Flaw in a Lottery Game and a Second Opportunity

Question

In my country there is a Lottery Game called "Melate". It works like this:

Get a Melate flyer, pick up 6 Numbers from 1 to 56, it doesn't matter the order, each combination costs $15, then wait for the raffle where a tombola is going to draw 6 balls, these are going to be called "The Natural Numbers" and then it is going to draw another ball, this is going to be called "The Additional Number"

Check if you win any prize depending on the number of balls matched:

• 1st Prize= 6 "Natural Numbers" (Variable Prize)
• 2nd Prize= 5 "Natural Numbers" and the "Additional Number" (Variable Prize)
• 3rd Prize= 5 "Natural Numbers" (Variable Prize)
• 4th Prize= 4 "Natural Numbers" and the "Additional Number" (Variable Prize)
• 5th Prize= 4 "Natural Numbers" (Variable Prize)
• 6th Prize= 3 "Natural Numbers" and the "Additional Number" ( \$161.29 )
• 7th Prize= 3 "Natural Numbers" ( \$43.01 )
• 8th Prize= 2 "Natural Numbers" and the "Additional Number" ( \$32.25 )
• 9th Prize= 2 "Natural Numbers" ( \$26.88 )

In order to compute the probabilities for every prize, the way I reasoned the problem was:

The total of possible combinations without repetition of 6 elements choosen between 56 is:

$ \binom{56}{6} = 32,468,436 $

The amount of possible combinations without repetition of x "Natural Numbers" is given by:

$ \binom{6}{x} $

The amount of possible combinations without repetition of "The Additional Number" is given by:

$ \binom{1}{1} $

The amount of possible combinations without repetition of the "y left spaces" that they can be chosen from any of the Remaning Numbers (56 - 6 -1 = 49) is given by:

$ \binom{49}{y} $

These three events are dependent, so:

P(X and Y and A) = P(X)P(Y after X)P(A after X)

So, this is the way I tried to compute the probabilities:

1st Prize

$ P(1st Prize)= \frac{\binom{6}{6}}{\binom{56}{6}} = \frac{1}{32,468,436} ≈ 0.000000030799 = 0.0000030799 \% $

2nd Prize

$ P(2nd Prize)= \frac{\binom{6}{5}}{\binom{56}{6}} \frac{\binom{1}{1}}{\binom{56}{6}} = \frac{6}{32,468,436} \frac{1}{32,468,436} = \frac{1}{5,411,406} ≈ 0.00000018479 = 0.000018479 \% $

3rd Prize

$ P(3rd Prize)= \frac{\binom{6}{5}}{\binom{56}{6}} \frac{\binom{49}{1}}{\binom{56}{6}} = \frac{6}{32,468,436} \frac{49}{32,468,436} = \frac{7}{773058} ≈ 0.000009055 = 0.0009055 \% $

4th Prize

$ P(4th Prize) = \frac{\binom{6}{4}}{\binom{56}{6}} \frac{\binom{49}{1}}{\binom{56}{6}} \frac{\binom{1}{1}}{\binom{56}{6}} = \frac{15}{32468436} \frac{49}{32468436} = \frac{35}{1546116} ≈ 0.000022637 = 0.0022637 \% $

5th Prize

$ P(5th Prize) = \frac{\binom{6}{4}}{\binom{56}{6}} \frac{\binom{49}{2}}{\binom{56}{6}} = \frac{15}{32,468,436} \frac{1176}{32,468,436} = \frac{70}{128843} ≈ 0.00054329 = 0.054329 \% $

6th Prize

$ P(6th Prize) = \frac{\binom{6}{3}}{\binom{56}{6}} \frac{\binom{49}{2}}{\binom{56}{6}} \frac{\binom{1}{1}}{\binom{50}{6}} = \frac{20}{32,468,436} \frac{1176}{32,468,436} \frac{1}{32,468,436} = \frac{280}{386529} ≈ 0.000724396 = 0.0724396 \% $

7th Prize

$ P(7th Prize) = \frac{\binom{6}{3}}{\binom{56}{6}} \frac{\binom{49}{3}}{\binom{56}{6}} = \frac{20}{32,468,436} \frac{18424}{32,468,436} ≈ 0.011348 = 1.1348 \% $

8th Prize

$ P(8th Prize) = \frac{\binom{6}{2}}{\binom{56}{6}} \frac{\binom{49}{3}}{\binom{56}{6}} \frac{\binom{1}{1}}{\binom{50}{6}} = \frac{15}{32,468,436} \frac{18424}{32,468,436} \frac{1}{32,468,436} ≈ 0.0085116 = 0.85116 \%$

9th Prize

$ P(9th Prize) = \frac{\binom{6}{2}}{\binom{56}{6}} \frac{\binom{49}{4}}{\binom{56}{6}} = \frac{15}{32,468,436} \frac{211,876}{32,468,436} ≈ 0.097884 = 9.7884 \% $

My first question is, the probabilities that I computed are correct? if they are, I think this lottery game has a flaw since P(7th Prize) > P(8th Prize) and 7th Prize > 8th Prize . In my opinion it should be 7th Prize < 8th Prize

I was watching a webpage where they computed the probabilities of the same lottery and they obtained simmilar results (altough they don't show the computing method):

Prize and Probability

• 1st Prize= 1:32.468.436
• 2nd Prize= 1:5.411.406
• 3rd Prize= 1:110.437
• 4th Prize= 1:44.175
• 5th Prize= 1:1.840
• 6th Prize= 1:1.380
• 7th Prize= 1:88
• 8th Prize= 1:117
• 9th Prize= 1:10

So, am I right? Is there a flaw in the 7th Prize?

If my probabilities are computed correctly, it is also possible to say:

$ P(nth Prize) = \frac{\binom{6}{6- \big\lfloor\frac{n}{2}\big\rfloor} \binom{49}{\big\lfloor\frac{n-1}{2}\big\rfloor} }{\binom{56}{6}} $

for $n=1,2,3,4,5,6,7,8,9$, where $\lfloor x\rfloor$ is the floor-function

After that, you can have a second opportunty of playing by sticking with your previous combination used in Melate, this second chance is called "Revancha", each combination costs $10

The Prizes are:

• 1st Prize= 6 "Natural Numbers" (Variable Prize)
• 2nd Prize= 5 Natural Numbers (Variable Prize)
• 3rd Prize= 4 Natural Numbers (Variable Prize)
• 4th Prize= 3 Natural Numbers ( \$26.88 )
• 5th Prize= 2 Natural Numbers ( \$10.75 )

Following the same logic, the probabilities that I computed are:

1st Prize

$ P(1st Prize)= \frac{\binom{6}{6}}{\binom{56}{6}} = \frac{1}{32,468,436} ≈ 0.000000030799 = 0.0000030799 \% $

2nd Prize

$ P(2nd Prize)= \frac{\binom{6}{5}}{\binom{56}{6}} \frac{\binom{50}{1}}{\binom{56}{6}} = \frac{6}{32,468,436} \frac{50}{32,468,436} = \frac{25}{2705703} ≈ 0.000009240 = 0.0009240 \% $

3rd Prize

$ P(3rd Prize) = \frac{\binom{6}{4}}{\binom{56}{6}} \frac{\binom{50}{2}}{\binom{56}{6}} = \frac{15}{32,468,436} \frac{1225}{32,468,436} ≈ 0.000565934 = 0.0565934 \% $

4th Prize

$ P(4th Prize) = \frac{\binom{6}{3}}{\binom{56}{6}} \frac{\binom{50}{3}}{\binom{56}{6}} = \frac{20}{32,468,436} \frac{19600}{32,468,436} ≈ 0.012073 = 1.2073 \% $

5th Prize

$ P(5th Prize) = \frac{\binom{6}{2}}{\binom{56}{6}} \frac{\binom{50}{4}}{\binom{56}{6}} = \frac{15}{32,468,436} \frac{230,000}{32,468,436} = \frac{70}{128843} ≈ 0.10639 = 10.639 \% $

Does the probability of winning with the same combination changes? Is it better to buy a new combination and play "Melate", or should I stick with the old one and play "Revancha"?

After that, there is also a third oppotunity, where you can continue sticking with the same combination again, this third opportunity is called "Revanchita", it costs $5 for each combination ..

The Prizes are:

• 1st Prize= 6 "Natural Numbers" (Variable Prize)

The probability that I computed is:

1st Prize

$ P(1st Prize)= \frac{\binom{6}{6}}{\binom{56}{6}} = \frac{1}{32,468,436} ≈ 0.000000030799 = 0.0000030799 \% $

should I stick with the old combination again?

This looks something similar to the "Monty Hall Problem", should I use the Bayes Theorem? Do probabilities change, or stay the same? and why? I am inclined to believe the probabilities are the same, but it is a little bit weird that "Revancha" only offers 5 prizes and "Revanchita" only one prize insted of the original 9 prizes, as if it was easier to win any prize .. if you know that a previous combination didn't win anything in the past, does the probability that the same combination wins anything increases? I don't think so, but if I reverse the question, What are the odds that the same combination wins twice?, it seems more tempting to believe that the past affects the current raffle but I am new on Probability Theory, so I don't really know

Answer 1

I didn't look until the end but the probabilities that you're computing for every prize are wrong.

I took only the 7th Prize case, in that one you have the following: $$\frac{\binom{6}{3}\binom{49}{3}}{\binom{56}{6}}$$

But you're dividing with $\binom{56}{6}$ twice, which is not needed in all the cases. If you recompute your probabilities correctly you're going to see that they differ from what they are specifying on the website.