> 3same <- length(same) - length(unique(same))

> 3same <- length(same) - length(unique(same))

Is there some way to index the group sizes systematically? On Feb 20, 2008, at 10:58 PM, Jeremy Hodgen wrote: but you want to check the group sizes systematically not randomly - so you need to count how many of your draws for each group size have 3 or more b'days the same. Then rather than being interested in the number as in your code: 3same <- length(same) - length(unique(same)) We're interested in simply in whether the second is less than the first. Jeremy On 21 Feb 2008, at 05:25, Keith Schnakenberg wrote: sample(x, sample(20:100)) generates random samples of random size. I checked to make sure. I'll get back to you on the last question. On Feb 20, 2008, at 9:12 PM, Joseph Williams wrote: I didn't try for such an elegant solution. I just plugged in different values for "people" until I got a value for "sameday" that gave me the desired probability. It only takes a couple of guesses to get the numbers right. I am not sure you can write sample(20:100) inside "room" Where is your if statement tallying the number of rooms? Joe -----Original Message----- From: gov2001-l-bounces at lists.fas.harvard.edu [mailto:gov2001-l-bounces at lists.fas.harvard.edu<gov2001-l-bounces at lists.fas.harvard.edu>] On Behalf Of Keith Schnakenberg Sent: Wednesday, February 20, 2008 11:21 PM To: gov2001-l at lists.fas.harvard.edu Subject: [gov2001-l] problem 2 Classmates, I am trying to avoid over-using the list, but I am not working with anyone and I don't know people who know this stuff, so this is the only form of collaboration available to me. I am working on problem 2, and I'm struggling with how to vary the sample size appropriately. Here is what I've been doing: sims <- 100000 alldays <- seq(1, 365, 1) for (i in 1:sims){ room <- sample(alldays, sample(20:100), replace=TRUE) size <- length(room) same <- x[duplicated(room)] 3same <- length(same) - length(unique(same)) } I am getting syntax errors, but you can see the logic of what I'm trying to do--I'm drawing random samples of random size and then storing the length and the number of samples with three or more same birthdays in vectors so that I can calculate the probabilities for each sample size. This is my third or fourth approach, but I am having difficulty getting the syntax to implement my ideas. I thought of indexing the room size so that it increased by 1 with each new simulation, but I could not figure out how to make that happen in terms of syntax. Is there a less cumbersome approach that I am missing? Thanks, Keith _______________________________________________ gov2001-l mailing list gov2001-l at lists.fas.harvard.edu http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l _______________________________________________ gov2001-l mailing list gov2001-l at lists.fas.harvard.edu http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l _______________________________________________ gov2001-l mailing list gov2001-l at lists.fas.harvard.edu http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l Dr Jeremy Hodgen Senior Lecturer in Mathematics Education King's College London Department of Education and Professional Studies Franklin-Wilkins Building Waterloo Bridge Wing 150 Stamford Street London SE1 9NH Tel: 020 7848 3102 Fax: 020 7848 3182 E-mail: jeremy.hodgen at kcl.ac.uk _______________________________________________ gov2001-l mailing list gov2001-l at lists.fas.harvard.edu http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l _______________________________________________ gov2001-l mailing list gov2001-l at lists.fas.harvard.edu http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l

=3, 1, 0))

=3, 1, 0))

Sorry, I replied to this a few hours ago with the wrong email again. This should help: Hi Keith, A few things: 1) 100,000 samples is probably unnecessary. Try 1,000 in the final version, and maybe 10 or 100 while you're testing. 2) The for-loop police might hunt me down for saying this, but maybe you should consider nesting loops to do this in a more rigorous/predictable way. (I think I mean predictable...). Basically, you could create a vector the length of the largest group you want to test (say, 100 elements) and store the correct probability for each group size in the element equal to the group size (i.e., element 1 corresponds to the prob. for group size 1, element 37 to the prob. for a group of 37, etc). You just need to cycle through the list of elements and calculate the probability for each one. That may have been ridiculously unclear, so here's an outline of the code: for(i in 1:maxgroupsize){ for(j in 1:sims){ [[calculate the probability]] } probs[i] <- [[the probability for the group]] **notice that this is in the OUTER loop } Fill in the blanks, decide what the max group size should be, and you're set. Good luck and feel free to send me an email if this is unclear. (Perhaps a private email, so we don't give away the whole answer on the list...) J On Thu, Feb 21, 2008 at 2:11 AM, Keith Schnakenberg < keith.schnakenberg at gmail.com> wrote:

>> Keith Schnakenberg <keith.schnakenberg at gmail.com> 2/21/2008 2:11 AM >>>

but you want to check the group sizes systematically not randomly - so you need to count how many of your draws for each group size have 3 or more b'days the same. Then rather than being interested in the number as in your code: We're interested in simply in whether the second is less than the first. Jeremy On 21 Feb 2008, at 05:25, Keith Schnakenberg wrote: > sample(x, sample(20:100)) generates random samples of random size. I > checked to make sure. I'll get back to you on the last question. > > On Feb 20, 2008, at 9:12 PM, Joseph Williams wrote: > >> I didn't try for such an elegant solution. I just plugged in >> different >> values for "people" until I got a value for "sameday" that gave >> me the >> desired probability. It only takes a couple of guesses to get the >> numbers >> right. >> >> I am not sure you can write sample(20:100) inside "room" >> >> Where is your if statement tallying the number of rooms? >> >> Joe >> >> -----Original Message----- >> From: gov2001-l-bounces at lists.fas.harvard.edu >> [mailto:gov2001-l-bounces at lists.fas.harvard.edu] On Behalf Of Keith >> Schnakenberg >> Sent: Wednesday, February 20, 2008 11:21 PM >> To: gov2001-l at lists.fas.harvard.edu >> Subject: [gov2001-l] problem 2 >> >> Classmates, >> >> I am trying to avoid over-using the list, but I am not working with >> anyone and I don't know people who know this stuff, so this is the >> only form of collaboration available to me. I am working on problem >> 2, and I'm struggling with how to vary the sample size >> appropriately. >> Here is what I've been doing: >> >> sims <- 100000 >> alldays <- seq(1, 365, 1) >> for (i in 1:sims){ >> room <- sample(alldays, sample(20:100), replace=TRUE) >> size <- length(room) >> same <- x[duplicated(room)] >> } >> >> I am getting syntax errors, but you can see the logic of what I'm >> trying to do--I'm drawing random samples of random size and then >> storing the length and the number of samples with three or more same >> birthdays in vectors so that I can calculate the probabilities for >> each sample size. This is my third or fourth approach, but I am >> having difficulty getting the syntax to implement my ideas. I >> thought >> of indexing the room size so that it increased by 1 with each new >> simulation, but I could not figure out how to make that happen in >> terms of syntax. >> >> Is there a less cumbersome approach that I am missing? >> >> Thanks, >> Keith >> _______________________________________________ >> gov2001-l mailing list >> gov2001-l at lists.fas.harvard.edu >> http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l >> >> _______________________________________________ >> gov2001-l mailing list >> gov2001-l at lists.fas.harvard.edu >> http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l > > _______________________________________________ > gov2001-l mailing list > gov2001-l at lists.fas.harvard.edu > http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l Dr Jeremy Hodgen Senior Lecturer in Mathematics Education King's College London Department of Education and Professional Studies Franklin-Wilkins Building Waterloo Bridge Wing 150 Stamford Street London SE1 9NH Tel: 020 7848 3102 Fax: 020 7848 3182 E-mail: jeremy.hodgen at kcl.ac.uk _______________________________________________ gov2001-l mailing list gov2001-l at lists.fas.harvard.edu http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l

Hey guys, While I admire everyone's attempts to create a complicated algorithm to test multiple room sizes and return a vector a probabilities, I think that approach is wildly inefficient for this particular problem. The problem is that, for your program not to take a whole afternoon to run, you will need to restrict yourself to a small number of iterations for each room size. Since the variation of your estimate will be high with so few trials, it's going to be difficult to pinpoint the right number---esp. when you don't have a good idea where to start. I would recommend just guessing and testing different numbers with medium size samples (maybe 1,000) to hone in on the correct answer. As you get closer to 0.5, you can increase the number of iterations dramatically (perhaps to 1,000,000) to get the right answer with whatever level of precision desired. This method should dramatically reduce the difficulty of this problem, esp. if you're new to R programming. Best, Jon 2008/2/21, Quan Li <quanli at mail.ucf.edu>: error to it. I run a double loop starting with a group of size 3, sampling from this group and calculate the probability of 3 or more people sharing a birthday, if this probability is less than 0.5, group size is increased by 1, sampling and calculating the probability again. The loop stops when the probability reaches 0.5 and the final group size is the minimum group size I need. same -- Jon Bischof Graduate Student Department of Government Harvard University _______________________________________________ gov2001-l mailing list gov2001-l at lists.fas.harvard.edu http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l