Calculating Average of 3 Ball Numbers Drawn from Bag

[ParisSpart]

In a bag there are 30 identical balls numbered from 1 to 30. Choose one after the other three balls (without Off Reset). What is the average value of the sum of the numbers of three balls chosen?

I am not sure on how i am going to solve this so i think that we will have a variable X
where X1 will be the number of first ball and X2 for the second and X3 for the third ball.,
so X=X1+X2+X3 and we want the average ,E(X)=E(X1)+E(X2)+E(X3) but i don't know how to estimate this. any ideas?

 

[jedishrfu]

Perhaps by looking at a smaller system like say throwing two die with sides 1 thru 6.

The sums wold range from 2 to 12 even though the die can fall 6 * 6 / 2 = 18 possible ways.

For 2: 1+1
For 3: 1+2
For 4: 1+3, 2+2
For 5: 1+4, 2+3
...
For 12: 6+6

 

[HallsofIvy]

It's very easy to find E(X1), it is just (1/30)(1+ 2+ 3+ ...+ 30)= (1/30)(30(31)/2= 31/2= 15.5 as we would expect. E(X2) is a little harder to find. There are now 29 balls left so the probability of choosing anyone of them is 1/29.

If the first ball drawn was 1, the expected value for the second ball is (1/29)(2+ 3+ 4+ ...+ 30)= (1/29)(1+ 2+ 3+ ...+ 30- 1)= (1/29)(30(31)/2- 1)= (1/29)(465- 1). Since the probability the first ball was 1 was 1/30, that contributes (1/30)(1/29)(465- 1)= 465/((29)(30))- 1/((29)(30)).

If the first ball drawn was 2, the expected value for the second ball is (1/29)(1+ 3+ 4+ ...+ 30)= (1/29)(1+ 2+ 3+ ...+ 30- 2)= (1/29)(30(31)/2- 2)= (1/29)(465- 2). Since the probability the first ball was 2 was 1/30, that contributes (1/30)(1/29)(465- 2)= 465/((29)(30))- 2/((29)(30)).

Do you see the pattern? Generally, if the first ball drawn was n, the expected value for the second ball is (1/29)(465- n). Since the probability the first ball was n was 1/30, that contributes 465/((29)(30))- n/((29)(30)).

For all 29 balls, since the first term is a constant, that adds to 465(29)/((29)(30)= 465/30= 93/6= 31/2= 15.5. The second term depends on n. Its sum will be (1/((29)(30))[1+ 2+ ...+ 30]= (1/((29)(30))[(30)(31)/2)= 31/29 (the sum is taken from 1 to 30, not 1 to 29 since ball 30 may still be in there).

So the expected value of E2 is 31/2- 31/29 which is about 14.3. The expected value of the sum of two balls is about 15.2+ 14.3= 29.5.

Can you do the same for the third ball?

 

[ParisSpart]

yea ok but my way is right for finding this?

 

[HallsofIvy]

What "way" are you talking about? All you did was state that "E(X)= E(X1)+ E(X2)+ E(X3)" which is true. And I used that in my solution.

 

[ParisSpart]

ok i will post here the E(X3) when i will find it

 

[jedishrfu]

It's very easy to find E(X1), it is just (1/30)(1+ 2+ 3+ ...+ 30)= (1/30)(30(31)/2= 31/2= 15.5 as we would expect. E(X2) is a little harder to find. There are now 29 balls left so the probability of choosing anyone of them is 1/29.

If the first ball drawn was 1, the expected value for the second ball is (1/29)(2+ 3+ 4+ ...+ 30)= (1/29)(1+ 2+ 3+ ...+ 30- 1)= (1/29)(30(31)/2- 1)= (1/29)(465- 1). Since the probability the first ball was 1 was 1/30, that contributes (1/30)(1/29)(465- 1)= 465/((29)(30))- 1/((29)(30)).

If the first ball drawn was 2, the expected value for the second ball is (1/29)(1+ 3+ 4+ ...+ 30)= (1/29)(1+ 2+ 3+ ...+ 30- 2)= (1/29)(30(31)/2- 2)= (1/29)(465- 2). Since the probability the first ball was 2 was 1/30, that contributes (1/30)(1/29)(465- 2)= 465/((29)(30))- 2/((29)(30)).

Do you see the pattern? Generally, if the first ball drawn was n, the expected value for the second ball is (1/29)(465- n). Since the probability the first ball was n was 1/30, that contributes 465/((29)(30))- n/((29)(30)).

For all 29 balls, since the first term is a constant, that adds to 465(29)/((29)(30)= 465/30= 93/6= 31/2= 15.5. The second term depends on n. Its sum will be (1/((29)(30))[1+ 2+ ...+ 30]= (1/((29)(30))[(30)(31)/2)= 31/29 (the sum is taken from 1 to 30, not 1 to 29 since ball 30 may still be in there).

So the expected value of E2 is 31/2- 31/29 which is about 14.3. The expected value of the sum of two balls is about 15.2+ 14.3= 29.5.

Can you do the same for the third ball?

Nice explanation and solution.

 

[ParisSpart]

for the third ball i will say that if the first and second ball was 1 and 2 ...
2 and 3..
?

 

[Ray Vickson]

It's very easy to find E(X1), it is just (1/30)(1+ 2+ 3+ ...+ 30)= (1/30)(30(31)/2= 31/2= 15.5 as we would expect. E(X2) is a little harder to find. There are now 29 balls left so the probability of choosing anyone of them is 1/29.

If the first ball drawn was 1, the expected value for the second ball is (1/29)(2+ 3+ 4+ ...+ 30)= (1/29)(1+ 2+ 3+ ...+ 30- 1)= (1/29)(30(31)/2- 1)= (1/29)(465- 1). Since the probability the first ball was 1 was 1/30, that contributes (1/30)(1/29)(465- 1)= 465/((29)(30))- 1/((29)(30)).

If the first ball drawn was 2, the expected value for the second ball is (1/29)(1+ 3+ 4+ ...+ 30)= (1/29)(1+ 2+ 3+ ...+ 30- 2)= (1/29)(30(31)/2- 2)= (1/29)(465- 2). Since the probability the first ball was 2 was 1/30, that contributes (1/30)(1/29)(465- 2)= 465/((29)(30))- 2/((29)(30)).

Do you see the pattern? Generally, if the first ball drawn was n, the expected value for the second ball is (1/29)(465- n). Since the probability the first ball was n was 1/30, that contributes 465/((29)(30))- n/((29)(30)).

For all 29 balls, since the first term is a constant, that adds to 465(29)/((29)(30)= 465/30= 93/6= 31/2= 15.5. The second term depends on n. Its sum will be (1/((29)(30))[1+ 2+ ...+ 30]= (1/((29)(30))[(30)(31)/2)= 31/29 (the sum is taken from 1 to 30, not 1 to 29 since ball 30 may still be in there).

So the expected value of E2 is 31/2- 31/29 which is about 14.3. The expected value of the sum of two balls is about 15.2+ 14.3= 29.5.

Can you do the same for the third ball?

When I do it I get $E(X_2) = E(X_1),$ because if $S = \sum_{k=1}^{30} k$, then
E(X_2) = \frac{1}{30} \sum_{n=1}^{30} \left( \frac{S - n}{29}\right)<br /> = \frac{1}{30} \frac{1}{29} ( 30 S - S) = \frac{1}{30} \frac{29 S}{29} = E(X_1).
However, the same result can be obtained essentially without calculations. In fact, if we draw all 30 balls one-by-one without replacement, and if $X_j$ is the number of the jth drawn ball, we have $E(X_j) = E(X_1)$ for all $j \geq 2$. Note that we even have $E(X_{30})= E(X_1).$

 

[ParisSpart]

and E(X3) is equal with them?

 

[haruspex]

and E(X3) is equal with them?

If you mean the expected value of the third ball, yes. Ray's point is that the probability distribution for the value of a ball does not depend on where in the sequence of draws it is. Therefore the expected value is the same for all. What's not quite so obvious is that the expected value of the sum is the sum of the expected values, even though the drawn values are not independent.

 

[Bacle2]

If you mean the expected value of the third ball, yes. Ray's point is that the probability distribution for the value of a ball does not depend on where in the sequence of draws it is. Therefore the expected value is the same for all. What's not quite so obvious is that the expected value of the sum is the sum of the expected values, even though the drawn values are not independent.

I think expectation is always linear.