Please help me solve this problem...?

Question

A die is continuously rolled until the total sum of all rolls exceeds 200. What is the probability that at least 65 rolls are necessary?

Please solve this problem and explain the steps for me...

Answer 1

I've thought of a way to do it, which is easier than what I was thinking before. You need to use the discrete probability distribution for multiples rolls of a die.

The probability is given by:

Prob(s,i,k) = 1/s^i * Sum[ (-1)^n * Com[i,n] * Com[k-s*n-1,i-1], n=0, n=(k-i)/s ]

or, in English, 1/s^i, multiplied by the Riemann sum over n, for n from 0 to (k-i)/s, of the quantity (-1)^n times the combination of i and n, times the combination of k-s*n-1 and i-1.

where s = how many sides the die has, i = the number of rolls, k = the sum of all the rolls, and n = a variable to be summed over.

Recall that
Com[x,y] = x! / (y! * (x-y)!)
where ! indicates the factorial.

So, the probability of rolling a 6-sided die 65 times and getting a total of exactly 200 is:
Prob[6,65,200] = 0.003959

However, we're interested in 65 rolls or MORE, so we need to do a Riemann sum:
Sum[Prob[6,r,200],r=65,r=200]
We know it cannot require more than 200 rolls, so we sum from r=65 to r=200.

However, there is the possibility of rolling a total of 201, 202, 203, 204, or 205, so we need to do ANOTHER Riemann sum, to include all possible totals.

Sum[ Sum[ Prob[6,r,t], r=65, r=200 ], t=200, t=205 ]

When this calculation is performed (I used Mathematica to do it), the answer obtained is 0.08391.

I don't like this answer, because it doesn't take into account the fact that rolling a 205 is impossible unless the very last roll is a 6. This means that 5/6 of the sequences that add up to 205 are not possible, seeing as how the rules indicate that we stop as soon as we hit 200. Therefore, I would modify the equation as follows:

Sum[ Sum[ Prob[6,r,t] * (206-t)/6, r=65, r=200 ], t=200, t=205 ]

which shows that ALL sequences totaling 200 should be counted, but only 5/6 of sequences totaling 201 should be counted, 4/6 of sequences totaling 202 should be counted, 3/6 of sequences totaling 203 should be counter, etc.

The number obtained from this, and my FINAL answer, is...

...drum-roll please...

0.0428366

I cannot guarantee this is correct, but I think it is. Time will tell if anyone can spot a flaw. I can tell you for certain that the guy who answered above me did not answer the question you were asking. You may be put off by my lengthy analysis, or even my silhouette avatar, but my answer is founded on solid theory, whereas his result is simply "your chances are very good that 65 rolls will be enough". Mine is an actual number that did not rely on opinion or guesswork.

I used Wikipedia to dig up the probability distribution:
http://en.wikipedia.org/wiki/Dice#Probability

In response to the guy below me, I would counter that humans (myself included) are notoriously bad at using intuition to solve probability questions. Just look at how many people don't understand the Monty Hall problem!

Additionally, consider that if you used a normal distribution approximation of 65+ events, the standard deviation would be very small. Stdev goes as 1/sqrt(N). This means that for a large number of trials (rolls of the die), the standard deviation is very small. The distribution is concentrated heavily around the mean, with only small amounts spread out far away. 65 could very well be more than 1 standard deviation from the mean.

Answer 2

Sherry, Sherry Baby, Girl you make me lose my mind. Why don't you come out with that Red dress on?

The average number of a die is 3.5.
1+2+3+4+5+6 = 21
21/6 = 3.5
200/3.5 = 57.14 times to add up to 200.
57.14/65 = 0.879, or an 87.9% probability.

So, your chances are really good that 65 rolls will do the trick, since 57.14 times is all that is needed, on the average.

Statistically speaking, you're in the money, with 7.86 rolls to spare!

Contrary to the following answer, I stand by my analysis. My friend Wiki-Pete may dazzle you with his formulas and his self-proclaimed mathematical knowledge, but he can't even explain what he is looking up.

The fact remains, if you use only one die, (you did state one die, by not stating the plural, dies or dice), and roll it, the number you will get will average out to 3.5. And by rolling one die at least 58 times, on average, you will aquire a sum of 200, or more, with an 87.9% probability.

I think he must have studied some kind of new math that they've been teaching these past few years. I learned the hard way on a slide-rule. By the way, a slide-rule got us to the Moon and back.

Answer 3

Though lithiumdeuteride's analysis is impressive, I simply cannot fathom that the answer would be as low as 4%. It would seem to me that the answer would be higher as approximately 58 rolls would leave you with an (approximate) 50% chance of receiving a total sum with a value of more than 200. Therefore the answer would be in the lower percentile range (I think).

Response to lithiumdeuteride: Eloquently put; I concede the point. Additionally, I never took statistics so I really wouldn't know.