Thu 02 August 2012 - 12:52:00 CDT
A few weeks ago the big news about the Higgs boson came out, and I read an interesting article that talked about the experimental process. Now, I am not a particle physicist. If I’m any kind of physicist, I am a laser physicist—and yes, I know, photons are particles, but I deal more with the wave model. Anyway, more than the particles, what caught my attention right at the beginning of the article was the discussion about probabilities.
A lot of the discussion about Higgs also talks about matter vs. anti-matter and why we exist at all. If a particular particle has an even chance of decaying into, say, a proton and an anti-proton (and some other stuff), and when a proton and anti-proton meet they annihilate, why do we have protons?
This raises the probability question that goes back to the apocryphal fair coin: if you flip a fair coin ten times, what is the chance of getting five heads and five tails (the zero-sum, total annihilation, proton-free nothingness)? As it happens, the chance is a bit less than one in four. How do we calculate this?
Forget about ten flips for now. Putting on my computer scientist hat, ten flips with two possible outcomes each gives a ten-bit representation with 2^10=1024 combinations. Let’s go with four flips, sixteen possible combinations. These are easily written as the four-bit binary values with a round zero as a head and, for lack of a good analogy, a one for a tail:
Of these 16, six have two heads and two tails. 37.5%.
Remember when I said to forget about ten flips? Now forget about that, but go a step further. How do we calculate the probability of total annihilation for M heads in N flips?
First, what is the chance of M heads followed by (N - M) tails? This is just the chance of a head multiplied by itself M times, and that multiplied by the chance of a tail (N - M) times. For a fair coin, head and tail both appear with a probability of ½, so the overall chance is (½)^N. (For the four flips in the table, only the fourth row out of sixteen rows is two heads followed by two tails.)
But we don’t care about the order, so how many different ways can we distribute those M heads in N flips? This is where the binomial coefficient (or sometimes the “choose” notation) comes in. You have N slots. How many ways can you choose M of them?
If you have two slots and want to choose one, you have two possibilities. Three slots, three possibilities. With three slots and picking two, it’s also three possibilities because it is like picking one and then taking the two that you didn’t pick. In general? N-choose-M is written as:
$$\binom {N}{M} = \frac{N!}{M!\left(N-M\right)!}$$
So if you take the probability of getting two heads followed by two tails, 1/16, and multiply that by all the different ways you can pick two slots to put the heads in, 4-choose-2 = 6, you get 6/16=37.5% which is what we simply counted earlier.
Ten flips?
$$\left(\frac{1}{2}\right)^{10} \cdot \left( \frac{10!}{5! \cdot 5!} \right) = 24.6\%$$
Ten billion flips? A hundred billion? 1E1000? Exercise left for the reader.
So based on this math, the chance of getting an even number of protons and anti-protons becomes vanishingly small over eons of decay events. When you factor in that the probabilities of the results are actually unequal—the coin isn’t even fair—the overall imbalance becomes larger still. When you factor in that the universe is expanding, and the chance of x/anti-x annihilation decreases due to just having a lot more space, even more imbalance.
And this is—at least a part of—why there is something.
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