Bubble Gum CardsDate: 09/28/97 at 14:57:22 From: Rebecca Skrine Subject: Probability The bubble gum company decides to promote its gum by including in each pack the photo of one of six rock music stars. Assuming that there are equal numbers of photos of each of the six stars and that when you buy a pack of gum your chances of getting any of the six photos are the same, about how many packs of gum would you expect to buy to get all six photos? Date: 09/28/97 at 18:59:43 From: Doctor Anthony Subject: Re: Probability To get the first new photo you only have to buy one pack of gum. Now the probability of getting a new photo is 5/6 with each pack you buy. The expected number of purchases to get the second photo is calculated from the sum of the series: 1 x prob of needing only 1 purchase + 2 x prob of needing two purchases, and so on to infinity since in theory you could go on getting the same old photo for evermore. Expected puchases to get second new photo = 1(5/6) + 2(1/6)(5/6) + 3(1/6)^2(5/6) + 4(1/6)^3(5/6) + ........... = (5/6)[1 + 2(1/6) + 3(1/6)^2 + 4(1/6)^3 + .......... The series inside the square bracket is of the form: S = 1 + 2x + 3x^2 + 4x^3 + ...... where x = 1/6 Multiply S by x xS = x + 2x^2 + 3x^3 + ...... ----------------------------------- and subtract the two series (1-x)S = 1 + x + x^2 + x^3 + ..... which is an infinite GP (1-x)S = 1/(1-x) and so S = 1/(1-x)^2 = 1/(5/6)^2 = (6/5)^2 And now the expected number of purchases is (5/6)(6/5)^2 = 6/5 Having got two photos, the probability of a fresh photo is 4/6 with each new purchase, and by an argument identical to that above, the expected number of purchases to next new photo is 6/4. Then with probability 3/6 of a new photo, the expected number of purchases will be 6/3, then 6/2 and finally 6/1. Total expected purchases to get complete set of photos is: 1 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 6[1/6 + 1/5 + 1/4 + 1/3 + 1/2 + 1/1] = 6 x 49/20 = 14 and 7/10 So after about 15 purchases you should have a complete set of photos. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 09/29/97 at 21:46:42 From: Anonymous Subject: Re: Probability Thanks so much for your help. Now I understand! |
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