Monte Carlo Club

(This week) Feb 10, 2017 "Dating" (Valentine's Day Special; Due 2/17/2017)

• Source: Paul Nahin (2008), "Digital Dice". Problem #20: "An Optimal Stopping Problem"
• Problem: What is the optimal time point when one should stop dating more people and settle on a mate choice (and live with the decision)
• Your best strategy is to date an initial sample of N individuals, rejecting all, and marry the next one ranked higher than any of your N individuals. The question is what is the optimal number for N.

1. The problem could be investigated by simulating a pool of 10 individuals, ranked from 1-10 (most desirable being 1) and then take a sample of N
2. You may only date one individual at a time
3. You cannot go back to reach previously rejected candidates
4. Simulate N from 0 to 9 (zero means marrying the first date, a sample size of zero)
5. For each N, obtain the probability of finding the perfect mate (i.e., ranked 1st) by running simulation 1000 times
6. Plot barplot of probability versus sample size N.
7. Expected answer: N=4

Submitted Codes

By Weigang pick.candidate <- function(min, array) { for (i in 1:length(array)) { if (array[i] < min) { return(array[i]) } else {next} } return(0) # No 1. has been sampled and rejected } candidates <- 1:10; output <- sapply(0:9, function(x) { ct <- 0; for(k in 1:1000) { if (x==0) { # no sample, marry the 1st guy sampled <- sample(candidates, 1); if (sampled == 1) {ct <- ct+1} } else { sampled <- sample(candidates, x); not.sampled <- candidates[-sampled]; not.sampled <- sample(not.sampled); if (pick.candidate(min = min(sampled), array = not.sampled) == 1) {ct <- ct+1} } } return(ct); }) barplot(output/1e3, names.arg = 0:9, xlab = "number of sampled dates", las=1, main = "Optimal stopping for dating (N=10 candidates)", ylab = "Prob(marrying No.1)")

(Last week) Feb 3, 2017 "US Presidents" (Due 2/10/2017)

• Download
  File:Presidents.txt
  : 1st column is the order, 2nd column is the name, the 3rd column is the year of inauguration; tab-separated
• Your job is to create an R, Perl, or Python script called “us-presidents”, which will

1. Read the table
2. Store the original/correct order
3. Shuffle/permute the rows and record the new order
4. Count the number of matching orders
5. Repeat Steps 3-4 for a 1000 times
6. Plot histogram or barplot (better) to show distribution of matching counts
7. Hint: For R, use the sample() function. For Perl, use the rand() function.

Submitted Codes

By Mei pres.list <- lapply(1:1000, function(x) pres[sample(nrow(pres)),]) cts <- sapply(pres.list, function(x) { ct.match <- 0; for (i in 1:45){ if (pres$order[i] == x[i,1]){ #cat(as.character(x[i,2]), x[i,1], "\n") #as.character to avoid the factor info ct.match <- ct.match + 1; #cat(ct.match) } } ct.match #return }) barplot(table(cts),xlab = "Number of matches per shuffle", border = "hotpink", col = "pink", ylab = paste("Frequency total of:",length(pres.list)), main = "US Presidents") By John import pandas as pd import numpy as np from pandas import DataFrame import matplotlib.pyplot as plt df = pd.read_table("presidents.txt", names=["num", "name", "presidency"]) name_list = list(df.name) # create a list to store matches after each shuffle shuffle_record = [] # create a function in which the first argument is the original dataset; second argument is number of shuffles def shuffler2(original, n):     record = []     for i in range(n):         num = 0         each_shuffle = {}         temp = original.reindex(np.random.permutation(original.index)) # do shuffling for each         compare = original.num == temp.num         matched_df = original.ix[compare[compare == True].index]         for i in matched_df.index:             each_shuffle[i] = matched_df.name[i]         shuffle_record.append(each_shuffle)         try:             num = compare.value_counts()[1]         except:             pass         record.append(num)     return(record) result2 = shuffler2(df, 1000) plt.hist(result2, color="yellow") plt.title("Histgram for President Data") plt.show() By Weigang p <- read.table("Presidents.txt", sep="\t", header=F) colnames(p) <- c("order", "name", "inaug.year") # Use "sapply" or "lapply" for loops: no need to pre-define a vector to store results p.sim <- sapply(1:10000, function (x) { length(which(sample(p$order) == p$order)) }) barplot(table(p.sim)/1e4, las=1) p.exp <- rpois(1e4, 1) # draw 10000 Poisson random deviates mp <- barplot(table(p.exp)/1e4, las=1, xlab = "Num of matching presidents") # mid-point on x-axis lines(mp[1:7], table(p.sim)/1e4, type="b", col=2) # add a line (Poisson-expected) to the barplot (simulated) legend("topright", c("Simulated", "Poisson expectation"), col=1:2, lty=1)

(Future weeks) Feb 17, 2017 "Birthday" (Due 2/24/2017)

1. Randomly select N individuals and record their B-days
2. Count the B-days NOT shared by ANY two individuals
3. Repeat (for each N) 1000 times, and obtain probability by averaging the previous counts divided by 1000
4. Vary N from 10 to 100, increment by 10
5. Plot probability of no-shared B-Day (Y-axis) versus N (x-axis), with either a stripchart or boxplot, or both