lamford Posted December 21, 2012 Report Share Posted December 21, 2012 In Atlanta for Xmas. In a packet of cereal which my nephew and niece eat, one gets a small plastic model. For simplicity we shall say that it is one of the seven dwarfs. The chance of each dwarf being in a random packet is 1/7. I mused (yes I know there are nine of them) how many packets I would need to buy to be favourite to get a set of all seven dwarfs. I am sure that you can work it out after you make excuses for not playing board games with relatives over Xmas. Excel helps. Quote Link to comment Share on other sites More sharing options...
Fluffy Posted December 21, 2012 Report Share Posted December 21, 2012 you should state clearly that the number of packets is suposed to be infinite so the 1/7 remains constant, I guess you actually know but are ignoring on purpose the fact that all of this collections never has the same probability for all figures and one or 2 of them are always extremelly hard to find Quote Link to comment Share on other sites More sharing options...
lamford Posted December 21, 2012 Author Report Share Posted December 21, 2012 you should state clearly that the number of packets is suposed to be infinite so the 1/7 remains constant, I guess you actually know but are ignoring on purpose the fact that all of this collections never has the same probability for all figures and one or 2 of them are always extremelly hard to findYes that approximation is necessary. In practice I agree that the cereal manufacturers make one dwarf scarcer than the others, so that the dopey customers are not happy too soon. Quote Link to comment Share on other sites More sharing options...
Antrax Posted December 21, 2012 Report Share Posted December 21, 2012 The time until you get the ith dwarf for the first time is distributed geometrically with parameter (8-i)/7. So, the expected to get all seven is the sum of i from 1 to 7 on 7/(8-i), which is 363/20, or 18 and a bit. According to Markov's theorem (I think), the odds it'll take more than twice that are < 0.5, i.e. you're a favourite to get all seven once you purchase the 37th packet.I think. Quote Link to comment Share on other sites More sharing options...
lamford Posted December 21, 2012 Author Report Share Posted December 21, 2012 The time until you get the ith dwarf for the first time is distributed geometrically with parameter (8-i)/7. So, the expected to get all seven is the sum of i from 1 to 7 on 7/(8-i), which is 363/20, or 18 and a bit. According to Markov's theorem (I think), the odds it'll take more than twice that are < 0.5, i.e. you're a favourite to get all seven once you purchase the 37th packet.I think.I think it is a lot less than that. I get7 0.0061198998 0.0244795969 0.05770190510 0.10491255511 0.16309620112 0.2284524413 0.29730654514 0.36657492415 0.43391882616 0.49771982317 0.556972711as the probability of getting all 7 dwarfs in n packets. The first two agree when checked using brute force, so I hope they are all right. Quote Link to comment Share on other sites More sharing options...
Cascade Posted December 21, 2012 Report Share Posted December 21, 2012 I think it is a lot less than that. I get7 0.0061198998 0.0244795969 0.05770190510 0.10491255511 0.16309620112 0.2284524413 0.29730654514 0.36657492415 0.43391882616 0.49771982317 0.556972711as the probability of getting all 7 dwarfs in n packets. The first two agree when checked using brute force, so I hope they are all right. I concur with these. 7 0.0061198998 0.0244795969 0.05770190510 0.10491255511 0.16309620112 0.2284524413 0.29730654514 0.36657492415 0.43391882616 0.49771982317 0.556972711 Quote Link to comment Share on other sites More sharing options...
Cyberyeti Posted December 21, 2012 Report Share Posted December 21, 2012 I think it is a lot less than that. I get7 0.0061198998 0.0244795969 0.05770190510 0.10491255511 0.16309620112 0.2284524413 0.29730654514 0.36657492415 0.43391882616 0.49771982317 0.556972711as the probability of getting all 7 dwarfs in n packets. The first two agree when checked using brute force, so I hope they are all right.I agree with your numbers. Basically the working is: After 2 packets you have 1/7 chance of only one dwarf and 6/7 chance of 2 different. When you add a third packet you have 1/7 x 1/7 chance of only one dwarf, 1/7 x 6/7 (chance you had 1 and got a different one) + 6/7 x 2/7 (chance you had 2 different and matched one) chance of 2, 6/7 x 5/7 chance of 3 You continue this process onwards and you have an iterative formula which basically says that the chance of n dwarves from m+1 packets is (the chance of n dwarves from m packets) x n/7 + (the chance of n-1 dwarves from m packets) x (8-n)/7 Iterating this on a spreadsheet gives the numbers above. Quote Link to comment Share on other sites More sharing options...
gwnn Posted December 21, 2012 Report Share Posted December 21, 2012 http://en.wikipedia.org/wiki/Coupon_collector's_problem Quote Link to comment Share on other sites More sharing options...
Antrax Posted December 22, 2012 Report Share Posted December 22, 2012 Yeah, I woke up this morning with my heart racing: "Markov's inequality is not a tight bound!". Alas, the post had already been made :( 2 Quote Link to comment Share on other sites More sharing options...
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