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How do!

I’m hoping some maths whizz Dakkanaut can help settle an utterly pointless non-argument that arose between a colleague and I today.

We were, for reasons, discussing the odds of a given number appearing in the UK National Lottery Draw as one of the six main numbers.

Now, the raw numbers because I guess those will help.

The pool is 1-59. And a draw involves 6 numbers, and a Bonus Ball number. All numbers are drawn from the same, single pool of 59 balls.

Now, I’m maths aware enough (I used to be really, really, really good. Then I got bored and grew old and massively out of practice) to know that it can’t be 1 in 59. And that’s because we’re drawing 6 numbers at a time to make up the winning draw. And as each number is drawn, it can’t be drawn again.

We got as far as thinking what we need to do is add up the big numbers “1 in 59, 1 in 58” and so forth, then divide by 6. And then our brains fell out and we both agreed we no longer have the maths skills to work it out.

But maybe a fellow Dakkanaut is? Remember this isn’t trying to “prove” anything, so much as just sheer interest. And yes I do play the lottery, yes I know the odds are astronomical, but I’m always aware a 1 in 1,000,000,000 or whatever chance doesn’t need 1,000,000,000 entires to actually happen.

Right, over to you, people that haven’t largely forgotten what they were taught in school!

 Mad Doc Grotsnik wrote:
How do!

I’m hoping some maths whizz Dakkanaut can help settle an utterly pointless non-argument that arose between a colleague and I today.

We were, for reasons, discussing the odds of a given number appearing in the UK National Lottery Draw as one of the six main numbers.

Now, the raw numbers because I guess those will help.

The pool is 1-59. And a draw involves 6 numbers, and a Bonus Ball number. All numbers are drawn from the same, single pool of 59 balls.

Now, I’m maths aware enough (I used to be really, really, really good. Then I got bored and grew old and massively out of practice) to know that it can’t be 1 in 59. And that’s because we’re drawing 6 numbers at a time to make up the winning draw. And as each number is drawn, it can’t be drawn again.

We got as far as thinking what we need to do is add up the big numbers “1 in 59, 1 in 58” and so forth, then divide by 6. And then our brains fell out and we both agreed we no longer have the maths skills to work it out.

But maybe a fellow Dakkanaut is? Remember this isn’t trying to “prove” anything, so much as just sheer interest. And yes I do play the lottery, yes I know the odds are astronomical, but I’m always aware a 1 in 1,000,000,000 or whatever chance doesn’t need 1,000,000,000 entires to actually happen.

Right, over to you, people that haven’t largely forgotten what they were taught in school!

The odds of it appearing in the first number is 1/59.
The odds of it appearing in the next are 1/58, so on and so forth, to 1/54.

You don't actually need to divide by six-since you're looking for the number to appear at least once, you should be able to literally just add the probabilities.

A quick calculator math gives us 10.629187105% chance.
I just plugged (1/59)+(1/58)+(1/57)+(1/56)+(1/55)+(1/54) into google to get that.

Just checking-once a number is pulled, it cannot be pulled again, right?

Yup. Each batch of numbered balls has a single instance of the numbers 1-59. So once a given draw has drawn a given number, there’s a 0% chance of that number turning up again.

Now, on your calculation? I was with you up to to “add them together”, then my brain fell out. So without questioning your result? I don’t understand how we got to the near-enough 10% chance of a given number being drawn.

Multiply the number received from adding the fractions by 100.

But now I have a question. Since there's 6 balls drawn, and it doesn't matter which of them is the required number, shouldn't the odds just be straight up 6/59?

Bran Dawri wrote:
Multiply the number received from adding the fractions by 100.

But now I have a question. Since there's 6 balls drawn, and it doesn't matter which of them is the required number, shouldn't the odds just be straight up 6/59?

I think it would be that if the ball pool isn't reduced with each draw.

 JNAProductions wrote:

 Mad Doc Grotsnik wrote:
How do!

I’m hoping some maths whizz Dakkanaut can help settle an utterly pointless non-argument that arose between a colleague and I today.

We were, for reasons, discussing the odds of a given number appearing in the UK National Lottery Draw as one of the six main numbers.

Now, the raw numbers because I guess those will help.

The pool is 1-59. And a draw involves 6 numbers, and a Bonus Ball number. All numbers are drawn from the same, single pool of 59 balls.

Now, I’m maths aware enough (I used to be really, really, really good. Then I got bored and grew old and massively out of practice) to know that it can’t be 1 in 59. And that’s because we’re drawing 6 numbers at a time to make up the winning draw. And as each number is drawn, it can’t be drawn again.

We got as far as thinking what we need to do is add up the big numbers “1 in 59, 1 in 58” and so forth, then divide by 6. And then our brains fell out and we both agreed we no longer have the maths skills to work it out.

But maybe a fellow Dakkanaut is? Remember this isn’t trying to “prove” anything, so much as just sheer interest. And yes I do play the lottery, yes I know the odds are astronomical, but I’m always aware a 1 in 1,000,000,000 or whatever chance doesn’t need 1,000,000,000 entires to actually happen.

Right, over to you, people that haven’t largely forgotten what they were taught in school!

The odds of it appearing in the first number is 1/59.
The odds of it appearing in the next are 1/58, so on and so forth, to 1/54.

You don't actually need to divide by six-since you're looking for the number to appear at least once, you should be able to literally just add the probabilities.

A quick calculator math gives us 10.629187105% chance.
I just plugged (1/59)+(1/58)+(1/57)+(1/56)+(1/55)+(1/54) into google to get that.

Just checking-once a number is pulled, it cannot be pulled again, right?

I get the same result, but remember that’s just the chance of a number (let’s say 42, because we’re nerds) appearing once within those 6 numbers.