On average how many rolls will it take to roll a 6?

[rudolfstr]

We have a simple, fair, 6 sided dice, and we ask the question - how many times will i have to roll it (on average) to get a 6? I made an experiment (although with only 50 rolls) and i got, that the average was around 5.4.

 

[daveb]

If you want to do it mathematically, let n = the number of rolls it takes to get a 6. What is it for n = 1, 2, etc? Can you figure it out from there?

 

[rudolfstr]

If you want to do it mathematically, let n = the number of rolls it takes to get a 6. What is it for n = 1, 2, etc? Can you figure it out from there?

Yes that is very simple . The probability to get a six in the first n rolls=sum from i=1 to n of 5^i-1/6ⁱ

 

[rcgldr]

It might be easier to express this as 1 - (odds of not getting at least one 6 in n rolls) = 1 - (5/6)ⁿ.

 

[Norwegian]

Immediate intuition tells me the answer should be 6, and also generalize in the obvious way to an N-sided dice. Summation of the appropriate series shows that this is true, but is there a more conceptual proof? I will think about this while trying to sleep, and challenge you all to beat me to it Googling is not allowed. Good night!

 

[skiller]

If you are asking the highest probability of rolls it will take to get your first 6, then the most likely outcome is the 1st roll, as that only has a probability of 1/6. The probability of any roll before the first 6 has to be multiplied by 5/6.

But I think you were asking the former, so 1/6.

 

[rudolfstr]

any progress on that? Still strugaling with this :(

 

[micromass]

The average rolls to gets a 6 is exactly 6. This is a special case of a geometric distribution: http://en.wikipedia.org/wiki/Geometric_distribution

In general: let the probability of event A be p, and let the probability of event B be 1-p. Then the average number of trials before getting A, will be 1/p.

 

[skiller]

any progress on that? Still strugaling with this :(

Not sure if you're talking to me, but if you are, then actually, yes, I am still struggling. I've got an A in O-Level and A-Level and 3 degrees in Maths, but for some reason my mind has gone blank for this! ...but then I am very, very... drunk!

The highest probability to get the first 6 is in the first throw; and the highest probability to get a 6 at all is for an infinitely many throws.

But that goes against my intuition for how many to get the first 6!

To be fair, my maths skills were never in the probability area!

 

[rudolfstr]

Not sure if you're talking to me, but if you are, then actually, yes, I am still struggling. I've got an A in O-Level and A-Level and 3 degrees in Maths, but for some reason my mind has gone blank for this!

The highest probability to get the first 6 is in the first throw; and the highest probability to get a 6 at all is for an infinitely many throws.

But that goes against my intuition for how many to get the first 6!

To be fair, my maths skills were never in the probability area!

Yeah, although I'm only a student in school I wouldn't say I' m bad at maths, I thought this problem would be pretty straight forward, but there seems to be a lot of problems with this. Another problem is that experimentally i allways get something below 6, like 5.6, 5.5, 5.4.

 

[daveb]

The average rolls to gets a 6 is exactly 6. This is a special case of a geometric distribution: http://en.wikipedia.org/wiki/Geometric_distribution

In general: let the probability of event A be p, and let the probability of event B be 1-p. Then the average number of trials before getting A, will be 1/p.

True, but I assumed the OP wasn't aware of this, so was going from first principles, developing the probability distribution, then taking the expectation value of the distribution.

 

[jwatts]

micromass has the right answer. The average number of rolls will be 6. Your experiment is perfectly described by a geometric random variable. It says the probability (in this case) of rolling a 6 in σ rolls is ((1-(1/6))^σ-1)(1/6). As you can see, as you increase the number of times tried the probability of rolling a six increases. If you then take the expectation of that probability ( in other words how many times you expect to roll the die before you get a 6) is 1/p where p is the probability of rolling a 6. The probability of rolling a 6 will always be 1/6 since the experiment is independent. So the expected number of rolls will be 1/1/6=6. I think you may be having trouble with the difference between the probability of rolling a 6 and the probability of rolling a 6 in more than one roll. As I said before the individual probability will always be 1/6. Just like no matter what you do the probability of flipping a Head as opposed to a Tail is always 1/2. But let's say now that you flip the coin twice. Now there are 4 possibilities (H,H),(H,T),(T,H),(T,T). The probability of getting anyone of these is 1/4 for two similar reasons. First off there are 4 equal possibilities and 1 outcome you want, therefore 1/4. Also the probability of getting a Head or Tail is 1/2 for the first flip and for the second flip so you get (1/2)(1/2)=1/4. But here's the difference so pay attention. Even though the probability of getting a Head is always 1/2 you can clearly see you expect to get a head 3/4 of the time. So as you increase the number of trials you increase the number of times you expect to see a particular outcome. Make sense?

 

[haruspex]

but is there a more conceptual proof?

Perhaps the easiest way is to let the answer be x and consider the first roll. If you don't get a 6, then the average number of additional rolls is still x:
x = 1 + 0*1/6 + x*5/6

 

[I like Serena]

Perhaps the easiest way is to let the answer be x and consider the first roll. If you don't get a 6, then the average number of additional rolls is still x:
x = 1 + 0*1/6 + x*5/6

Nice!