# easy riddle

July 11, 2017
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From the current Riddler, a problem that only requires a few lines of code and a few seconds of reasoning. Or not.

N households each stole the earnings from one of the (N-1) other households, one at a time. What is the probability that a given household is not burglarised? And what are the expected final earnings of each household in the list, assuming they all start with \$1?

The first question is close to Feller’s enveloppe problem in that

$\left(1-\frac{1}{N-1}\right)^{N-1}$

is close to exp(-1) for N large. The second question can easily be solved by an R code like

N=1e3;M=1e6
fina=rep(1,N)
for (v in 1:M){
ordre=sample(1:N)
vole=sample(1:N,N,rep=TRUE)
while (min(abs(vole-(1:N)))==0)
vole[abs(vole-(1:N))==0]=sample(1:N,
sum(vole-(1:N)==0))
cash=rep(1,N)
for (t in 1:N){
cash[ordre[t]]=cash[ordre[t]]+cash[vole[t]];cash[vole[t]]=0}
fina=fina+cash[ordre]}


which returns a pretty regular exponential-like curve, although I cannot figure the exact curve beyond the third burglary. The published solution gives the curve

${\frac{N-2}{N-1}}^{999}\times 2+\frac{1}{N-1}\right)^{t-1}\times{\frac{N-1}{N}}^{N-t}\times\frac{N}{N-1}$

corresponding to the probability of never being robbed (and getting on average an extra unit from the robbery) and of being robbed only before robbing someone else (with average wealth N/(N-1)).

Filed under: Books, Kids, R Tagged: FiveThirtyEight, mathematical puzzle, R, simulation, The Riddler, William Feller

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