# Throwing a ball

#### lgon

##### New Member
Imagine a game where you throw a ball in to a box where are three sticks; the probability of the ball to throw down one stick is 75%, two sticks is 20% and the three sticks is 5%
Additionally, when you throw the ball, it passes through a random accelerating mechanism that can make the ball pass in the box one, two or three times, with probabilities of 30%, 50% and 20%; each time thee ball re-enters the box, all the sticks are up, independently of the previous time
What is the probability of, in one shot, the maximum number of sticks being thrown down is 6? And 2?
And what is the probability of, in one shot, the minimum number of sticks being thrown down is 2?

#### Dason

Hi! :welcome: We are glad that you posted here! This looks like a homework question though. Our homework help policy can be found here. We mainly just want to see what you have tried so far and that you have put some effort into the problem. I would also suggest checking out this thread for some guidelines on smart posting behavior that can help you get answers that are better much more quickly.

#### lgon

##### New Member
I see... Well... is not a homework because I've been there for 20+ years. I was just trying to confirm my reasoning.
My results are: 99%, 66% and 34%.
Hope it is correct and sorry for the annoyance

#### lgon

##### New Member
And my next question would be:
Assuming that the ball as passed once in the box and has thrown down 1 stick, what is the new probability of throwing down at least 6 sticks?

#### lgon

##### New Member
And my answer would be 1%; that means the first outcome doesn’t change the probability (in this case).

In the first scenario we have the following outcomes

(1) (2) (3)
(1) 1 2 3
(2) 2 4 6
(3) 3 6 9

With the probabilities given by the matrix multiplication

[75%] x [30% 50% 20%] =
[20%]
[5% ]

[22.5% 37.5% 15.0%]
[ 6.0% 10.0% 4.0%]
[ 1.5% 2.5% 1.0%]

Resulting in cumulative probabilities of (probability of the outcome being less or equal to):
1 22.5%
2 66.0%
3 82.0%
4 92.5%
6 99.0%
9 100.0%

In the second scenario, we already know that the first time the ball entered the box it has thrown down 1 stick. But the probability of the ball enters again in the box is the same, as it is the probability in that turn to thrown 1, 2 or 3 sticks. So IMHO, the 6.0% and 1.5% probabilities in the matrix above given to the outcomes 2 and 3 in the first round of the ball will be now probabilities assigned to the outcome “1”.

Therefore, the new matrix would be:

[30.0% 37.5% 15.0%]
[ 0% 10.0% 4.0%]
[ 0% 2.5% 1.0%]

Resulting in cumulative probabilities of (probability of the outcome being less or equal to):
1 30.0%
2 67.5%
3 82.5%
4 92.5%
6 99.0%
9 100.0%

#### Mean Joe

##### TS Contributor
And my answer would be 1%; that means the first outcome doesn’t change the probability (in this case).

In the first scenario we have the following outcomes

(1) (2) (3)
(1) 1 2 3
(2) 2 4 6
(3) 3 6 9
This doesn't look right. You are thinking that if a ball goes through twice, then the possible scores are only 2, 4, or 6. But actually 3 or 5 is also possible. This table is entirely missing the possibility of 7 or 8 as well.

My solution wouldn't be so elegant. It would be like:
Let Probability of 5 = P(5)
Then
P(5) = P(5 when going thru twice) + P(5 when going thru three times)
= P(2 and 3 or 3 and 2) + P(1,1,3 or 1,2,2 or 1,3,1 or 2,1,2 or 2,2,1 or 3,1,1)
= 0.20 * (0.50*0.20 + 0.20*0.50) + 0.05 * (0.30*0.30*0.20 + 0.30*0.50*0.50 + 0.30*0.20*0.30 + 0.50*0.30*0.50 + 0.50*0.50*0.30 + 0.20*0.30*0.30)

#### lgon

##### New Member
You're right. My reasoning was flawed.
But how can we put the solution in an analytical way? If instead of 3^3=27 outcomes we had 5^5=3125 or 10^10= 10 billion outcomes, how could we calculate the odds?