A return to Problems of the Week.
So I’ve been quite quiet on the blogging front recently. But, having recently recovered from one of those pesky travel-related illnesses (which I was vaccinated for back in Australia; it seems the vaccine didn’t work too well) I am now back and saying hello to anyone reading this again.
Last night a neat little mathematics problem entered my head, so I’ve decided to post it up here. But before I get to that, I’ll also include a quick English problem:
Should the title of this post be “Problems of the Weeks”? I mean, one problem of the week is simply the Problem of the Week. If I had many problems, all in the same week, then they would be Problems of the Week. But, as is the case here, where I have multiple problems spread out over multiple weeks, what is the correct grammatical description?
Now that the English-nerds have something to keep them occupied, I’ll move on to the much more entertaining mathematics problem.
Suppose I am collecting basketball cards. There are n cards in the set, and every time I purchase a card I have an equal chance (1 in n) of receiving any particular card. The first time I get a double (ie receive a card that I already have a copy of) is when I receive my 50th card. What is the most likely (integer) value for n?
Stated another way, the problem is this: Positive integers between 1 and n inclusive are uniformly randomly generated with replacement. The first time a number is generated that has already been generated is on the 50th generation. What is the MLE (maximum likelihood estimate) for n?
Good luck! I have no idea what the answer is myself, I haven’t stopped to think about it yet. This weekend I am heading off to Los Angeles for a week’s worth of holidays (woohoo!) so I’ll try to provide an answer by the end of that. Cheers.
Last night a neat little mathematics problem entered my head, so I’ve decided to post it up here. But before I get to that, I’ll also include a quick English problem:
Should the title of this post be “Problems of the Weeks”? I mean, one problem of the week is simply the Problem of the Week. If I had many problems, all in the same week, then they would be Problems of the Week. But, as is the case here, where I have multiple problems spread out over multiple weeks, what is the correct grammatical description?
Now that the English-nerds have something to keep them occupied, I’ll move on to the much more entertaining mathematics problem.
Suppose I am collecting basketball cards. There are n cards in the set, and every time I purchase a card I have an equal chance (1 in n) of receiving any particular card. The first time I get a double (ie receive a card that I already have a copy of) is when I receive my 50th card. What is the most likely (integer) value for n?
Stated another way, the problem is this: Positive integers between 1 and n inclusive are uniformly randomly generated with replacement. The first time a number is generated that has already been generated is on the 50th generation. What is the MLE (maximum likelihood estimate) for n?
Good luck! I have no idea what the answer is myself, I haven’t stopped to think about it yet. This weekend I am heading off to Los Angeles for a week’s worth of holidays (woohoo!) so I’ll try to provide an answer by the end of that. Cheers.
5 Comments:
Well you people out there are no fun at all. The answer I just discovered is 1208, and the problem only took about 10 minutes with the help of excel.
However I shall give you all a chance to redeem yourselves with a second problem, a little similar but probably easier.
Most people who have studied probability are familiar with the classic birthday problem; that is how many random people need to be in a room until there is a greater than 50% chance that there is at least one shared birthday. The answer is 23. My problem is related to this.
Say we are start with 1 person in a room, and are looking for shared birthdays. We keep adding people to the room, one by one, and stop when we finally get a shared birthday. What's the most likely number of people in the room when we stop? (and no the answer is not 23). For simplicity's sake, assume Feb 29 doesn't exist, and that birthdays are uniforlmy distributed across the remaining 365 days.
And if you can do that, tell me what the average number of people in the room are when we stop (using the same assumptions).
Is it 183?
No Roslyn, it ain't. Perhaps you were thinking that we needed to find someone who shared a birthday with the first person, as opposed to any pair of people (such as the 3rd and 5th people entering sharing a birthday).
And since I can tell that everyone is just sweating on the answer, I shall provide it now. The most likely number of people in the room when we stop is 20 (with a 3.23% chance) and the average number is 24.6166.
*shrug*
at least i had a go!
now: update!
Yes, you did :) But I would have expected you to answer the original english problem that I posed, not the maths one!
I shall endeavour to update later today.
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