m= 10^6; n = 5; x = y = numeric(m)
for (i in 1:m) {
  t = rnorm(n); t.max = max(t)
  new = rnorm(3)
  x[i] = sum(new > t.max)                # nr new above t.max
  y[i] = sum(diff(c(t.max, new) > 0))  } # 3 means incr from t.max on
mean(x == 3);  mean(y == 3)
## 0.017736  # aprx prob all 3 new exceed max of 5 historical
## 0         # aprx prob max.t < t.6 < t.7 < t.8