[Guest]

What is the expected number of times a fair die must be thrown until all scores appear at least once?

[Guest]

What is the expected number of times a fair die must be thrown until all scores appear at least once?

If it is indeed a FAIR die, then the number of times thrown would only need to equal the number of sides on the die

[Guest]

Well appear is a tricky word, because all the numbers except for one will appear.

This is for a 6-sided die.

If we are assuming that appear means visible, then only one side is not visible on each normal throw. If this is the case, then 2 throws would yield a 97.22% chance that you would see all the numbers.

If we are assuming appear means that only one number appears, it would not be the number of sides = times.

First thrown, will result in a good new number. Second thrown will result in 5/6 chance, third 4/6 chance, etc...

1 * 5/6 * 4/6 * 3/6 * 2/6 * 1/6 = 1.5% of it occuring that way.

[Guest]

It seems to me that the number of sides is irrelevant, as is whether or not you look at the only the top number, or all faces visible. The answer is always the same: infinite.

The problem asks

What is the expected number of times a fair die must be thrown until all scores appear at least once?

At the start of any tossing, there is an infinitesmal chance that one side will not turn up. For example, with a 6-sided die - assuming that we are looking at all possible faces, it is possible that "5" will not show after 1 throw (chances are 1:6). For 2 throws the chances are 1:36. For 3 throws, 1:216.. ad nauseum... For my 1,204,575,475th try, the chances are really slim, but not 0.

No matter how many times I plan to toss the die, I cannot guarantee that at least one face will not always be stubborn and be face down. So the answer is to "how many times to do you have to toss a die to see all the faces" is infinite.

[Guest]

It seems to me that the number of sides is irrelevant, as is whether or not you look at the only the top number, or all faces visible. The answer is always the same: infinite.

The exception to this would be a Mobius strip with the a number written on it! In that case the answer is 1 throw.

[Guest]

Just a clarification to my question.

The die has of course 6 faces ... and when I say visible, I mean the top one.

By fair-die I mean all the numbers have equal probability of appearing on the top face.

I am asking for the expected number of throwing the die so that all the numbers (1-6) will appear (top face) at least once.

[bonanova]

What is the expected number of times a fair die must be thrown until all scores appear at least once?

expectation means...']To be certain of seeing all six numbers requires an infinite number of throws.

However, you might see all six numbers after only six throws.

If you count the throws it takes to see all six numbers, and average that over a large number of trials,

you get what's called the expectation value.

Expectation gives an average of the things that might happen by chance.

It does not guarantee anything, but it gives you an idea of what will probably happen.

To clarify what expectation is, and why it's important: you'll lose money if you bet against it.

how you calculate it']To see one number you only have to roll the die once.

The next roll might give you a new number, but there is a 1/6 chance it wont.

You can say the next roll has a new-number effectiveness of only 5/6.

So you need [on average] 6/5 additional rolls to see a 2nd new number.

The next roll might give you a new number, but now there is a 2/6 chance it wont.

You can say the next roll has a new-number effectiveness of only 4/6.

So you need [on average] 6/4 additional rolls to see a 3rd new number.

The next roll might give you a new number, but now there is a 3/6 chance it wont.

You can say the next roll has a new-number effectiveness of only 3/6.

So you need [on average] 6/3 additional rolls to see a 4th new number.

The next roll might give you a new number, but now there is a 4/6 chance it wont.

You can say the next roll has a new-number effectiveness of only 2/6.

So you need [on average] 6/2 additional rolls to see a 5th new number.

The next roll might give you a new number, but now there is a 5/6 chance it wont.

You can say the next roll has a new-number effectiveness of only 1/6.

So you need [on average] 6/1 additional rolls to see a 6th new number.

So in total, you need

6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 = [60 + 72 + 90 + 120 + 180 + 360]/60 = 882/60 =

14.7 rolls of the dice to expect to see all six numbers.

If you bet even odds that you'll get all six numbers in 15 rolls of the dice, [and make the bet often enough] you'll win money.

Make it 16 rolls, and you'll win more.

Make it 14 rolls, and over the long run you'll lose.

[Guest]

expectation means...']To be certain of seeing all six numbers requires an infinite number of throws.

However, you might see all six numbers after only six throws.

If you count the throws it takes to see all six numbers, and average that over a large number of trials,

you get what's called the expectation value.

Expectation gives an average of the things that might happen by chance.

It does not guarantee anything, but it gives you an idea of what will probably happen.

To clarify what expectation is, and why it's important: you'll lose money if you bet against it.

how you calculate it']To see one number you only have to roll the die once.

The next roll might give you a new number, but there is a 1/6 chance it wont.

You can say the next roll has a new-number effectiveness of only 5/6.

So you need [on average] 6/5 additional rolls to see a 2nd new number.

The next roll might give you a new number, but now there is a 2/6 chance it wont.

You can say the next roll has a new-number effectiveness of only 4/6.

So you need [on average] 6/4 additional rolls to see a 3rd new number.

The next roll might give you a new number, but now there is a 3/6 chance it wont.

You can say the next roll has a new-number effectiveness of only 3/6.

So you need [on average] 6/3 additional rolls to see a 4th new number.

The next roll might give you a new number, but now there is a 4/6 chance it wont.

You can say the next roll has a new-number effectiveness of only 2/6.

So you need [on average] 6/2 additional rolls to see a 5th new number.

The next roll might give you a new number, but now there is a 5/6 chance it wont.

You can say the next roll has a new-number effectiveness of only 1/6.

So you need [on average] 6/1 additional rolls to see a 6th new number.

So in total, you need

6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 = [60 + 72 + 90 + 120 + 180 + 360]/60 = 882/60 =

14.7 rolls of the dice to expect to see all six numbers.

If you bet even odds that you'll get all six numbers in 15 rolls of the dice, [and make the bet often enough] you'll win money.

Make it 16 rolls, and you'll win more.

Make it 14 rolls, and over the long run you'll lose.

I do see a problem here...

I haven't taken you to Vegas yet.

[Guest]

expectation means...']To be certain of seeing all six numbers requires an infinite number of throws.

However, you might see all six numbers after only six throws.

If you count the throws it takes to see all six numbers, and average that over a large number of trials,

you get what's called the expectation value.

Expectation gives an average of the things that might happen by chance.

It does not guarantee anything, but it gives you an idea of what will probably happen.

To clarify what expectation is, and why it's important: you'll lose money if you bet against it.

how you calculate it']To see one number you only have to roll the die once.

The next roll might give you a new number, but there is a 1/6 chance it wont.

You can say the next roll has a new-number effectiveness of only 5/6.

So you need [on average] 6/5 additional rolls to see a 2nd new number.

The next roll might give you a new number, but now there is a 2/6 chance it wont.

You can say the next roll has a new-number effectiveness of only 4/6.

So you need [on average] 6/4 additional rolls to see a 3rd new number.

The next roll might give you a new number, but now there is a 3/6 chance it wont.

You can say the next roll has a new-number effectiveness of only 3/6.

So you need [on average] 6/3 additional rolls to see a 4th new number.

The next roll might give you a new number, but now there is a 4/6 chance it wont.

You can say the next roll has a new-number effectiveness of only 2/6.

So you need [on average] 6/2 additional rolls to see a 5th new number.

The next roll might give you a new number, but now there is a 5/6 chance it wont.

You can say the next roll has a new-number effectiveness of only 1/6.

So you need [on average] 6/1 additional rolls to see a 6th new number.

So in total, you need

6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 = [60 + 72 + 90 + 120 + 180 + 360]/60 = 882/60 =

14.7 rolls of the dice to expect to see all six numbers.

If you bet even odds that you'll get all six numbers in 15 rolls of the dice, [and make the bet often enough] you'll win money.

Make it 16 rolls, and you'll win more.

Make it 14 rolls, and over the long run you'll lose.

Thanks for posting that, It was exactly what i was thinking. I deleted what I was going to post when the sides were unknown and I didn't want to take the time to figure out a formula while at work.

[bonanova]

Thanks for posting that, It was exactly what i was thinking. I deleted what I was going to post when the sides were unknown and I didn't want to take the time to figure out a formula while at work.

Ha! One of the benefits of being retired.