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Fun little statistics problem..

S

Syringer

Lifer
No it's not for homework or anything..classes don't start for me until Thusday..soo here it is:

There's a game where you roll one dice at at a time, and you keep rolling until you roll a 1. Your score is the sum of your rolls..if you roll a 1 on your first try, you get 0. For instance if you roll 5, 3, 4, 1, your score is 12.

What will be your long term average score?
 
I haven't taken statistics but you can calculate the probability of rolling one on the nth turn as
1/6*(5/6)^(n-1).

Also, since it's a fair die, all outcomes are equiprobable so when you don't lose you'll average a score of 4/turn.

Now I assume (and this is the part I'm not sure of since I haven't taken a stats class) that the average numbers of rolls you'll get before you lose is n for which you have a 50% probability of winning

so we can do a quick sum and solve the equation

0.5 = 1/6*sum((5/6)^n,n=0..N)

and we find that N ~= 2.80 (I did it with Maple, but you can solve this manually without too much trouble since it's a sum of a geometric series).

So your average score will be 4*2.80 = 11.2

Thats my solution, is it right?
 
Short version:

The average of 2, 3, 4, 5, 6 (i.e. the rolls where the game continues) is 4, so you can assume you get a 4 on every turn in which the game continues. The expected number of rolls before you roll a 1 is 6 (1/(1/6)). Therefore, you expect to roll 5 times in which the game continues and then roll a 1. Therefore, the average score is 5*4=20.



Long version:

Let n be the number of rolls before you roll a 1. Once again, you can assume you roll a 4 whenever you don't roll a 1. Therefore, the expected value of your score is an infinite sum as such:

E = 4*1*5/6*1/6+4*2*(5/6)^2*1/6+4*3*(5/6)^3*1/6+....

= SUM(4*1/6*n*(5/6)^n,n,1,infinity) = 2/3*SUM(n*(5/6)^n,n,1,infinity)

That summation is kinda weird, but you can compute it like such (let X be the sum):

X = 1*(5/6) + 2*(5/6)^2 + 3*(5/6)^3 + 4*(5/6)^4 + ....
5/6*X = 1*(5/6)^2 + 2*(5/6)^3 + 3*(5/6)^4 + ...

Subtracting those two equations gives us:

1/6*X = (5/6)^1+(5/6)^2+(5/6)^3+(5/6)^4+.....

which is just an infinite geometric series with ratio 5/6.

Therefore 1/6*X = (5/6)/(1-5/6) = 5, so X=30.

Therefore the expected score is E=2/3*30 = 20.

Any questions?
 
Originally posted by: chuckywang
Short version:

The average of 2, 3, 4, 5, 6 (i.e. the rolls where the game continues) is 4, so you can assume you get a 4 on every turn in which the game continues. The expected number of rolls before you roll a 1 is 6 (1/(1/6)). Therefore, you expect to roll 5 times in which the game continues and then roll a 1. Therefore, the average score is 5*4=20.

EDIT: Long version in progress.
Click to expand...

That part definitely isn't right.
you're misinterpreting the 1/6, 1/6 is the average proportion of 1's you will roll compared to other numbers, what you're looking for is the average number of rolls it takes before it pops up.

Let me put it this way. Imagine you roll once at a time but regroup those rolss into 6's.
So you'd say roll 1-6 are group 1, 7-12 is group 2, 13-18 is group 3, etc. Logically you'd assume that if you average a large number of these groups, you'd get one occurance of 1 per group. What you're implying these 1's would all appear on the sixth roll, which is clearly see is wrong.

 
Originally posted by: chuckywang

The expected number of rolls before you roll a 1 is 6 (1/(1/6)).
Click to expand...

What I mean by this statement is this:

Say you conduct an experiment when you record the number of rolls it takes before you see a "1".
If you conduct a large number of these experiments (say 100 to 1000), and take the average of all your recorded number of rolls before you see a "1", you would get something close to 6.

While it is true that every six rolls, you would expect to get one "1", that isn't relevant to the problem. In reality, sometimes every six rolls you would get two "1"'s and sometimes you would get no "1"'s. But on average, it would take six rolls to get a "1".

The difference in what we are saying is precisely the difference between the Binomial random variable and the Geometric random variable. It is the Geometric random variable that is relevent here, not the Binomial.

Geometric r.v. asks how many rolls it would take before you see a "1"
Binomial r.v. asks if you roll n times, how many times would you see a "1".
 
your summation is of the wrong form
your probability of rolling a 1 on your first roll is 1/6
your probability of rolling twice and getting a 1 only on the 2nd roll is 5/6*1/6
your probability of rolling 3 times a 1 only on the 3rd roll is (5/6)^2*1/6
etc.

So your probability of rolling a 1 within N rolls is 1/6*SUM((5/6)^n, n=0..N-1)
 
Originally posted by: RaynorWolfcastle
your summation is of the wrong form
your probability of rolling a 1 on your first roll is 1/6
your probability of rolling twice and getting a 1 only on the 2nd roll is 5/6*1/6
your probability of rolling 3 times a 1 only on the 3rd roll is (5/6)^2*1/6
etc.

So your probability of rolling a 1 within N rolls is 1/6*SUM((5/6)^n, n=0..N-1)
Click to expand...


You are absolutely correct. To calculate the expected value, you have to multiply that probability with the score you would get in each case, and then sum over all cases.
 
ahhh I see what you did now. Your explanation that was a bit off and confused me. What you actually did is take a weighted average of the scores for the (infinite set of) probabilities. That makes sense 🙂

Nonetheless, the average tosses before a 1 shows up si something like 3.8, not 6.
 
Solve the summation for an N for which you have 50% probability of having gotten a 1

Nevermind, I'm wrong. 😱 I wrote a quick Java program that simulated it to check it out.
 
You can get a 1 on the 1st roll with prob. 1/6
You can get a 1 on the 2nd roll with prob. 5/6*1/6
You can get a 1 on the 3rd roll with prob. (5/6)^2*(1/6)
.
.
.

Therefore the expected value of rolls to get a 1 is SUM(n*(5/6)^(n-1)*1/6,n,1,infinity) = 6.

Perhaps the confusion lies in the fact that "average" does not mean "have a 50% probability of"

 
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