Hard probability question

"A dice it thrown until the sum of the outcomes is greater than 300. Find, approximately, the probability that at least 80 experiments are required."

I have the suspicion that if
X = number of throws required until sum>300

then X has normal distribution with mean 85 and ... some variance http://www.thestudentroom.co.uk/imag...olondollar.gif

How could I prove that?

Also, I calculated the probability using another method and I got 0.765 as a result, but when I simulated the experiment I got 0.89 as a result ... :(

I doubt it's normal. We know that between 50 and 300 experiments are required, and the expected number of experiments is about 300/3.5, 85.7.

Edit: Between 51 and 301 (if you are considering "strictly greater than 300").

Quote:

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Originally Posted by smalltalk

"A dice it thrown until the sum of the outcomes is greater than 300. Find, approximately, the probability that at least 80 experiments are required."

I have the suspicion that if
X = number of throws required until sum>300

then X has normal distribution with mean 85 and ... some variance http://www.thestudentroom.co.uk/imag...olondollar.gif

How could I prove that?

Also, I calculated the probability using another method and I got 0.765 as a result, but when I simulated the experiment I got 0.89 as a result ... :(

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Hi Smalltalk,

Let's say the outcome of one die throw is . Then you want to find . For brevity, let's say .

Y is approximately normally distributed, by the Central Limit Theorem. You need to compute the mean of Y (it's not 85), and you need to find its standard deviation before you can look up the sought-for probability in a table.

Hint: .
Can you find the variance of ?