# the problem of the 3 balls

#### thanro

##### New Member
Hello everyone. I have a problem, or at least, I am stuck to find a formula that would work and apply all the time to find the probability for each ball to be possessed by a certain person. Here the explanation:

I have 3 bags. Red, blue and green bag.
Red and Blue bags have 6 balls each.
Red’s balls are numbered 1 to 6
Blue’s balls are numbered 7 to 12
Green bag has 9 balls, numbered 13 to 21.

First thing I do is I take out one ball from each of the bags and the 3 balls are put in a small purple bag. All other balls are put all together into a bigger white bag.

Then 6 people pick 3 balls each - without seeing them so you cannot tell how many per color (you can end up with 3 red, or 2 blue and a green for example) - from the white bag.

Now, let’s think at ball 1, originally in the red bag.
I can say that the probability for it being in the purple bag is 16.66% (1/6).

But what is the probability for it to belong to person number 1?

It’s here that I am lost.
If I knew (but I don’t!) that this ball is for sure in the white bag, and I only consider the probability to be picked by one person, then am I right to say that each ball has a 3/18th of chance (so still 16.66%) to be picked.

But because first of all I can't tell if ball 1 is in white bag and if that is not the case, then it can be picked by someone else, how can I tell each ball’s probability to belong to each person?

I hope I made sense. I would love to see a formula for it, cause I want to see how this % changes if I change the variables
variables are for example,
I know that the ball is for sure in the white bag or
I know that the first two people did not pick it or
Person number 4 has already picked two balls that are not number 1 (during the game, balls are slowly revealed)

I think it's a quite easy problem for you experts, possibly so a very big thank you in advance to whoever can help me!

#### obh

##### Active Member
Hi Thanro,

I can say that the probability for it being in the purple bag is 16.66% (1/6).
But what is the probability for it to belong to person number 1?

1. As you wrote: the probability that ball #1 will be in the white bag is: 5/6 , (1-1/6)
2. If the ball in the white bag and 6 persons peak up 3 balls from the 18 in the white bag.
So the probability any ball will be peaked up from the white bag by person x is 1/6.

So the probability that ball #z will be picked up by person #x is 5/6 * 1/6
Since you are talking about ball number it doesn't really matter what is the color.

Is that make sense?

#### thanro

##### New Member
thank you obh

ok, above makes sense.

question 1) the 1/6 can change to 1/5 for example if I know that one person surely did not pick #z.

(during the game I will know some of the balls people pick...)

now, question 2) I understand this is the general possibility for a ball to be held by a specific person, but let's say, I know that person A has for sure ball #2, so only 2 picks to find #1... but person B has only 1 pick left. (and C,D,E,F no picks left).

Can I say the below? Or am I mixing things up?

probability that ball #1 will be picked up by person #A: (5/6 * 1/2) * 1/2
probability that ball #1 will be picked up by person #B: (5/6 * 1/2) * 1/3

?

And what if I knew that balls #3 & #4 from red bag are for sure not in the purple bag as they have been picked up by someone.

So the formula will be (still for ball #1)

probability that ball #1 will be picked up by person #A: (3/4 * 1/2) * 1/2
probability that ball #1 will be picked up by person #B: (3/4 * 1/2) * 1/3

Am I correct?

#### obh

##### Active Member
Hi Thanro,

1. Correct 1/5
2. I'm not sure I understand the question... if the only options for ball #1 are Person1: 2 picks, Person2: 1 pick.
so the probability that person1 picks the ball is: 2/(2picks + 1pick)=2/3
and the probability that person1 picks the ball is: 1/(2picks + 1pick)=1/3

#### thanro

##### New Member
thank you obh

about question 2... what I am trying to say is, can I combine the two probabilities?

I believe the number of picks left influence the probability for a person to have a specific ball. Or am I wrong?

we said:
probability that ball #z will be picked up by person #x is 5/6 * 1/6
probability that person1 picks the ball is: 2/(2picks + 1pick)=2/3

probability for person1 to pick ball#1 is then 5/6 * 1/6 * 2/3?
probability for person2 to pick ball#1 is then 5/6 * 1/6 * 1/3?

is this a correct assumption?

#### obh

##### Active Member
maybe try to write is as words instead of numbers

P(person1 to pick ball#1) = P(ball1 in white bag)*p(person1 pick ball#1 | (ball in white bag)

If you know that person1 or person2 will pick the ball the 1/6 is not relevant.
Or did I not understand your question?

#### thanro

##### New Member
Hi, yes well in the example I knew that person1 or person2 will pick the ball and indeed the 1/6 is not relevant, but let's assume I do not know that...

Let's say I don't know yet which balls are in the white or purple bag and I start with all people with the 3 picks to be done.

I can say that each probability for each "combination" ball/person is:

1-1/(number of total balls in the original bag - number of balls surely in white bag) * 1/ (6 - number of people that for sure can't have it) * (picks left per person/total picks left by all people).

I would like to set up an excel table where each cell is a combination ball/person, so I need a formula as starting point which will change each time for example I exclude a certain person has a specific ball or I know they picked a specific ball. So the variables change continuously.

is the above formula correct?