# A probability to 6 in a dice

I have a question about probability of a dice:
I'm reading know about the history of probability and I'm reading about the chance to get $$6$$ at dice of $$\{1,2,3,4,5,6\}$$.

If you get $$6$$ you win, otherwise - you lose....
The question is, how many times you need to roll the dice to get a chance for a win that is greater then $$\frac{1}{2}$$.

The solution I read is: $$1-\left(\frac{5}{6}\right)^n$$ (and the answer is $$n=4$$...)

My question is:
Why the solution is not $$\frac{n}{6}$$ Because here also $$n=4$$ will give chance of getting win with more then $$\frac{1}{2}$$.

If you can explain me where is my mistake it will be great.

::EDIT::
I see what I missed! I need to use the Inclusion–exclusion principle
Now I'm understand where I wrong... :)

Thank you!

• "The question is, how many times you need to roll the dice to a chance ..." - To a chance of what exactly? What chance/event are you trying to get a valuation for? – Eevee Trainer May 12 '19 at 8:32
• I fix my question :-) – CS1 May 12 '19 at 8:32
• Please don't self-delete your question after getting an answer. – GNUSupporter 8964民主女神 地下教會 May 13 '19 at 20:31
• I understand my mistake... So if it's OK to delete it - I'll be glad :) – CS1 May 13 '19 at 23:52
• As an aside, the answer of $\frac{n}{6}$ is the correct answer to a different question. $\frac{n}{6}$ counts the expected value of the number of sixes thrown in $n$ throws of a fair six-sided die. This is, as mentioned already, not the same thing as the probability of getting at least one six. – JMoravitz May 13 '19 at 23:59

An easy way to see your solution is wrong is to evaluate it for $$n = 6$$ or $$n = 7$$.
The deviation between your answer and the correct answer has to do with the fact that you can throw a 6 more than once. If we want to check the chance for $$n = 2$$, we can count the amount of cases where we threw a 6 on the first throw - $$\frac{1}{6}$$-th of them. Similarly, in one sixth of the cases we will throw a 6 on the second throw. However, the amount of cases where we threw a 6 on the first or second throw is not $$\frac{1}{6} + \frac{1}{6}$$, because we counted the case where we threw a six on both the first throw and second throw twice. So
$$P = \frac{1}{6} + \frac{1}{6} - \frac{1}{36} = \frac{11}{36} = 1 - (\frac{5}{6})^2.$$