[bonanova]

On average, how many times must I roll a fair six-sided

die before all the numbers 1, 2, 3, 4, 5, 6 show up?

[Guest]

On average, how many times must I roll a fair six-sided

die before all the numbers 1, 2, 3, 4, 5, 6 show up?

1/ (6/5 * 5/6 * 4/6 * 3/6 * 2/6 * 1/6 ) = 64.8 times?

[Guest]

1/ (6/5 * 5/6 * 4/6 * 3/6 * 2/6 * 1/6 ) = 64.8 times?

Close, but I think it should be

6/6+6/5+6/4+6/3+6/2+6/1=14.7

[Guest]

probability says 14.7 times. practicality says 15 times - seeing that it is difficult to roll a die .7 times....

edited to wrap in spoiler..

Edited September 20, 2008 by AAAsn888s

[Guest]

There is no average. It is a trick question (possibly unintentionally). Probability may say 14.7 (I haven't actually done the math[i used CR ans AAAsn888s' answers]) but that is in a theoretically perfect world. In the real world you may throw 1-6 every time you try and therefore the average is 100% or you may never get it giving an average of 0%. This is because each roll has it's own chance; it doesn't get more likely every time you miss. This is also assuming the dice roll will result in a random outcome which isn't possible. I don't beleive in a random number. How could it happen? If you roll a die, the outcome depends on the velocity you throw the die, the angle it is thrown at, which side is facing up, the surface it will roll on, air resistance, die size, obstacles, die weight, gravitational affects, initial die height (sorry, i think like a scientist). Same occurs with coin flips and I won't even start on 'inny, minny, miney, mo'.

[Guest]

There is no average. It is a trick question (possibly unintentionally). Probability may say 14.7 (I haven't actually done the math[i used CR ans AAAsn888s' answers]) but that is in a theoretically perfect world. In the real world you may throw 1-6 every time you try and therefore the average is 100% or you may never get it giving an average of 0%. This is because each roll has it's own chance; it doesn't get more likely every time you miss. This is also assuming the dice roll will result in a random outcome which isn't possible. I don't beleive in a random number. How could it happen? If you roll a die, the outcome depends on the velocity you throw the die, the angle it is thrown at, which side is facing up, the surface it will roll on, air resistance, die size, obstacles, die weight, gravitational affects, initial die height (sorry, i think like a scientist). Same occurs with coin flips and I won't even start on 'inny, minny, miney, mo'.

To the best of current scientific knowledge, random numbers do exist in real life. They are most difficult to avoid when you are observing the universe on a very, very small scale. http://en.wikipedia.org/wiki/Quantum_physi...al_consequences. What most scientists believe is that, even if you knew absolutely everything about our universe to perfect precision, it would still behave in a random, probabilistic way.

I should also point out that, even if the rolling of a dice were completely deterministic (ie, the result could be determined exactly if you knew all the inputs exactly), the results would be indistinguishable from a random event because there is no practical way to precisely control all those variables you mentioned when throwing a dice by hand.

[Guest]

It is not possible to calculate such an avarage. Because you may need infinite rolling to get each number show up. If an event is a compound of many events, and if probability of one of them is infinite, then avarage probability of main event goes up to infinite. Sure not infinite in practice, but it can not be estimated. (I think so )

[Guest]

21

[bonanova]

For those who have difficulty with what "on average" means, use this meaning.

Roll a single die until you have seen all six numbers, and write down the number R of rolls it took.

Repeat the process a large number of times and take the average of R.

That gives you a good estimate of "on average".

And you can see from this description that the average [not necessarily an integer] exists and is finite.

The average can be computed based on probabilities for a fair die, as Chuck Rampart did.

Some responses answered this question "How many times must one roll a single fair die to be certain all six numbers show?"

That's a different question.

No finite number of rolls will insure that.

The die could come up "1" on every roll, for example.

The difference between certainty and expectation of an outcome based on chance is basically this.

If you bet money on outcomes that have a favorable expectation, then over time you will win, and v.v.

In this case, take CR's answer [it's not an integer].

If you bet that you would show all six numbers in the next higher integer rolls of a die, you will win money over time.

If you bet that you would show all six numbers in the next lower integer rolls of a die, you will lose money over time.

Hope that helps explain logical expectation in practical terms.

[Guest]

It is not possible to calculate such an avarage. Because you may need infinite rolling to get each number show up. If an event is a compound of many events, and if probability of one of them is infinite, then avarage probability of main event goes up to infinite. Sure not infinite in practice, but it can not be estimated. (I think so )

Of course an average can be calculated. Do you not think there is a certain probability that red will be the next color a roulette ball lands on? There is. On a standard roulette wheel with two green spots the probability that the ball will land on red is 47.37%. The fact that the ball may land on black for every spin for all of eternity is irrelevant. If such an average or probability could not be calculated, no one would be foolish enough to spend the massive amounts of money it takes to run a casino. The house edge keeps them in business.

[bonanova]

It is not possible to calculate such an avarage. Because you may need infinite rolling to get each number show up. If an event is a compound of many events, and if probability of one of them is infinite, then avarage probability of main event goes up to infinite. Sure not infinite in practice, but it can not be estimated. (I think so )

Hi nobody,

Here's a way to see past the "may need infinite" difficulty that you allude to.

It's counterbalanced by the equally infinite number of trials you will have to perform before the infinitely long trial occurs.

That is, the infinite difficulty is tempered by the infinitessimal probability that the difficulty will occur.

So averaging over all possible trial outcomes can still give you a finite result even when some outcomes are very unfavorable.

Also, take the argument that "if you bet against someone who claims he can

show all six numbers in say 20 rolls of the die you will lose money" to show

that the average number has to be less than 20.

p.s. don't take that bet.

Hope that helps.

- bn

[Prime]

Now, I begin to feel insecure. The OP labels this problem "simple" and Chuck Rampart gives a formula for solving without giving any rational. Is the problem really so trivial and obvious? Is the solution so evident, it requires no proof?

Although 14.7 seems to be the solution, I feel an urgency to obfuscate the issue.

RATIONAL 1:

On the first roll any number will do, so the probability to roll needed number is 1 and the average number of rolls it takes is 1.

After that, there are 5 numbers out of 6 that serve our purpose, and the probability to roll one is 5/6. The average number of trials to get the event with the probability 5/6 is 6/5. I am going to leave out the proof that I see for now, as it involves geometric series. (Or is there a simpler derivation?)

After that 4 out of 6 numbers fitting our purpose are left. The average number of rolls to get one is 6/4.

Then 3/6 with the average number of rolls to get it – 6/3.

Then 2/6 with the average number of rolls 6/2.

And, finally, 1/6 which takes on average 6 rolls to get. (Same geometric series are needed to prove that seemingly obvious point.)

Adding up the average numbers of rolls for each consecutive step:

1 + 6/5 + 6/4 + 6/3 + 6/2 + 6 = 14.7.

That seems as a plausible rational, but I don't see it as a proof. Is overall average truly a sum of individual averages thusly defined?

Here is an attempt at obfuscation. Although, calculations in spreadsheet do not support the following line of reasoning:

RATIONAL 2:

It takes on average 6 rolls to roll any individual number. Same as in RATIONAL 1, I will leave out the proof.

So we need an average of 6 rolls to roll number 1, for example.

Then we also need 6 rolls for the number 2 on the die. But since we already have made 6 rolls in the step above and used only one of those six, we have 5 rolls, which we can re-apply. Thus we need just one additional roll to get the required average number of rolls for the number 2 to show up.

If we apply the same reasoning to all six numbers, we get the total number of rolls we need: 6+5=11.

The true formula, which I would consider, proved is a sum of infinite series:

6*P₆ + 7*P₇ + 8*P₈ + …

Where P_n is the probability to have all six numbers to show up exactly after n rolls.

How does it add up to the same result as in RATIONAL 1, I don't see yet.

[bonanova]

Now, I begin to feel insecure. The OP labels this problem "simple" and Chuck Rampart gives a formula for solving without giving any rational. Is the problem really so trivial and obvious? Is the solution so evident, it requires no proof?

Although 14.7 seems to be the solution, I feel an urgency to obfuscate the issue.

RATIONAL 1:

On the first roll any number will do, so the probability to roll needed number is 1 and the average number of rolls it takes is 1.

After that, there are 5 numbers out of 6 that serve our purpose, and the probability to roll one is 5/6. The average number of trials to get the event with the probability 5/6 is 6/5. I am going to leave out the proof that I see for now, as it involves geometric series. (Or is there a simpler derivation?)

After that 4 out of 6 numbers fitting our purpose are left. The average number of rolls to get one is 6/4.

Then 3/6 with the average number of rolls to get it – 6/3.

Then 2/6 with the average number of rolls 6/2.

And, finally, 1/6 which takes on average 6 rolls to get. (Same geometric series are needed to prove that seemingly obvious point.)

Adding up the average numbers of rolls for each consecutive step:

1 + 6/5 + 6/4 + 6/3 + 6/2 + 6 = 14.7.

That seems as a plausible rational, but I don't see it as a proof. Is overall average truly a sum of individual averages thusly defined?

Here is an attempt at obfuscation. Although, calculations in spreadsheet do not support the following line of reasoning:

RATIONAL 2:

It takes on average 6 rolls to roll any individual number. Same as in RATIONAL 1, I will leave out the proof.

So we need an average of 6 rolls to roll number 1, for example.

Then we also need 6 rolls for the number 2 on the die. But since we already have made 6 rolls in the step above and used only one of those six, we have 5 rolls, which we can re-apply. Thus we need just one additional roll to get the required average number of rolls for the number 2 to show up.

If we apply the same reasoning to all six numbers, we get the total number of rolls we need: 6+5=11.

The true formula, which I would consider, proved is a sum of infinite series:

6*P₆ + 7*P₇ + 8*P₈ + …

Where P_n is the probability to have all six numbers to show up exactly after n rolls.

How does it add up to the same result as in RATIONAL 1, I don't see yet.

If the probability of an event for a single trial is p, then it takes, on average 1/p trials to achieve the event.

The desired events are, in turn,

1. any number to appear. p = 1. Trials needed = 1/1 = 1.
   
2. a different number to appear. p = 5/6. Trials needed = 6/5.
   
3. a different number to appear. p = 4/6. Trials needed = 6/4.
   
4. a different number to appear. p = 3/6. Trials needed = 6/3.
   
5. a different number to appear. p = 2/6. Trials needed = 6/2.
   
6. a different number to appear. p = 1/6. Trials needed = 6/1.
   

But your interpretation of this is that it's somehow adding up averages to get another average.

You say:"Is overall average truly a sum of individual averages thusly defined?"

Think of it, instead, of adding up the expected effort of performing six individual tasks: 1+6/5+6/4+6/3+6/2+6/1 = 14.7

In your second rationale, you seem to require a particular number to appear first,

requiring 6 rolls, then somehow conclude that in the next 5 rolls all the others will appear.

I tried but failed to follow your reasoning. So I can't reasonably discuss it. Sorry.

You mention some geometric series. Care to elaborate?

[Prime]

...

If you bet that you would show all six numbers in the next higher integer rolls of a die, you will win money over time.

If you bet that you would show all six numbers in the next lower integer rolls of a die, you will lose money over time.

...

I am going to pick an issue here. I presume, you mean an even money bet. But the average number of trials for an event to occur and 50/50 chance are not the same thing.

In a fair coin toss, you need on average two tosses for it landing heads up. And yet a single toss has probability 1/2 for heads.

We know that it takes on average 6 rolls of a fair die to roll 6. I'll gladly take an even money bet on 6 showing up within just 5 consecutive rolls, or even just four rolls.

[Guest]

Now, I begin to feel insecure. The OP labels this problem "simple" and Chuck Rampart gives a formula for solving without giving any rational. Is the problem really so trivial and obvious? Is the solution so evident, it requires no proof?

Although 14.7 seems to be the solution, I feel an urgency to obfuscate the issue.

RATIONAL 1:

On the first roll any number will do, so the probability to roll needed number is 1 and the average number of rolls it takes is 1.

After that, there are 5 numbers out of 6 that serve our purpose, and the probability to roll one is 5/6. The average number of trials to get the event with the probability 5/6 is 6/5. I am going to leave out the proof that I see for now, as it involves geometric series. (Or is there a simpler derivation?)

After that 4 out of 6 numbers fitting our purpose are left. The average number of rolls to get one is 6/4.

Then 3/6 with the average number of rolls to get it – 6/3.

Then 2/6 with the average number of rolls 6/2.

And, finally, 1/6 which takes on average 6 rolls to get. (Same geometric series are needed to prove that seemingly obvious point.)

Adding up the average numbers of rolls for each consecutive step:

1 + 6/5 + 6/4 + 6/3 + 6/2 + 6 = 14.7.

That seems as a plausible rational, but I don't see it as a proof. Is overall average truly a sum of individual averages thusly defined?

Here is an attempt at obfuscation. Although, calculations in spreadsheet do not support the following line of reasoning:

RATIONAL 2:

It takes on average 6 rolls to roll any individual number. Same as in RATIONAL 1, I will leave out the proof.

So we need an average of 6 rolls to roll number 1, for example.

Then we also need 6 rolls for the number 2 on the die. But since we already have made 6 rolls in the step above and used only one of those six, we have 5 rolls, which we can re-apply. Thus we need just one additional roll to get the required average number of rolls for the number 2 to show up.

If we apply the same reasoning to all six numbers, we get the total number of rolls we need: 6+5=11.

The true formula, which I would consider, proved is a sum of infinite series:

6*P₆ + 7*P₇ + 8*P₈ + …

Where P_n is the probability to have all six numbers to show up exactly after n rolls.

How does it add up to the same result as in RATIONAL 1, I don't see yet.

When people talk about an "average" of probabilities, the strictly correct term is "expected value." The expected value of a random variable X is denoted E(X). One of the properties of an expected value is that it behaves linearly - ie, E(aX+bY)=aE(X)+bE(Y). See http://en.wikipedia.org/wiki/Expected_value#Properties

So E(# rolls to get one number + #rolls to get a different number + ... sixth number) = E(one number)+...+E(sixth number).

Finding E(nth number) from a geometric series is pretty simple, to give you the answer.

[Guest]

I'd have to argue that the answer should be 100%. the question was worder that the numbers only must show up, it said nothing about stopping on them. I'm fairly certain I have never thrown a dice and not had every number be on top/appear at least once.

[Prime]

...If the probability of an event for a single trial is p, then it takes, on average 1/p trials to achieve the event.

...

You mention some geometric series. Care to elaborate?

Bona,

You and CR rely on theorems of probability, which I don't see as a common knowledge on this forum. Case in point: average trials to achieve an event of probability P. If you ask how many rolls of die on average does it take to roll "six" (or any other individual number), most people will answer -- 6. But can you prove it? (That's where geometric series comes in.)

With respect to this problem, I stand my ground: the basic formula for the average is infinite series:

6*P₆ + 7*P₇ + 8*P₈ + …

Where P_n is the probability to have all six numbers to show up exactly after n rolls.

This formula relies only on the basic definition of average. It must add to 14.7 in this case. However, each individual P_n is hard to estimate.

[bonanova]

Bona,

You and CR rely on theorems of probability, which I don't see as a common knowledge on this forum. Case in point: average trials to achieve an event of probability P. If you ask how many rolls of die on average does it take to roll "six" (or any other individual number), most people will answer -- 6. But can you prove it? (That's where geometric series comes in.)

With respect to this problem, I stand my ground: the basic formula for the average is infinite series:

6*P₆ + 7*P₇ + 8*P₈ + …

Where P_n is the probability to have all six numbers to show up exactly after n rolls.

This formula relies only on the basic definition of average. It must add to 14.7 in this case. However, each individual P_n is hard to estimate.

Totally agree that proofs are more useful than opinions.

Let's do one.

You refer to the following as a theorem of probability:

If the probability of a favorable outcome after one trial is p, it will take on average 1/p trials to obtain it.

Fair enough. Let's see if we can prove it, at least within the context of this puzzle.

After showing say a '2' on the first roll we want a 1, 3, 4, 5 or 6 to show next.

There are 6 possible outcomes; 5 are favorable.

If we were to roll the die 6 times we would expect 5 favorable results [5 of 6 equally likely events are favorable].

Number of trials: n=6.

Number of favorable outcomes: f=5.

There are two things we can now say:

The probability of a favorable outcome = f/n = 5/6

The number of trials for a favorable outcome = n/f = 6/5.

The first quantity is p. The second quantity is 1/p.

Thus, if the probability of a favorable result after one trial is p, then it takes 1/p trials to expect the favorable result.

[bonanova]

I'll gladly take an even money bet on 6 showing up within just 5 consecutive rolls, or even just four rolls.

You're on.

I just did ten trials of four rolls. You got your 6 on four of the trials.

I just repeated the ten trials. You got your 6 on five of them.

That's 9 wins on 20 bets. I think that means I won.

By the way,

if you'd bet on 1 you'd have won 7 out of 20 times.

if you'd bet on 2 you'd have won 7 out of 20 times.

if you'd bet on 3 you'd have won 11 out of 20 times.

if you'd bet on 4 you'd have won 10 out of 20 times.

if you'd bet on 5 you'd have won 11 out of 20 times.

On 17 of the 20 trials a number came up more than once.

Keep in mind that if you bet on 6 and 6 comes up twice - say the four rolls give 5 3 6 6 - you get paid only once.

But if a number you didn't bet on comes up more than once, your odds get longer.

[bonanova]

I'd have to argue that the answer should be 100%. the question was worder that the numbers only must show up, it said nothing about stopping on them. I'm fairly certain I have never thrown a dice and not had every number be on top/appear at least once.

The question was how many rolls would it take for all numbers to show?

The answer should be a number; your answer is a percentage.

Are you saying that on 100% of the rolls of a fair die all six numbers show?

[Prime]

...

If we were to roll the die 6 times we would expect 5 favorable results [5 of 6 equally likely events are favorable].

...

The probability of a favorable outcome = f/n = 5/6

The number of trials for a favorable outcome = n/f = 6/5.

...

I was referring to proving the above statements. Anyway, we could make a separate post of that as a basic (not necessarily simple) probability problem.

It could be formulated something like: Prove that it takes on average 6 rolls of a die to roll "6". (I see it as considerably easier problem than the one given here.)

I'm about to post, what I think is rather difficult probability problem. If you haven't encountered it before, you'll enjoy it.

[Guest]

Bona,

You and CR rely on theorems of probability, which I don't see as a common knowledge on this forum. Case in point: average trials to achieve an event of probability P. If you ask how many rolls of die on average does it take to roll "six" (or any other individual number), most people will answer -- 6. But can you prove it? (That's where geometric series comes in.)

With respect to this problem, I stand my ground: the basic formula for the average is infinite series:

6*P₆ + 7*P₇ + 8*P₈ + …

Where P_n is the probability to have all six numbers to show up exactly after n rolls.

This formula relies only on the basic definition of average. It must add to 14.7 in this case. However, each individual P_n is hard to estimate.

I agree that I used theorems of probability to arrive at my answer which a lot of people on this forum won't know. In this case, I think the result is somewhat intuitive, but in other cases it isn't. In many of those other cases, that's the point (ie Monty Hall Problem).

You're also right with your 6*P₆ + 7*P₇ + 8*P₈ + … formula - that is another way to solve the problem, albeit much harder.

[Prime]

You're on.

I just did ten trials of four rolls. You got your 6 on four of the trials.

I just repeated the ten trials. You got your 6 on five of them.

That's 9 wins on 20 bets. I think that means I won.

By the way,

if you'd bet on 1 you'd have won 7 out of 20 times.

if you'd bet on 2 you'd have won 7 out of 20 times.

if you'd bet on 3 you'd have won 11 out of 20 times.

if you'd bet on 4 you'd have won 10 out of 20 times.

if you'd bet on 5 you'd have won 11 out of 20 times.

On 17 of the 20 trials a number came up more than once.

Keep in mind that if you bet on 6 and 6 comes up twice - say the four rolls give 5 3 6 6 - you get paid only once.

But if a number you didn't bet on comes up more than once, your odds get longer.

The bet is simply: if you roll dice 4 times in a row and "6" didn't come up -- you win. If "6" came up -- I win. (No need to complete 4 rolls, when "6" comes up).

The probability not to roll "6" in 4 consecutive rolls is simply (5/6)⁴, which is less than 1/2. (Even though it takes 6 rolls on average to roll "6").

[Guest]

I used a different approach, and thought I would get the same answer as Chuck. I did not....

I calculated the number of times rolling a k sided die n times does, and does not, yield all k values. I then solved for the value of n that yields equal probability of either outcome.

Total number of outcomes of n rolls of a k sided die: k^n

Number of times at least one number is not rolled: k(k-1)^n

The difference is the number of times all numbers are rolled: k^n - k(k-1)^n

If you equate the last two, and solve for k=6, the answer is 13.6, not 14.7.

O_o

[Guest]

It could be considered this way...

To determine simple averages, (forget the die for a moment) you add up all possible outcomes and divide by the number of outcomes. In theory, you could roll a die any number of times without getting all six numbers. You could roll forever and only get 3's! With infinity (as well as every integer) in the averaging, on average you would never roll all six. It really depends on your definition of "average" in the question.

[Prime]

...

Number of times at least one number is not rolled: k(k-1)^n

...

That formula is incorrect.

Formula for the number of times at least one number is not rolled is more complex and yields smaller number of combinations.