1. What is the probability of there being a single molecule of active ingredient in a 30C homeopathic remedy? Assume no contamination and pure water.

Molar mass of water = 18.01528 grams.

1 drop of water approx = 0.05 grams.

=> 1 drop of water = 0.05/18 = 0.0027 moles

=> 1 drop of water = 6 * 10^23 (Avogadro constant) * 0.0027 = 1.66*10

^{21}molecules

Number of molecules in a 30C preparation:

= 1.7x10

^{21}(molecules in a drop) / 100

^{30}(level of dilution) = 1.7*10

^{-39}

As this is less than one, it is the probability of finding a molecule.

Answer: 1.7*10

^{-39}

2. What is the probability of winning the National Lottery 5 weeks in a row. You get one ticket per week of course.

There are 6 balls drawn out of 49 and you can pick 6 numbers.

Probability that your 1st number will match any one of the 6 drawn = 6/49

Probability that your 2nd number will match any one of the remaining 5 drawn = 5/48

Probability that your 3rd number will match any one of the remaining 4 drawn = 4/47

Probability that your 4th number will match any one of the remaining 3 drawn = 3/46

Probability that your 5th number will match any one of the remaining 2 drawn = 2/45

Probability that your 6th number will match any one of the remaining 1 drawn = 1/44

Probability that all 6 match = 6/49 * 5/48 * 4/47 * 3/46 * 2/45 * 1/44 = 7.15 x 10-8

Or, about 1 in 14 million.

Probability of winning 5 weeks in a row = (7.15x10

^{-8})

^{5}= 1.87 x 10

^{-36}

3. Which is more likely?

Comparing 1.7*10

^{-39}and 1.87 x 10

^{-36}, you can see that winning the national lottery 5 times in a row is about 1,000 times more likely.

Unless of course, there was some contamination during the process.

4. How much would it cost to buy enough pills that you'd expect to have one molecule of active ingredient? Note: At Boots, they're about £5 for 84 pillules.

To get one molecule, you'd need 100

^{30}water molecules (10

^{60}).

Molar mass of water = 18 grams.

10

^{60}molecules would therefore weigh 18x10

^{60}/6x10

^{23}= 3.0 × 10

^{37}grams

One drop of water weighs approx 0.05 grams

So you'd need 3.0 × 10

^{37}/ 0.05 = 6 x 10

^{38}pills

At £5 for 84, that will cost 5/84 * 6 x 10

^{38}= £3.57 * 10

^{37}

A Trillion is 10

^{12}. Trillion Trillion Trillion = 10

^{36}

So that's about 35 Trillion Trillion Trillion pounds.

Other Notes

David P made a good point when he pointed out that the mother tincture may not be pure. Obviously the mathematics above assumes it is.

Another point I realised while I was doing this is that there is another argument here against the "memory of water" bollocks" spouted by homeopaths.

At 12C, there are no molecules of the mother tincture remaining. So the only molecules in the final preperation that could have been in the same mix as any of the original ingredient must come out of this mix.

So by the time you get to 24C, there are not only no molecules of original ingredient left, but there are no molecules of water that have ever been in the same mix as the original ingredient left.

Even the water has memory argument therefore breaks down at dilutions beyond 24C. At best, homeopaths would need to argue that water not only has memory, but that water is capable of passing this memory on.

Are homeopaths going to come to the conclusion that a 30C remedy works because the water contains the memory of a rumour passed on by other water molecule that remembers the mother tincture?

## 8 comments:

"Number of molecules in a 30C preparation:

= 1.7x1021 (molecules in a drop) / 10030 (level of dilution) = 1.7*10-39

As this is less than zero, it is also the probability."

1.7*10^-39 is very almost zero, but is still infinitesimally minutely bigger than zero. Not enough bigger to make homeopathy remotely believable though....

@Anne Fay.

Whoops! I meant to say less than one, not zero. Changed it now.

Scott Hurst has just emailed me with a couple of points:

To accurately figure what you want to figure, you must realize that it would be different for each and every ‘remedy’. You are interested in the molar mass of what you diluting, not the solvent. You also need to specify how the mother tincture is prepared.

A better solution would look like:

Mother tincture is X ml of solvent with Y grams of REMEDY dissolved in it.

Figure the number of molecules of active remedy: Molar mass of REMEDY / grams of Y * molecules per mole = Z (molecules of remedy)

Figure the number of molecules per X ml of final solution: Z / 100^30 (for 30C) = F (molecules of remedy per unit X of final solvent)

Figure the number of molecules per pill: volume of solution per pill P ml / X ml = R (molecules of remedy per pill)

If you dissolve the same number of grams of Lithium into one mother tincture as grams of Uranium into another, the odds of finding an atom of either in their final solutions, prepared the same way would vary factor of about 35.

This should work whether your solvent is water, alcohol, benzene, you name it….. What say you?

So my interpretation of Scott's message is that:

(a) My calculations assume the mother tincture is 100% pure.

(b) My calculations assume that the molar mass of any chemical used my the homeopaths has the same molar mass as water.

So I guess my calculations are more of a best-case scenario but the exact quantities will vary.

Still, at least it would be cheap to make, I mean the cost of the ingredients would be negligable

I know you see these all the time but I fell over this today http://www.huffingtonpost.com/dana-ullman/how-homeopathic-medicines_b_389146.html Aargh.

I'm sure they will think of something- don't forget they are already assuming that the water passes this memory onto the lactose in the pillule.

Hey! Thanks for this great example! Forgive me if I missed this, but could you please explain where the number 1.7 (molecules in a drop) came from?

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