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# Probability & Statistics/winning the lottery

Question
Hi Clyde,

Here in my country, when you buy lottery tickets, you try to guess the six (6) combination of numbers out of 42 numbers. Unlike in Sweepstakes, the order of numbers does not really matter.

http://www.philippinepcsolotto.com/6-42-lotto-result-summary

Some say that the probability of winning the jackpot is simply 6/42 = 14.29%. That is high!

Since I was in high school, that formula seemed to make sense for me. After all, what is the probability of guessing one of the six combinations? 1/42. That is the probability of guessing each number in the combination of six. That is the result when you add the probability for each number.

(1/42) X 6 = 6/42

1/42 + 1/42 + 1/42 + 1/42 + 1/42 + 1/42 = 6/42

But I've come across another possibility today. Without repetitions of numbers, you can create 5,245,786 combinations of six digits from 1 to 42. What is the probability that you will guess the right combination out of more than 5 million combinations? 1 out of 5,245,786. That is 1.906 X 10 raised to negative 7.

I'm confused now because both solutions seem to make sense, but this second idea makes more sense. Each one is a computation based on a different point of view.

I hope you can enlighten me on this one.

Thanks,
John

If the probability of winning the lottery was 6/42 = 1/7, then by the end of two months, most of your population would be lottery winners. It would also be impossible for the lottery to pay out more than 7x the original ticket price, rather than huge quantities of money.

The figure 6/42 makes very little sense. A national lottery should be very hard to win, like one in one million.

The major mistake you are making is that you have to get all the numbers correct. In order to compute the probability that you get the first one, and the second, and the third... you have to use multiplication , not addition.

It would be almost correct to say that the probability is:

1/42 × 1/42 × 1/42 × 1/42 × 1/42 × 1/42 = 1/5489031744 ≈ 1.82181×10^-10

However, that does not account for the fact that (1) the order is irrelevant and (2) numbers cannot repeat. Instead, the correct answer is:

6!( 1/42 × 1/41 × 1/40 × 1/39 × 1/38 × 1/37) = 720/3776965920 = 1/5245786 ≈ 1.90629×10^-7

The decreasing numbers (42, 41, ..., 37) are due to the removal of one number at each pick (each number chosen cannot be reused, so there is one fewer at each step). The "6!" is a factorial, 5! = 6×5×4×3×2×1 = 720. It accounts for the possible rearrangements of the numbers on a winning ticket, since order does not matter. For example, if 5-7-13-20-28-30 is the winning ticket, so is any of the 719 other ways to rearrange those six numbers.

The figure 5245786 is indeed the total number of lottery picks. The probability of winning a lottery will always be one out of the total number of picks, since only one combination will win. That number is known as a "binomial" quantity, the number of ways to choose 6 things from 42 ("42 choose 6"). For more, see:

http://en.wikipedia.org/wiki/Binomial_coefficient#Combinatorics_and_statistics
http://en.wikipedia.org/wiki/Lottery#Probability_of_winning
Questioner's Rating
 Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment Thank you very much, Clyde.

Probability & Statistics

Volunteer

#### Clyde Oliver

##### Expertise

I can answer all questions up to, and including, graduate level mathematics. I do not have expertise in statistics (I can answer questions about the mathematical foundations of statistics). I am very much proficient in probability. I am not inclined to answer questions that appear to be homework, nor questions that are not meaningful or advanced in any way.

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I am a PhD educated mathematician working in research at a major university.

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AMS

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Various research journals of mathematics. Various talks & presentations (some short, some long), about either interesting classical material or about research work.

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BA mathematics & physics, PhD mathematics from a top 20 US school.

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Various honors related to grades, various fellowships & scholarships, awards for contributions to mathematics and education at my schools, etc.

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In the past, and as my career progresses, I have worked and continue to work as an educator and mentor to students of varying age levels, skill levels, and educational levels.