Mog Bonanza

Originally, I wanted to blog about the huge update today, but since my download rate has dropped to almost nil, I have decided to blog about the upcoming bonanza instead.

Here is how it works: you can create five-digit tickets using the numbers zero through nine, allowing repeats, by talking to the Bonanza Moogles. Each character is allowed a maximum of ten tickets, and you can hold identical tickets as well. The winning number is announced two months later, and a multitude of prizes are available based on the number of digits you guessed correctly.

As one can tell, the odds of winning the grand prize seem low, but with 100M gil at stake, I bet most of you are fairly interested in waking up with a new relic weapon or Armada Hauberk.










But just how low of a chance does one have at winning these coveted items? To solve the question, I have decided to apply some high school mathematics, and we begin by looking at some mathematical definitions:

Permutation: A selection of "things" where the order they are selected in matters.

Combination: A selection of "things" where the order does not matter.

By examining how the prizes are determined, clearly a ticket is a permutation of numbers rather than a combination. After all, 12345 is different from 54321. 

Ex. 1) The rank 1 prize requires you to guess all five digits correctly. What is the probability of striking it rich with a single ticket?

Think of getting each digit correct as a separate event:

_   _   _   _   _

A  B  C  D  E 

The probability that all the events will occur simultaneously is the product of the individual probabilities. If we need all five digits to match, then there is only one case per event where this can happen:

_ _ _ _ _
1 1 1 1 1

Since there are ten choices per digit to begin with, the probability is:

Probability = (1/10)(1/10)(1/10)(1/10)(1/10)
Probability = (1/10)^5
Probability = 1/100,000
Probability = 0.00001
Percent Chance  = 0.001%

If you have multiple tickets (and none of these are duplicates), multiply the number of tickets by the probability above to calculate your average chance of winning. With fifty unique tickets, I have a 0.05% chance at striking it rich. 

Khaki, what about the other prizes?

It turns out that the grand prize is the easiest scenario to calculate one's probability for. For ranks 2 through 5, you will need less than five correct digits, and this is where a little thinking is needed.









Ex. 2) What are the odds of winning a rank 2 prize with a single ticket?
Let us go back to our case model. We need the last four digits correct, so that leaves ten possible digits for the first slot and one digit for the others.

_ _ _ _ _
10 1 1 1 1

This means there are only ten cases that satisfy the rank 2 requirements; however, since it is also possible to win the rank 1 prize in this scenario, we must subtract the number of rank 1 cases from the rank 2 cases to get our overall probability.

Probability = [ (Total Rank 2 Cases) - (Total Rank 1 Cases) ] / (Total Cases)
Probability = [ (10*1*1*1*1) - (1*1*1*1*1) ] / (10*10*10*10*10)
Probability = [ 10 - 1 ] / (10)^5
Probability = 9 / 100,000
Probability = 0.00009
Percent Chance = 0.009%

In general, the probability of winning a rank 'n' prize, where 'n' does not equal 1, is:

Probability = [ (Total rank 'n' cases) - (Total rank "n-1" cases) - (Total rank "n-2" cases) - ... - (Total Rank 1 cases) ] / 100,000

Again, just multiply the probability by the number of unique tickets to get average probability. 



Notes: 
EDIT: 2010 UPDATE
a) The Mog Bonanza worked slightly different than originally described. Since each rank can have different winning numbers, one must carefully consider the chance of winning a higher-tier prize for ranks 2 through 5.

Prob(Rank 2 not Rank 1) = Prob(Rank2) - Prob(Rank 2 upgraded to Rank 1)
Prob(Rank 2 not Rank 1) = (1/10,000) - [(1/10,000)(1/100,000)]
Prob(Rank 2 not Rank 1) = 0.00009999999
Percent Chance = 0.009999999%