(This article was originally published at Statistical Modeling, Causal Inference, and Social Science, and syndicated at StatsBlogs.)
In comments, Rick Schoenberg wrote:
One thing I tried to say as politely as I could in [the book, "Probability with Texas Holdem Applications"] on p146 is that there’s a huge error in Chen and Ankenman’s “The Mathematics of Poker” which renders all the calculations and formulas in the whole last chapter wrong or meaningless or both. I’ve never received a single ounce of feedback about this though, probably because only like 2 people have ever read my whole book.
Jerrod Ankenman replied:
I haven’t read your book, but I’d be happy to know what you think is a “huge” error that invalidates “the whole last chapter” that no one has uncovered so far. (Also, the last chapter of our book contains no calculations—perhaps you meant the chapter preceding the error?). If you contacted one of us about it in the past, it’s possible that we overlooked your communication, although I do try to respond to criticism or possible errors when I can. I’m easy to reach; firstname.lastname@yale.edu will work for a couple more months.
Hmmm, what’s on page 146 of Rick’s book? It comes up if you search inside the book on Amazon:
So that’s the disputed point right there. Just go to the example on page 290 where the results are normally distributed with mean and variance 1, check that R(1)=-14%, then run the simulation and check that the probability of the bankroll starting at 1 and reaching 0 or less is approximately 4%.
I went on to Amazon but couldn’t access page 290 of Chen and Ankenman’s book to check this. I did, however, program the simulation in R as I thought Rick was suggesting:
waiting <- function(mu,sigma,nsims,T){
time_to_ruin <- rep(NA,nsims)
for (i in 1:nsims){
virtual_bankroll <- 1 + cumsum(rnorm(T,mu,sigma))
if (any(virtual_bankroll<0)) {
time_to_ruin[i] <- min((1:T)[virtual_bankroll<0])
}
}
return(time_to_ruin)
}
a <- waiting(mu=1,sigma=1,nsims=10000,T=100)
print(mean(!is.na(a)))
print(table(a))
Which gave the following result:
> print(mean(!is.na(a))) [1] 0.0409 > print(table(a)) a 1 2 3 4 5 6 8 9 218 107 53 13 9 7 1 1
These results indicate that (i) the probability is indeed about 4%, and (ii) T=100 is easily enough to get the asymptotic value here.
Actually, the first time I did this I kept getting a probability of ruin of 2% which didn't seem right--I couldn't believe Rick would've got this simple simulation wrong--but then I found the bug in my code: I'd written "cumsum(1+rnorm(T,mu,sigma))" instead of "1+cumsum(rnorm(T,mu,sigma))".
So maybe Chen and Ankenman really did make a mistake. Or maybe Rick is misinterpreting what they wrote. There's also the question of whether Chen and Ankenman's mathematical error (assuming they did make the mistake identified by Rick) actually renders all the calculations and formulas in their whole last chapter, or their second-to-last chapter, wrong or meaningless or both.
P.S. According to the caption at the Youtube site, they're playing rummy, not poker, in the above clip. But you get the idea.
P.P.S. I fixed a typo pointed out by Juho Kokkala in an earlier version of my code.
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