**A** quick riddle from The Riddler that, when thinned to the actual maths problem, ends up asking for

which is equal to 2/3…. Anticlimactic.

an attempt at bloggin, nothing more…

**A** quick riddle from The Riddler that, when thinned to the actual maths problem, ends up asking for

which is equal to 2/3…. Anticlimactic.

**T**he latest riddle from The Riddler was both straightforward: given four iid Normal variates, X¹,X²,X³,X⁴, what is the probability that X¹+X²<X³+X⁴ given that X¹<X³ ? The answer is ¾ and it actually does not depend on the distribution of the variates. The bonus question is however much harder: what is this probability when there are 2N iid Normal variates?

I posted the question on math.stackexchange, then on X validated, but received no hint at a possible simplification of the probability. And then erased the questions. Given the shape of the domain where the bivariate Normal density is integrated, it sounds most likely there is no closed-form expression. (None was proposed by the Riddler.) The probability decreases roughly in N³ when computing this probability by simulation and fitting a regression.

> summary(lm(log(p)~log(r))) Residuals: Min 1Q Median 3Q Max -0.013283 -0.010362 -0.000606 0.007835 0.039915 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -0.111235 0.008577 -12.97 4.11e-13 *** log(r) -0.311361 0.003212 -96.94 < 2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.01226 on 27 degrees of freedom Multiple R-squared: 0.9971, Adjusted R-squared: 0.997 F-statistic: 9397 on 1 and 27 DF, p-value: < 2.2e-16

**A** Riddler’s riddle on breaking the unit interval into 4 random bits (by which I understand picking 3 Uniform realisations and ordering them) and finding the length of the bit containing ½ (sparing you the chore of converting inches and feet into decimals). The result can be found by direct integration since the ordered Uniform variates are Beta’s, and so are their consecutive differences, leading to an average length of 15/32. Or by raw R simulation:

simz=t(apply(matrix(runif(3*1e5),ncol=3),1,sort)) mean((simz[,1]>.5)*simz[,1]+ (simz[,1]<.5)*(simz[,2]>.5)*(simz[,2]-simz[,1])+ (simz[,2]<.5)*(simz[,3]>.5)*(simz[,3]-simz[,2])+ (simz[,3]<.5)*(1-simz[,3]))

Which can be reproduced for other values than ½, showing that ½ is the value leading to the largest expected length. I wonder if there is a faster way to reach this nice 15/32.

**A** very standard (one-line) question on X validated, namely whether min(X,Y) could enjoy a finite mean when both X and Y had infinite means [the answer is yes, possibly!] brought a lot of traffic, including an incorrect answer and bringing it to be one of the “Hot Network Questions“, for no clear reason. Beside my half-Cauchy example, some answers pointed out the connection between mean and cdf, as integrated cdf on the negative half-line and integrated complement cdf on the positive half-line, and between mean and quantile function, as

since it nicely expands to

but I remain bemused by the excitement..! (Including the many answers and the lack of involvement of the OP.)

A challenge found on the board of the coffee room at CEREMADE, Université Paris Dauphine:

When sampling with replacement three numbers in {0,1,…,N}, what is the probability that their average is (at least) one of the three?

With a (code-golfed!) brute force solution of

mean(!apply((a<-matrix(sample(0:n,3e6,rep=T),3)),2,mean)-apply(a,2,median))

producing a graph pretty close to 3N/2(N+1)² (which coincides with a back-of-the-envelope computation):