Expected Value Help

[livn4s]

If there are 7 balls in a machine, one ball is green and the others white, and one ball is drawn once a day for seven days over a period of 30 days, what will be the # of expected green balls for the month? Note: If a white ball is drawn it is removed from the machine for the rest of the week. If a green ball is drawn it is returned to the machine.

I have tried calculations and keep coming up with 4 but a friend is coming up with 7 and I don't know how they are arriving at 7. Any assistance will be helpful.

My calculation = 1/7 * 30 = 4.28

 

[Dr.Peterson]

We can start by observing why your simple answer isn't right: The probability of drawing a green ball on a given day is not always 1/7. In fact, since white balls are not returned, after 6 white balls have been drawn, you are certain to get a green ball every day from then on! So you can never get fewer than 23 green balls; it looks like your friend is wrong, too.

No, wait a minute ... what does it mean to draw a ball "once a day for seven days over a period of 30 days"??? Maybe the 30 is entirely irrelevant, and you just get 7 draws, not 30; you're certain to get at least 1 green ball. But your answer is still wrong.

Your turn: Try a more careful calculation.

 

[livn4s]

We can start by observing why your simple answer isn't right: The probability of drawing a green ball on a given day is not always 1/7. In fact, since white balls are not returned, after 6 white balls have been drawn, you are certain to get a green ball every day from then on! So you can never get fewer than 23 green balls; it looks like your friend is wrong, too.

No, wait a minute ... what does it mean to draw a ball "once a day for seven days over a period of 30 days"??? Maybe the 30 is entirely irrelevant, and you just get 7 draws, not 30; you're certain to get at least 1 green ball. But your answer is still wrong.

Your turn: Try a more careful calculation.

 

[livn4s]

Every seven days, the white balls are placed back in. So the same scenario will start again on week 2, week 3 and week 4. The white balls are removed during the week if drawn to ensure that the green ball will be drawn at least once weekly.

I think I may have figured it out. 1/7 *30 = One green ball out of seven * 30 days = 4.28 Then take 30/4.28 =7 . Min num of green balls a month = 4 and max num of green balls a month = 30. Assuming 30 days

 

[Steven G]

I have tried calculations and keep coming up with 4 but a friend is coming up with 7 and I don't know how they are arriving at 7. Any assistance will be helpful.

My calculation = 1/7 * 30 = 4.28

I am confused. You keep coming up with 4, yet your calculation comes up with a number other than 4 ?????????

 

[Steven G]

Every seven days, the white balls are placed back in. So the same scenario will start again on week 2, week 3 and week 4. The white balls are removed during the week if drawn to ensure that the green ball will be drawn at least once weekly.

I think I may have figured it out. 1/7 *30 = One green ball out of seven * 30 days = 4.28 Then take 30/4.28 =7 . Min num of green balls a month = 4 and max num of green balls a month = 30. Assuming 30 days

You say that you figured it out but I do not see your answer unless you meant the answer is between 4 and 30?
Please take your own advice which is the same scenario will start again on week 2, week 3 and week 4.
So find the expected number of green balls for a week.

 

[JeffM]

If, on days 1, 8, 15, 22, and 29, the situation is the same as on day 1, you can calculate the expected numbers for week 1, multiply by 4 and add the expected number for the two-day stub because expected values add.

The two-day stub is easiest. The probability of no green balls is

[MATH]\dfrac{6}{7} * \dfrac{5}{6} = \dfrac{30}{42} = \dfrac{210}{294}.[/MATH]
The probability of 2 green balls is

[MATH]\dfrac{1}{7} * \dfrac{1}{7} = \dfrac{1}{49} = \dfrac{6}{294}.[/MATH]
Thus the probability of 1 green ball is

[MATH]1 - \dfrac{210 + 6}{294} = \dfrac{294 - 216}{294} = \dfrac{78}{294}.[/MATH]
Thus the expected value for the stub is

[MATH]\dfrac{0 * 210 + 1 * 78 + 2 * 6}{294} = \dfrac{90}{294} = \dfrac{15}{49}.[/MATH]
There is a bunch of calculations to find the expected value for a week, but nothing conceptually difficult.

 

[JeffM]

Upon reflection, the calculation for a week is much trickier than I thought because things depend on when balls of different colors are chosen. I have not been able to find a generating function, which of course does not mean that such a function does not exist. The very simplest cases are obvious:

[MATH]\text {P(selecting no greens)} = 0 \text { and P(selecting green every day)} = \dfrac{1}{7^7}.[/MATH]
But things quickly get complicated.

[MATH]\text {P(selecting green on exactly 1 day)} =[/MATH]
[MATH]\dfrac{1 * (6 * 5 * 4 * 3 * 2 * 1)}{7 * (7 * 6 * 5 * 4 * 3 * 2)} + \dfrac{6 * 1 * (5 * 4 * 3 * 2 * 1)}{7 * 6 * (6 * 5 * 4 * 3 * 2)} + \ ... \ \dfrac{(6 * 5 * 4 * 3 * 2 * 1) * 1}{(7 * 6 * 5 * 4 * 3 * 2) * 1} =[/MATH]
[MATH]\dfrac{6!}{7!} * \sum_{d=1}^7 \dfrac{1}{8 - d} = \dfrac{1}{7} *\sum_{d=1}^7 \dfrac{1}{8 - d}.[/MATH]
[MATH]\text {P(selecting green on exactly 6 days)} =[/MATH]
[MATH]\dfrac{6}{7} * \dfrac{1}{6^6} + \dfrac{1}{7} * \dfrac{6}{7} * \dfrac{1}{6^5} + \ ... \ \dfrac{1}{7^6} * \dfrac{6}{7} =[/MATH]
[MATH]6 * \sum_{d=1}^7 \dfrac{1}{7^d * 6^{(7-d)}}.[/MATH]
And things go downhill from there. I do not see a closed-form formula. I'd have to attack it by cases. There are
(64 - 1) * 2 + 1 = 127 cases to consider. Without a formula in hand, I'd write a looper au denis or a spreadsheet. May do so tomorrow