take a pill, man (pharmaceutical math)

Quote: backprop

Each day, you shake two units out of the bottle. One of the following things can happen:

- You get 2 whole pills. You cut one in half, take your dose, and put the other half pill back in the bottle. (Obviously this will happen on the first day).
- You get 1 whole pill and a half pill and take them
- You get 2 half pills, so you shake another out of the bottle, at which time:
--- You get a half pill and take the dose
--- You get a whole pill and break it in half, taking you dose and depositing the other half back into the bottle.

Quote: estebanrey

*You get a whole pill and break it in half, taking you dose and depositing the other half back into the bottle.

Why would you do that? Surely if you've got two halves then get another whole, the obvious thing would be to put one of the halfs back and just take the whole and half you have left. Why would you break the whole one in two?

Quote: MangoJ

Why would you put the second half part back into the bottle in the first if you will anyway need it the very next day ?

Quote: backprop

I only have one bottle, it's childproof, and I have a toddler.

Quote: backprop

Just bored enough to intellectually challenge myself with something slightly less than earth-shattering, I suppose :)

Quote: dwheatley

The states you could use to do Matrix Algebra are all different. There's no movement between states: you need to individually track the number of whole and half pills in the bottle. I agree, not elegant.

Quote: backprop

For the record, the simulation settles in at around 34.5% having a full/half on the last day, and the remainder with three half pills.

Quote: sodawater

Just take 9 pills at once every 6 days

Quote: backprop

For the record, the simulation settles in at around 34.5% having a full/half on the last day, and the remainder with three half pills.

Quote: MonkeyMonkey

An aspect of this that I haven't seen mentioned is the number of pills that could tumble out of the bottle. What sort of assumption is being made? If you could only ever get 2 where: a) both are whole, b) one is half, one is whole, then we're missing a possible instance where 3 halves could tumble out. if 3 pieces can come out there are a lot of other combinations, some of which would allow you to pick the necessary pieces to make the 1 1/2 you need and wouldn't that potentially skew the result at the end?

Quote: EdCollins

I think you have this backwards.

<script>

var numpills=45;
var numtrials=1000000;
var numdays=30;

// this counter tracks the number of cycles where there is a full/half pill left, vs only halfs
var totalNumFulls=0;

for(i=0;i<numtrials;i++){
// fill up the pill bottle with full pills
  var bottle=new Array();
  for(var p=0;p<numpills;p++){
  	bottle[p]=1;
  }
  
  for(d=0;d<numdays;d++){
  	// keep shaking out pills until we get to or past our 1.5 dosage
  	var dosagefound=false;
  	var pillsout=0;
	
	// if it's the last day, just figure out now if there's a full pill and add it to the tally
	if(d==numdays-1){					
			for(var j=0;j<bottle.length;j++){			   
				if(bottle
==1){
					totalNumFulls++;
				}				
			}
		}
	
	// keep shaking
  	while(!dosagefound){

		// get the first available random pill
		pilltocheck=Math.floor(Math.random()*bottle.length)
		// add it to what's out of the bottle for today.
		pillsout+=bottle[pilltocheck];
		// if you don't have enough, remove that pill from the bottle array and proceed
  		if(pillsout < 1.5){		
  			bottle.splice(pilltocheck,1);
  		}
		// if you have exactly what you need, stop shaking and remove that pill from the bottle array
  		else if(pillsout == 1.5){
  			dosagefound=true;
			bottle.splice(pilltocheck,1);
  		}
		// if you're past what you need, stop shaking and return 0.5 to the bottle position where you got the full
  		else if(pillsout > 1.5){
  			dosagefound=true;
			bottle[pilltocheck]=0.5;
  		}
  	}	
  }
}

document.write("Out of " + numtrials + " trials, " + totalNumFulls + " (" + totalNumFulls*1.0/numtrials*100.0 +  "%) had a full/half while " + (numtrials - totalNumFulls) + " (" + (numtrials - totalNumFulls)*1.0/numtrials*100.0 + "%) had three halfs.")


</script>

Quote: ThatDonGuy

Assuming that the ratio of the probabilities of drawing a whole pill to drawing a half pill is equal to the ratio of whole pills to half pills, I did a brute force recursive/state-mapped solution and got 0.688130417076 for having a whole and a half left, and 0.311869582925 for having three halves.

Quote: ThatDonGuy

Assuming that the ratio of the probabilities of drawing a whole pill to drawing a half pill is equal to the ratio of whole pills to half pills, I did a brute force recursive/state-mapped solution and got 0.688130417076 for having a whole and a half left, and 0.311869582925 for having three halves.

Here's the method:
Let (W,H) be the state with W whole pills and H half pills.
From (W,H):
(1) your first pill is a whole (probability W / (W+H))...
(1a) and your second pill is also whole (probability (W-1) / (W+H-1)); this is state (W-2,H+1)
(1b) and your second pill is a half (probability H / (W+H-1)); this is state (W-1,H-1)
(2) your first pill is a half (probability H / (W+H))...
(2a) and your second pill is whole (probability W / (W+H-1)); this is state (W-1,H-1)
(2b) and your second pill is also a half (probability (H-1) / (W+H-1))...
(2b1) and your third pill is whole (probability W / (W+H-2)); this is state (W-1,H-1) (two halves out, then a whole out, then a half back in)
(2b2) and your third pill is also a half (probability (H-2) / (W+H-2)); this is state (W,H-3)

The initial state is (45,0)
Since you can get from (W,H) to (W, something < H) but not (W, something > H) or (something > W, anything), work from W=45 down to W=0, and for each W, work from H = (45-W)/2 (since 2H + W <= 45) down to H = 0 (you need to work down because, for example, state (0,12) will contribute to state (0,9)).
From state (W,H):
add W / (W+H) x (W-1) / (W+H-1) to state (W-2, H+1);
add W / (W+H) x H / (W+H-1), H / (W-H) x W / (W+H-1), and H / (W+H) x (H-1) / (W+H-1) x W / (W+H-2) to state (W-1,H-1);
add H / (W+H) x (H-1) / (W+H-1) x (H-2) / (W+H-2) to state (W, H-3).
The eventual probabilities for one of each and three halves are in states (1,1) and (0,3), respectively.

Quote: backprop

This is really cool; thanks. I need to get my head around it by trying to attack it the same way. So it's basically recursing through all the (valid) states 'downstream' of (45,0) and then finding the total of (1,1) instances and (0,3) instances and dividing by the total # of states achieved?

Quote: ten2win

You should ask SOOPOO for the answer.

He's a Doctor.

Quote: sodawater

because that's 2 pills per day, not 1.5

Quote: ThatDonGuy

Assuming that the ratio of the probabilities of drawing a whole pill to drawing a half pill is equal to the ratio of whole pills to half pills, I did a brute force recursive/state-mapped solution and got 0.688130417076 for having a whole and a half left, and 0.311869582925 for having three halves.

Here's the method:

.....

<script>

	phalfs=0;
	pwholehalf=0;
	
	function drilldown(wp,hp,p,i){

		
		if (p<=0 || wp<0 || hp<0) return;
		
		if(wp==1 && hp==1){			
			pwholehalf+=p;
			return;
		}
		
		if(wp==0 & hp==3){			
			phalfs+=p;
			return;
		}
		
		// Now figure out the different combinations and probabilities of 'next' states.
		// 1.  next state is wp-2, hp+1
		//whole and whole
		if(wp>=2) drilldown(wp-2,hp+1,p*((wp/(wp+hp))*((wp-1)/(wp+hp-1))),i+1);
		
		
		// 2.  next state is wp-1, hp-1
		// add possibilities of whole/half, half/whole, half/half/whole
		phw=0;
		if(wp>=1 && hp>=1) phw+=(wp/(wp+hp))*(hp/(wp+hp-1))
		if(wp>=1 && hp>=1) phw+=(hp/(wp+hp))*(wp/(wp+hp-1))
		if(hp>=2 && wp>=1) phw+=(hp/(hp+wp)*(hp-1)/(hp+wp-1)*wp/(hp+wp-2));
		if(phw>0) drilldown(wp-1,hp-1,p*phw,i+1);
	
		
		// 3.  next state is hp-3
		//three halfs		
		if(hp>=3) drilldown(wp,hp-3,p*(hp/(wp+hp))*((hp-1)/(wp+hp-1))*((hp-2)/(wp+hp-2)),i+1)
		
	}
	
	drilldown(45,0,1.0,0);
	
	document.write("<br /><br />Probability of whole/half: " + pwholehalf*100.0 + "% <br />Probability of halves: " + phalfs*100.0 + "%");

</script>

Quote: backprop

I'm guessing the small difference is due to floating point operation errors but maybe I have something slightly off.

Quote: backprop

... Here's the script.....

Quote: zippyboy

While you're busy writing a script, why not have the doctor rewrite you a script for a different dosage pill? If you need 150mg per day, and he gave you 100mg dosage, how about getting a script for three 50mg pills rather than breaking the 100s apart? Or maybe there's a single 150mg you could take.

Quote: backprop

I followed this method and got very similar numbers:

Probability of whole/half: 68.81304169264088%
Probability of halves: 31.186958287814676%

I'm guessing the small difference is due to floating point operation errors but maybe I have something slightly off.

Here's the javascript. It was interesting that scenarios up to 36 pills went fairly quickly, but 45 took over an hour (on my old-ish laptop).

I didn't do absolutely pure recursion, as I kept the tallies for the probabilities of (1,1) and (0,3) in global variables. But otherwise it's very simple. Yesterday I missed the half/half/whole probability for about two hours and kept winding up with total probabilities < 100%. If not for that it would have been a 15 minute exercise.

Thanks for the pointers.

var data = [];
var size = 45;
data[size] = [];
data[size][0] = 1;


for (whole = data.length-1; whole >=0; whole--)
{
    if (!data[whole]) continue;
    for (half = data[whole].length-1; half >= 0; half--)
    {
        if (!data[whole][half]) continue;
        // test each possible state for each day;
        
        // if state has >=2 whole pills, can transition via whole/whole 
        if (whole >= 2)
        {
            if (!data[whole-2]) data[whole-2] = [];
            if (!data[whole-2][half+1]) data[whole-2][half+1] = 0;
            data[whole-2][half+1] += data[whole][half] * C(whole,2)/C(whole+half,2);
        }
                
        // if state has >=1 whole and >=1 half, can transition via whole/half
        if ((whole >= 1) && (half >= 1))
        {
            if (!data[whole-1]) data[whole-1] = [];
            if (!data[whole-1][half-1]) data[whole-1][half-1] = 0;
            data[whole-1][half-1] += data[whole][half] * C(whole,1)*C(half,1)/C(whole+half,2);
        }
                
        // if state has >=2 half and >=1 whole, can transition via h/h/whole
        // leads to the same end-state as whole/half
        if ((whole >= 1) && (half >= 2))
        {
            if (!data[whole-1]) data[whole-1] = [];
            if (!data[whole-1][half-1]) data[whole-1][half-1] = 0;
            data[whole-1][half-1] += data[whole][half] * (C(half,2)/C(whole+half,2)) * (C(whole,1)/C(whole+half-2,1));
        }
        
        // if state has >=3 half, can transition via h/h/h
        if ((half >= 3))
        {
            if (!data[whole][half-3]) data[whole][half-3] = 0;
            data[whole][half-3] += data[whole][half] * (C(half,3)/C(whole+half,3));
        }
        
    }
 
}
console.log("p(whole/half) = " +data[1][1]*100.0 + "%; p(half/half/half) = " + data[0][3]*100.0 + "%");

Quote: MathExtremist

I think your recursion is diving too deep and redundantly. I just did an iterative solution (also Javascript. Combinations function, C(n,r), omitted for clarity):

Quote: AcesAndEights

Why in holy hell would you use Javascript for something like this? My brain hurts.

Quote: backprop

That is really cool. It is taking me a few minutes to sink in but I think it's finally clicking. I think I need to write it to actually understand but it makes sense. Thanks!

btw I did javascript because I have effectively notepad and a browser on my netbook where I usually noodle with this stuff :)

Quote: MathExtremist

In that case, definitely check out Cloud9. You can still use Javascript (or Python, Ruby, or PHP) but you also get a proper code editor with syntax highlighting and auto-completion, plus debug mode with breakpoints, state inspection, etc.

www.c9.io

Nobody should ever need to code in Notepad. Not with all the fun stuff on the web. But if you must, at least download something like Notepad++ and Firebug.

Quote: MathExtremist

Because the OP did. The client is always right. :)

I would normally use C# for this kind of thing because I like the built in set manipulators, but I admit to not having learned any useful languages recently. The hardcore coding part of my career is long over.

Quote: ThatDonGuy

If somebody has an arbitrary-precision calculator (and if I crunched the numbers correctly), the probability of having one whole and one half is exactly:

(581103 x 15647406142778942242813) / (4194304 x 625 x 184877 x 37349 x 11191 x 65231)

Note that all of the numbers are relatively prime to each other.