1 year ago 1 year ago 2 years ago
Mega Millions Math: Probability of No Jackpot Winner

You would like to figure out the approximate probability that there are no jackpot winners for the Mega Millions. On March 27, 2012 4,715,569 tickets from the total pool of tickets were winners. Approximately 1 out of 40 randomly chosen tickets are winners. Therefore we can approximate that 40 * 4,715,569 = 188,622,760 total tickets were sold.

But wait, you have a 1 in 175,711,536 chance of hitting the jackpot. Approximately 188,622,760 were sold. So does that mean there’s a 100% chance of one of those tickets being the jackpot winner? No, simply because some tickets have the same numbers which increases the total ticket pool.

What is the probability that no one hit the Mega Millions jackpot last night? You have to know a little about Bernouilli trials. A Bernouilli trial is an experiment whose outcome is random and can be either of two possible outcomes, “success” and “failure”. So in our case “success” is winning the jackpot, “failure” is not winning the jackpot.

Say a gambler plays a slot machine that pays out with a probability of one in n and plays it n times. Then, for large n the probability that the gambler will lose every bet is approximately 1/e or about 37%. This is a Bernoulli trials process and you can approximate probabilities with respect to this process using the constant e.

We can apply this situation to Mega Millions. The Mega Millions is sort of like a slot machine that pays out the jackpot with a probability of 1/175,711,536. So according to our experiment, if you play 175,711,536 tickets there’s still a 37% chance that there would be no winner!

It’s not intuitive but the math is there and you can’t argue against it. Because it’s math.

Last night there were 188,622,760 tickets played. Use e again to approximate the probability of no jackpot. The Mega Millions went through a period where there was an expectation of 1.0735 jackpot winners (188,622,760/175,711,536 = 1.0735). The approximate probability that there were no jackpots in 1.0735 “cycles” is e^(-1.0735) or approximately 34%.

Cliffs:
-Lotteries and games of chance are usually not intuitive.
-The probability that there would be no jackpot winner last night is 34%. Consequently, the probability that there should’ve been AT LEAST ONE jackpot winner is 66%. Think about how I came up with that number.
-The mathematical constant e is awesome.
-They carded me last night when I tried to buy Quick Picks. 

2 years ago

James R. Murphy's La Guardia students with their string figures.

2 years ago 2 years ago 2 years ago
Packing problem: different ways to pack oranges in two dimensions (left and right); according to Johannes Kepler, the most efficient way to pack spheres in three dimensions is the face-centred cubic arrangement (centre).
Imagine filling a large container with small equal-sized spheres. The density of the arrangement is the proportion of the volume of the container that is taken up by the spheres. In order to maximize the number of spheres in the container, you need to find an arrangement with the highest possible density, so that the spheres are packed together as closely as possible.
Experiment shows that dropping the spheres in randomly will achieve a density of around 65%. However, a higher density can be achieved by carefully arranging the spheres as follows. Start with a layer of spheres in a hexagonal lattice, then put the next layer of spheres in the lowest points you can find above the first layer, and so on – this is just the way you see oranges stacked in a shop. At each step there are two choices of where to put the next layer, so this natural method of stacking the spheres creates an uncountably infinite number of equally dense packings, the best known of which are called cubic close packing and hexagonal close packing. Each of these arrangements has an average density of

The Kepler conjecture says that this is the best that can be done—no other arrangement of spheres has a higher average density.

Packing problem: different ways to pack oranges in two dimensions (left and right); according to Johannes Kepler, the most efficient way to pack spheres in three dimensions is the face-centred cubic arrangement (centre).

Imagine filling a large container with small equal-sized spheres. The density of the arrangement is the proportion of the volume of the container that is taken up by the spheres. In order to maximize the number of spheres in the container, you need to find an arrangement with the highest possible density, so that the spheres are packed together as closely as possible.

Experiment shows that dropping the spheres in randomly will achieve a density of around 65%. However, a higher density can be achieved by carefully arranging the spheres as follows. Start with a layer of spheres in a hexagonal lattice, then put the next layer of spheres in the lowest points you can find above the first layer, and so on – this is just the way you see oranges stacked in a shop. At each step there are two choices of where to put the next layer, so this natural method of stacking the spheres creates an uncountably infinite number of equally dense packings, the best known of which are called cubic close packing and hexagonal close packing. Each of these arrangements has an average density of

The Kepler conjecture says that this is the best that can be done—no other arrangement of spheres has a higher average density.

2 years ago
Why are numbers beautiful? It’s like asking why is Beethoven’s Ninth Symphony beautiful. If you don’t see why, someone can’t tell you. I know numbers are beautiful. If they aren’t beautiful, nothing is. Cite Arrow Paul Erdős
3 years ago
Welcome to another edition of Secretly Judging You

48÷2(9+3) = ????

3 years ago
A sum, proved impossible by the theorem, appears in an episode of The Simpsons, “Treehouse of Horror VI”. In the three-dimensional world in “Homer3”, the equation 178212 + 184112 = 192212 is visible, just as the dimension begins to collapse. The joke is that the twelfth root of the sum does evaluate to 1922 due to rounding errors when entered into most handheld calculators; notice that the left hand side is odd, while 192212 is even, so the equality cannot hold. Instead of 1922, it actually is 1921.999999995 (via Fermat’s Last Theorem in fiction)

A sum, proved impossible by the theorem, appears in an episode of The Simpsons, “Treehouse of Horror VI”. In the three-dimensional world in “Homer3”, the equation 178212 + 184112 = 192212 is visible, just as the dimension begins to collapse. The joke is that the twelfth root of the sum does evaluate to 1922 due to rounding errors when entered into most handheld calculators; notice that the left hand side is odd, while 192212 is even, so the equality cannot hold. Instead of 1922, it actually is 1921.999999995 (via Fermat’s Last Theorem in fiction)