# Hard probability question

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• July 21st 2012, 07:13 PM
smalltalk
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 ... :(
• July 22nd 2012, 12:09 AM
richard1234
Re: Hard probability question
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").
• July 22nd 2012, 06:33 AM
awkward
Re: Hard probability question
Quote:

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 ... :(

Hi Smalltalk,

Let's say the outcome of one die throw is $X_i$. Then you want to find $\Pr(\sum_{i=1}^{80} X_i \le 300)$. For brevity, let's say $Y = \sum_{i=1}^{80} X_i$.

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: $var(Y) = \sum_{i=1}^{80} var(X_i)$.
Can you find the variance of $X_i$?
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