[Guest]

On average, how many times do you need to roll six dice together to see all six different numbers turn up within a single such group roll?

can any1 solve this, and show working ?

[Molly Mae]

On average, how many times do you need to roll six dice together to see all six different numbers turn up within a single such group roll?

can any1 solve this, and show working ?

I think I can.

Let's roll them sequentially. The first die doesn't matter at all. You have a 6/6 chance of rolling a unique number. The next die, you have a 5/6 chance of rolling a unique number. Then 4/6 (2/3), 3/6 (1/2), 2/6 (1/3), and 1/6.

Just multiply 5/6*2/3*1/2*1/3*1/6.

A quick calc gets me .0154319. About 1.54%.

I rounded.

EDIT: I've assumed that you're asking for one of each number to turn up. I've calculated the odds of rolling once and turning up all unique numbers.

As far as your question, though, if you're asking how many times you would have to roll to see all six numbers...I don't think there can be an answer. You can roll a die a billion times. Statistically, you'll have rolled every number--but that doesn't mean that you did.

Edited September 8, 2010 by Molly Mae

[Guest]

can any1 solve this, and show working ?

This sounds like homework to me.

[Guest]

There are 46,656 possible outcomes(6^6) every roll. If I am not mistaken (and I probably am) all six numbers will come up 720 times (6*5*4*3*2*1) in differant orders. If we divide our totals 46,656/720=64.8 or 1.54%.

If you somehow rolled a unique combo everytime you would see all 6 numbers in 65 rolls.

Edited September 8, 2010 by kkehoe5

[Guest]

at 45 Rolls you have a 50.3% chance of rolling all 6 numbers - I won't show you how - you can do your own homework

[EventHorizon]

We can examine the dice one by one without changing the answer.

It doesn't matter what is rolled on the first die (probability of uniqueness = 6/6).

Since 1 value is taken now, there are only 5 possibilities for the next die (probability of uniqueness = 5/6).

Assuming die 2 is not equal to die 1 (prob=5/6)and since 2 values are now taken, there's 4 possibilities left for the next die (probability of the next die being unique = 4/6). Multiplying these results in (4*5*6)/(6*6*6) = 20/36.

Following the same logic, the final probability for a given roll to have all unique numbers showing is 6!/6^6 = 0.01543...

To get the average number of trials you simply take the reciprocal of the probability of the occurrence (1/0.01543...=64.8).

So 64.8 rolls are needed on average before you see a roll with all unique numbers showing.

We want each number to come up. The probability for each one is 1/6, so that's 1/(6^6).

We can permute which die has which number. The number of permutations is 6p6 = 6!/(6-6)! = 6!.

Multiplying the probability with specific dice by the number of permutations is 6!/(6^6) = 0.01543...

Take the reciprocal, etc.

Cheesner's answer is actually the median number of rolls needed.

We want the probability after the number of rolls to be 1/2. The probability that we do not get the event to occur is 1-0.01543....= .98456...

We need to see how many times we don't get the event before the probability reaches 1/2, so .98456^x=1/2.

log (.98456^x) = log (1/2)

x*log (.98456) = log (1/2)

x = log (1/2) / log (.98456)

x = 44.568...

So the median is about 45 rolls.

[Guest]

I did this as an experiment and set up an EXCEL spreadsheet to tabulate everything. After having the spreadsheet "roll the six dice" 1,000 times I got 2.9% as the "unique" rolls with an average 34 rolls between the "unique" rolls.

Edited September 8, 2010 by labrat

[Guest]

Eventhorizon, I didn't believe your post (its been about 25 years since college statistics) so I ran a simulation in excel using a random number generator. In 50 trials I got average 62 and median 46. Nice job!

Edited September 8, 2010 by Cheesner

[Guest]

The answer provided must be a number of rolls, not a percent or a probability.

So, we must calculate an average, which is equal to the sum of [outcome1]*[probability1] + [outcome2]*[probability2], etc.

p_success for one trial = .01543

p_fail for one trial = 1-p_success

Answer = 1*(probability of success on first roll) + 2*(probability of fail on first and success on second) + 3*(probability of 2 failures followed by success), etc to infinity;

Answer = sum(i from 1 to infinity) of terms (i*p_success*p_fail^(i-1)), which, when calculated all out, is 1/p_success = 64.8 rolls, which is troublesome because you cannot roll something that last .8 times.

[Guest]

The answer provided must be a number of rolls, not a percent or a probability.

So, we must calculate an average, which is equal to the sum of [outcome1]*[probability1] + [outcome2]*[probability2], etc.

p_success for one trial = .01543

p_fail for one trial = 1-p_success

Answer = 1*(probability of success on first roll) + 2*(probability of fail on first and success on second) + 3*(probability of 2 failures followed by success), etc to infinity;

Answer = sum(i from 1 to infinity) of terms (i*p_success*p_fail^(i-1)), which, when calculated all out, is 1/p_success = 64.8 rolls, which is troublesome because you cannot roll something that last .8 times.

[Guest]

No need to be troubled by non integer EV.

If you roll one die the expected value of the roll is 3.5

For the current case the probability that you achieve one or more succesful rolls in 65 tries is 1-(1-0.015432099)^65 =~ 0.636111

[EventHorizon]

Since it has been used a couple times in this thread, I thought I'd give the proof for why expected number of trials before a success is equal to the reciprocal of the probability of a success.

mean = p + 2p(1-p) + 3p(1-p)^2 + ... <---As strombone gave, average = sum of [outcome1]*[prob1] + [outcome2]*[prob2] + ...

mean / p = 1 + 2(1-p) + 3(1-p)^2 + ... <---divide both sides by p

mean / p = sum(i=1 to inf) i * (1-p)^(i-1) <--- summation notation

mean / p = d( integrate( sum(i=1 to inf) i * (1-p)^(i-1) ) )/dp <--- derivative of integral = original equation

mean / p = d( sum(i=1 to inf) -1* (1-p)^(i) )/dp <--- integrate

mean / p = -d( sum(i=1 to inf) (1-p)^(i) )/dp <--- factor out -1

mean / p = -d( (1-p)/(1-(1-p)) )/dp <---sum of geometric series = (first value)/(1-ratio)

mean / p = -d( (1-p)/p )/dp <--- simplify

mean / p = -d( (1/p) - 1 )/dp <--- pull out 1

mean / p = -(-1/(p^2)) <---derivative

mean / p = 1/(p^2) <---simplify

mean = 1/p <---multiply both sides by p

Done!