Revomaze Competition! 10 X Revomaze Purple Up For Grabs

[maztec]

Can any admin tell us how Many people entered the lotery ?????

1. You are the only person who entered! Congratulations.

2. Who knows, Joseph may tell us Saturday.

[QuantumCow]

ps... hopefully someone out there in Revoland is a real statistician and can referee my numbers

Actually I calculate the odds are much larger.

There are 49 balls, and 6 numbers drawn. The calculation is easier if you calculate what the probability is that you will get 0 numbers, and then subtract from 1.

Let's say you have N numbers (N=0 through 15).

On the first draw, there are 49 possibilities, but (49-N) of them are losing possibilities, so the probability of losing is (49 - N)/49.

On the second draw, there are 48 possibilities, but (48-N) of them are losing possibilities, etc....

so the probability of losing is

(49-N)(48-N)(47-N)(46-N)(45-N)(44-N)
---------------------------------------
49*48*47*46*45*44

or

(49-N)!*43!/((43-N)!*49!)
Therefore, the probability of winning a particular round is 1 minus that.
given a probability p of winning 1 round, the probability of winning m rounds is p^m

Here is a table of winning successive rounds, from 1 to 7:

1, 0.122449, 0.01993, 0.0018, 0.00022, 2.7E-05, 3.370E-06, 4.12749E-07
2, 0.232142, 0.05896, 0.0125, 0.00290, 0.00067, 0.00015650, 3.63318E-05
3, 0.330167, 0.10012, 0.0359, 0.01188, 0.00392, 0.001295398, 0.000427698
4, 0.417536, 0.17333, 0.0727, 0.03039, 0.01269, 0.005298681,0.002212394
5, 0.495198, 0.24224, 0.1214, 0.06013, 0.02977, 0.014746048, 0.00730222
6, 0.564035, 0.31139, 0.1794, 0.10121, 0.05708, 0.032198559, 0.018161115
7, 0.624867, 0.39451, 0.2439, 0.15245, 0.09526, 0.05952878, 0.037197591
8, 0.678457, 0.46306, 0.3122, 0.21188, 0.14375, 0.09752967, 0.066169758
9, 0.725512, 0.52367, 0.3818, 0.27706, 0.20101, 0.145837795, 0.105807171
10, 0.76668, 0.58809, 0.4506, 0.34551, 0.26490, 0.203097454, 0.155711932
11, 0.80258, 0.64413, 0.5169, 0.41491, 0.33299, 0.267258125, 0.214496101
12, 0.83375, 0.69143, 0.5795, 0.48322, 0.40288, 0.33590832, 0.280064172
13, 0.86071, 0.74824, 0.6376, 0.54881, 0.47237, 0.406578178, 0.349946303
14, 0.88392, 0.78322, 0.6906, 0.61046, 0.53960, 0.47697423, 0.421609836
15, 0.90382, 0.81894, 0.7383, 0.66732, 0.60314, 0.545134835, 0.492706083

The people with higher points will almost certainly win, depending on how many of them there are.
You have to be careful when extrapolating these values to the probabilities of how many rounds there will be, since winners are tied together. For example, if there are four people, and between them they have all 49 numbers, at least one of them is guaranteed to go to the next round. It's complicated, but I think the best approach is to do a monte carlo simulation a few million times (or more, until the numbers start to converge).

I predict the contests will go on a very long time, at least 7-10 rounds each, but it totally depends on the distribution of higher number people.
EDIT: The above numbers are for 6 numbers per game. If the bonus ball 7 is counted, all of the odds above would go up

[QuantumCow]

By the way, I have 10 numbers. So my only hope is that there are less than 10 people with more than 10 numbers.

One interesting fact is that ordinarily, every number would be equal to every other number, in terms of winning, however, people will likely choose their numbers non randomly. Personally, I found a random number generator online that generates random numbers based on cosmic rays. However, someone who is more of a statistician than me could probably calculate whether it hurts or helps to have numbers that other people have, in the long run, and can also use psychology to figure out which will be the most common numbers in order to use that to their advantage.

The real way to game the system would be to buy extra Revomazes just for this contest to get yourself up to 15, and then sell them afterwards. Having 15 numbers pretty much guarantees a purple, unless there are more than 10 people with 15 numbers. I couldn't bring myself to do that though, and I don't mind pointing it out now since the lottery entries are closed.

[Sejanus]

I am sure a good amount of people who entered have at least 7 or more numbers. I personally have 9.. and that may not be enough unless luck is on my side. That is ok though!

[MJShasko]

Thanks Quantum.

Your findings seem to support my observations against the last 3 months of data....so somewhere along the way my math tanked.

Looks like many draws (more than 3) will be required to determine any set of winners.

[QuantumCow]

Attached is an excel worksheet for the history of the lottery to figure out some statistics on how you would have done given the lottery results as downloaded from the UK Lottery website. Eventually I'm going to make one that simulates a random lottery so that you can get proper statistics...Keep in mind that this is only 60 or so runs, so this is likely not going to be a proper result like the theoretical result.

To use this, put your numbers where the -1 s are in row 2. Then you can look at the bottom to see the longest run, and the average run. My average run of lotteries is slightly below 3, which makes sense because I calculated my chances of winning a given night to be roughly 75% with 10 numbers. I'm happy to see I have at least a few runs of 8 or 9 in a row, although there are also losses of up to 3 in a row, which is to be expected.

Enjoy =]
Rich

[maztec]

8 entries of 15 numbers
3 of 14 numbers
2 of 12 numbers
2 of 10 numbers
6 of 9 numbers
5 of 8 numbers
1 of 6 numbers
2 of 2 numbers
http://revomaze.co.uk/phpBB3/viewtopic.php?f=7&t=1672

At the very least that is the floor. Assume that distribution is the normal distribution and that that represents roughly 2 out of 3 of the competitors.

Thus, based on a basic extrapolation, you are looking at
12 of 15
5 of 14
3 of 12
3 of 10
9 of 9
8 of 8
2 of 6
3 of 2

A total of 45 entrants with 479 entries between them - an average of 10.6 entries per person.

That seems like a reasonable estimate, unless everyone with a single maze actually entered, which is unlikely.

Of course, to get a ceiling you could look through the user lists and generate a reasonable ceiling for number of entrants and entries.

Assuming 479 entries for the first drawing each week that means each number entered has a chance of being drawn 9.776 times. With 6 numbers being drawn (there are 6 numbers + 1 extra number; are they going to go by the 6 or the 7 for this? I was never clear on that) you get 58.65 of the numbers entered drawn (ignoring statistical weighting of one number over another, assuming truly random distribution - which is unlikely). That means a single entry has a 12% chance of being picked - woo, right on with Quantum Cow's calculation but I did it the long way around. However, what is interesting to me is that you can do this to limit the pool. So ...

Week 1: 479 entries of which 58.65 are drawn.
Week 2: 58.65 entries of which each number has a chance of being drawn 1.196 times, leaving each entry having a 0.2% chance of being picked. (right on, still agree with Quantums math, although my error ratio is off by a bit).

Now, here is the problem, the entries at week 2 aren't simply the 58.65 entries, but that the pool is repopulated each week with the other undrawn numbers from the same entrant. That means that in week 1 where 58.65 numbers are drawn, you have to then calculate how many people that represents and then add their numbers back in. And there we get beyond my memory of statistics on how to calculate the distribution of people that would have an entry picked, put that person back in, shake them up, etc.


Thanks for the thought math Mr. Cow.

[maztec]

Hey fun spreadsheet. Thanks Rich!

Here's a statistical question for you. I have seen numerous claims over the years that when picking lottery numbers you should make sure your numbers add up to being near the center of the bellcurve on distribution of what numbers would add up to. That is, it's more likely for randomly distributed numbers to add up to things in the center than at the ends (i.e., in craps you are more likely to get a 7 than a 1 or a 12 and so forth). Do you think this holds true for the lottery?

[QuantumCow]

Attached is a monte carlo simulation of 10,000 lottery drawings. For simplicity, it doesn't take into account the fact that you can't redraw the same ball, since this is not a trivial thing to do in excel, as far as I know. This will slightly skew the results down.

Next thing to do (which I shouldn't do tonight, since I'm procrastinating real homework =]) is to do a monte carlo simulation using the theoretical statistics and maztec's assumptions on the entries (which are kind of sad since I expect they will keep numbers 10 and below from having much chance at a purple, but one can hope)

Edit: oops, 10,000 is too big for the forum =] I'll cut the rows down to 1000, and you can select them and pull them down if you want to make them 10,000.

[QuantumCow]

Here's a statistical question for you. I have seen numerous claims over the years that when picking lottery numbers you should make sure your numbers add up to being near the center of the bellcurve on distribution of what numbers would add up to. That is, it's more likely for randomly distributed numbers to add up to things in the center than at the ends (i.e., in craps you are more likely to get a 7 than a 1 or a 12 and so forth). Do you think this holds true for the lottery?

Hmm. I don't think this applies to the lottery because the numbers are never added together. It is definitely true for craps that 7 is the most likely number. There are 6 different ways to get a 7, and 36 total possibilities, so in any given roll there's a 1 in 6 chance of getting a 7. On the other hand, a 2 or a 12 (1 is impossible) have only 1 way to get them, so there is only a 1 in 36 chance of getting each.

In lotteries though, the numbers never get added together, so numbers at the ends are just as likely as numbers in the middle. That having been said, if the lottery depends on what other people have picked (as the revomaze competition does), people are psychologically more likely to pick numbers near the ends, I would think (or perhaps less likely, if they try to take that into account). People are terrible at choosing random numbers.

It would be neat to see a histogram of numbers that people chose, perhaps after the competition is over. My guess is that certain numbers are either more or less likely (3, 7, 1, 11, 42). This crowd is far from average though, unfortunately (or fortunately?), so probably this is much different than if you asked a proper distribution of people.