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However, I also was very worried about getting in this time and so applied to about 100 programs. Getting 13 interviews after applying to 100 programs is probably very different statistically from getting 13 interviews after applying to 13 programs, since having interviewed at more places increases chances both because you have more people who might accept you and because it is caused by you being a better candidate whom more people want to interview. I applied to somewhat more places than average so I don’t know how valuable those data would be.

On the other hand, my pet theory is that those curves are actually wrong; the real data show that international applicants with low-teens number of places on their rank-order lists do better than people with 16+, probably because some people get confused and just fill in all their rank order list slots (20) without having actually interviewed there (this has no effect). So it looks like getting 13 ranks may be somewhere better than what the curve would predict.

Right now I think my main strategy is just going to be to be terrified but find out one way or the other soon enough.

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I think 93% is too high. Let’s consider Scott’s previous comments: http://squid314.livejournal.com/307609.html 40 applications; http://squid314.livejournal.com/311111.html all rejected.

—

0/40 eh; this looks like a binomial distribution, the Bayesian version of which is the beta distribution (http://en.wikipedia.org/wiki/Beta_distribution). A uniform distribution is α = β = 1. Stealing shamelessly from http://learnbayes.blogspot.com/2007/09/r-computations-for-proportion-using.html , we can ask what happens if he applied to 40 last year and 40 this year too:


a=1; b=a; # uniform prior
n=40; y=0 # tries, successes
a1=a+y; b1=b+n-y # update; remind you of Laplace's rule?
qbeta(.5, a1, b1) # median
# [1] 0.01676
qbeta(c(.025,.975), a1, b1) # 95% credible interval
# 0.0006173 0.0860438
library(LearnBayes)
m=40; ys=0:m
pred.probs =1)) # at least 1 success out of 40
# [1] 0.4938
sum((pred.probs/2) * (ys>=1)) # halve all probabilities to penalize for being damaged goods
# [1] 0.2469

# redo with just 10 trials last year and 10 more this year
# since 40 does sound implausibly high
n=10; y=0; a1=a+y; b1=b+n-y
qbeta(.5, a1, b1) # median
# [1] 0.06107
qbeta(c(.025,.975), a1, b1)
# [1] 0.002299 0.284914
library(LearnBayes)
m=10; ys=0:m
pred.probs =1))
# [1] 0.4762
sum((pred.probs/2) * (ys>=1))
# [1] 0.2381


Some non-Bayesian calculations using the upper confidence interval instead:


n=40
binom.test(0,n)
# Exact binomial test
#
# data: 0 and 40
# number of successes = 0, number of trials = 40, p-value = 1.819e-12
# alternative hypothesis: true probability of success is not equal to 0.5
# 95 percent confidence interval:
# 0.0000 0.0881
# sample estimates:
# probability of success
# 0
1 - pbinom(1, n, binom.test(0,n)$conf.int[2]) # (1 - chance of less than 1) = chance of at least 1
# [1] 0.8784

n=10
binom.test(0,n)
# Exact binomial test
#
# data: 0 and n
# number of successes = 0, number of trials = 10, p-value = 0.001953
# alternative hypothesis: true probability of success is not equal to 0.5
# 95 percent confidence interval:
# 0.0000 0.3085
# sample estimates:
# probability of success
# 0
1 - pbinom(1, n, binom.test(0,n)$conf.int[2])
# [1] 0.8635


So one way gives me 30-50% and another way gives me well under your 90%.

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If done right, this sort of thing can be the start of a very promising career.

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I work at a medical journal. Lots of medical doctors here who are not in practice. Worth looking at if you need it?

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I know C,some R and some Haskell. I’ve helped some of my friends with their Uni classes and I enjoyed myself.

Some of my programs are on my profile, I’m not a stellar programmer but I should know the basics.

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Maybe you’re better off outside such a warped system.

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