[un]solved riddles

On the Riddler of last week, first a birthday puzzle:

Given a group of 23 persons, what is the probability of observing three pairs of identical birthdays?

which can be found by a quick simulation as

ave=0
for (t in 1:1e6){
 dupz=dates[duplicated(sample(1:365,23,rep=TRUE))]
 ave=ave+as.integer((length(dupz)==3)&
     (length(unique(dupz))==3))}}
ave/M

returning a value of 0.0183, but which combinatoric resolution I could not fully fathom without a little help from a friend (-ly blog). I had the multinomial coefficient

for the allocation of the 23 persons to one of the three pairs or none, as well as the probability

but I had forgotten the 3! in the denominator for the permutations of the three pairs, which leads again to

A question that also led to an unsolved question: in the even this probability was much smaller, what is an easy way into constructing a more efficient importance sampler?

The second riddle was just as easy to code in R:

A game of tag goes by the following rules: (i) anyone untagged can tag anyone untagged; (ii) anyone tagged by a player tagged gets untagged; (iii) the winner is the last untagged player. What is the expected number of runs for N players?

The outcome of

game=function(N=12){
  a=rep(0,N);T=0
  while (sum(a==0)>1){
   ij=sample((1:N)[a==0],2)
   a[ij[2]]=ij[1];a[a==ij[2]]=0
   T=T+1}
  return(T)}

leads to an average value of

but I had no clear quick explanation for the doubling phenomenon. Until I picked a pen and a sheet of paper and drew the last steps of the game: to decrease down to 1, the size of the untagged survivors has to get through …,3,2 and each time the eliminated player needs to have tagged no other player since otherwise the population grows again. This has to apply all the way to the second round, where N-1 players remain and the one tagged needs to be anyone but the one who tagged the first one. And so on…