume you mean the Batmobile], 
• A green and orange ukulele, 
• A dinosaur that can really move and roar and scare people like a monkey on a horse, 
• A helicopter (remote-controlled will do if Santa can't source a real one for you). 

Hence the following Santamaths model applies:

\[\begin{array}{l}{\left[ {{\eta _i} + \left( {1 - {\eta _i}} \right)} \right]^5} = {}^5{C_5}{\left( {{\eta _i}} \right)^5}{\left( {1 - {\eta _i}} \right)^0} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad {}^5{C_4}{\left( {{\eta _i}} \right)^4}{\left( {1 - {\eta _i}} \right)^1} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {}^5{C_3}{\left( {{\eta _i}} \right)^3}{\left( {1 - {\eta _i}} \right)^2} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {}^5{C_2}{\left( {{\eta _i}} \right)^2}{\left( {1 - {\eta _i}} \right)^3} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {}^5{C_1}{\left( {{\eta _i}} \right)^1}{\left( {1 - {\eta _i}} \right)^4} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {}^5{C_0}{\left( {{\eta _i}} \right)^0}{\left( {1 - {\eta _i}} \right)^5}\end{array}\]
and inputting the respective values for η_i and 1 – η_i...

\[\begin{array}{l}{\left[ {{\eta _i} + \left( {1 - {\eta _i}} \right)} \right]^5} = 1{\left( {0.46} \right)^5}{\left( {0.54} \right)^0} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad 5{\left( {0.46} \right)^4}{\left( {0.54} \right)^1} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 10{\left( {0.46} \right)^3}{\left( {0.54} \right)^2} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 10{\left( {0.46} \right)^2}{\left( {0.54} \right)^3} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 5{\left( {0.46} \right)^1}{\left( {0.54} \right)^4} + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 1{\left( {0.46} \right)^0}{\left( {0.54} \right)^5}\\\quad \quad \quad \quad \quad \quad \quad  = 0.0205... + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad 0.1208... + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0.2838... + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0.3332... + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0.1955... + \\\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0.0459...\end{array}\]
So as you can see, there is therefore only a 2.06% chance that you will receive all the presents you have asked Santa for this year — indeed, it is more than twice as likely that you will receive nothing.  The most likely thing to happen is that you will receive 2 of the 5 presents you have asked Santa for, but there is more than a 50% chance that you will receive 2 or 3 of the presents you have asked for. 

I am sharing this analysis in the detail I am with you so as to prepare you for any possible disappointment on Christmas morning.  I must of course emphasise that this is a purely theoretical model and thus that the experimental error that is life may get in the way, and result in you having all of the presents you have asked for, or, indeed, even more.  (Your sister, incidentally, has asked for 4 presents.  I wonder what proportion of her behaviour would need to have been deemed nice in order for her to have a 90% or more chance of receiving at least 3 of them?)

Night-night.