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Calculus/winning the lottery

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Question
Hi Frederick,

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

Answer
The first manner of calculating the required probability which you cited is flawed. If say, each of these 6 combinations can be recycled, the probability would be computed as (1/42)^6 = 1/5489031744.

However, I am assuming they cannot be recycled, as such there will be 42 possible combinations for the first number, 41 for the second, 40 for the third, so on and so forth. That said, the probability would be computed as (1/42)*(1/41)*(1/40)*(1/39)*(1/38)*(1/37)= 1/3776965920

You may therefore question: why isn't this equivalent to 1/5245786, which is considerably larger in fact? The reason is (1/42)*(1/41)*(1/40)*(1/39)*(1/38)*(1/37) actually examines the instance when sequencing of numbers matter, as opposed to 1/( 42 C 6) = 1/5245786 which doesn't pay any regard to the ordering of numbers. Clearly, the former would involve much greater stakes and thus even lower likelihood of winning.

Hope this clarifies. Peace.

Calculus

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Frederick Koh

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I can answer questions concerning calculus, complex numbers, vectors, statistics , algebra and trigonometry for the O level, A level and 1st/2nd year college math/engineering student.

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More than 7 years of experience helping out in various homework forums. Latest Presence is over at http://www.thestudentroom.co.uk/ . You can also visit my main maths website http://www.whitegroupmaths.com where I have designed "question locker" vaults to store tons of fully worked math problems. A second one is currently being built. Peace.

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Former straight As A level student from HCJC (aka HCI); scored distinctions in both C and Further Mathematics B Eng (Hons) From The National University Of Singapore (NUS) B Sc (Hons) From University of London External (Grad Route)

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