Free gifts
A breakfast cereal includes a free gift is included in each box. There are six different gifts. Assuming there is an equal chance of getting any one of the gifts, what is the expected number of boxes you need to buy in order to get all six gifts?
Labels: Probability
posted by Chris at
1:56 PM
16 Comments:
"six different gifts"
"to get all four gifts?"
I am confused.
Two gifts are the same?
Chris musta been in the scotch ...
Ooops. OK I've fixed it. I'm more used to the scotch being in me.
I think it could be 15
6/6+
6/5+
6/4+
6/3+
6/2+
6/1 =14.7, thus 15 boxes
But I may just not have a clue!!
Regards, Curtis
Well done Curtis. I was expecting that one to last longer. Only nit-pick is that you shouldn't round to 15. On average (i.e. over an infinite number of trials), it really is 14.7 (exactly) boxes.
Does anyone fancy explaining what Curtis's expression is saying?
hi Chris...i'll nit-pick a little more
Curtis might be right to round up to 15,as you did not say anything about "on average" in the question.you asked for the "expected" number,and one can not "expect" to but .7 boxs :)
Hi Pequod. Expectation is a formal term for average. I.e. I've just been a bit cheeky.
The question inherently involves averages. Each of the terms that Curtis used is the average number of boxes that have to be opened to get the nth present. There's no wriggling out of it, 15 is not the right answer, regardless of how vaguely my question was phrased.
I made the point because rounding up to an integer is the correct thing to do in some problems, but not this one.
Another take on it: you might get all 6 presents using just 6 boxes, or you might need an infinite number. So, if we're not dealing with the average, what is the meaning of 14.7 (or 15) as an answer?
It's because of this, that I asked if anyone would like to explain what Curtis's equation was all about. That he was looking at an average value would become clear.
thanks Chris...that should keep her quiet for a while
It's a pleasure to have been of service. LOL
I have used this one before. But the average roll of a fair die is 3.5 - not 3 and not 4.
... now I've lost the plot.
The probability of getting the 1st pressie in the 1st box is 1 = 6/6. There is now a 5/6 probability of getting the 2nd pressie in the second box, then 4/6 in 3rd, 3/6 in 4th, 2/6 in 5th and 1/6 in 6th.
So average number of boxes to open, using number of boxes = 1/probability of next pressie in next box (hence my previous "Probability" post) is:
1/(6/6) + 1/(5/6) + 1/(4/6) + 1/(3/6) + 1/(2/6) + 1/(1/6)
= 6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 = 14.7
Bit rough. But I'm lazy, and besides, who's taking any notice?
"Probability" blog was actually "Average Number" blog.
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