# We continuly throw one dice until we get a specific result for first time

I have a problem with exercise. Basically, I am not sure if I am right and how should it be. That's why if you can solve it so I will understand. I am reading this but I have difficulties and I am trying alone. The exercise says:

We throw continually one dice until we have succeeded for the first time a result below than 3.What is the possibility, to be more than 3 tries, until succeed the given result(the number to be smaller than 3)?

What I did I take this type from possibilities of Geometry: $$f(x)=p(1-p)^{x-1}$$ but I am not sure until now if this is the type or another. I believe this is it from what I have read

• If I understand you, you are asking for the expended number of tosses you have to make until you get a result less than $3$, yes? Are you including that last toss or not? That is, if you toss, in succession, $5, 6, 5, 2$ is the answer (in that case) four or three? Either way, this is an example of a Geometric Distribution. – lulu Sep 11 at 10:24
• Sentences start with capital letters. There is a space between each sentence. Sigh. – Klangen Sep 11 at 10:26
• i have been confused little bit with your comments and answers – s.m Sep 11 at 10:34
• What exactly don't you understand? If you have a question about a specific answer, please leave a comment below that answer. – Toby Mak Sep 11 at 10:36
• @TobyMak i did it :) – s.m Sep 11 at 10:41

You require the score on each of the first 3 tries to be in {3,4,5,6}.

The probability is therefore $$(\frac {4}{6})^3$$.

• The question says "What is the possibility, to be more than 3 tries, until succeed the given result(the number to be smaller than 3)" .Example it can be 8 times the dice i throw it i am not saying only the 3 times .I am not looking for the first 3 tries.I am looking for only the first try i got a number less than 3.When i got 1 the number or 2 the number then i will have success.It maybe be 8 trial .I hope you got me now.I think you counted the numbers of the dice 3,4,5,6 . – s.m Sep 11 at 10:41
• No, you only care about the first 3 tries. Whatever happens after those 3 tries don't matter. For example, it would not matter if you roll a die 4 times or 8 times, since both of them have taken over 3 tries. – Kyky Sep 11 at 10:55
• thanks a lot man ,so it isnt Geometric Distribution as i thought? – s.m Sep 11 at 10:56
• Thanks @Kyky, I was trying to think of a way to explain that which wouldn't cause any confusion! – S. Dolan Sep 11 at 11:00
• @S.Dolan man sorry for asking again but what you thought to solve this ?(i mean type ) etc .It isn't Geometric Distribution what it is i mean – s.m Sep 11 at 11:08

You want the probability that you don't succeed (in getting a number below $$3$$) in the first three trials. This means you want the first three outcomes to be $$3,4,5$$ or $$6$$. The answer is $$(\frac 4 6)(\frac 4 6)(\frac 4 6)=(\frac 2 3)^{3}$$

• why should i want 3,4,5,6? and not until 10 for example? – s.m Sep 11 at 10:45