Revert "Tighten the error tolerance requirement by 100x (#26588)" (#26593)

This reverts commit f835f3f97c.

test/cpp/end2end/xds_end2end_test.cc

@@ -1615,25 +1615,28 @@ grpc_millis NowFromCycleCounter() {
   return grpc_cycle_counter_to_millis_round_up(now);
 }
 
-// Returns the number of RPCs needed to pass error_tolerance at 99.99994%
-// chance. Rolling dices in drop/fault-injection generates a binomial
-// distribution (if our code is not horribly wrong). Let's make "n" the number
-// of samples, "p" the probabilty. If we have np>5 & n(1-p)>5, we can
-// approximately treat the binomial distribution as a normal distribution.
+// Returns the number of RPCs needed to pass error_tolerance at 99.995% chance.
+// Rolling dices in drop/fault-injection generates a binomial distribution (if
+// our code is not horribly wrong). Let's make "n" the number of samples, "p"
+// the probabilty. If we have np>5 & n(1-p)>5, we can approximately treat the
+// binomial distribution as a normal distribution.
 //
 // For normal distribution, we can easily look up how many standard deviation we
 // need to reach 99.995%. Based on Wiki's table
-// https://en.wikipedia.org/wiki/68%E2%80%9395%E2%80%9399.7_rule, we need 5.00
-// sigma (standard deviation) to cover the probability area of 99.99994%. In
-// another word, for a sample with size "n" probability "p" error-tolerance "k",
-// we want the error always land within 5.00 sigma. The sigma of binominal
-// distribution and be computed as sqrt(np(1-p)). Hence, we have the equation:
+// https://en.wikipedia.org/wiki/Standard_normal_table, we need 4.00 sigma
+// (standard deviation) to cover the probability area of 99.995%. In another
+// word, for a sample with size "n" probability "p" error-tolerance "k", we want
+// the error always land within 4.00 sigma. The sigma of binominal distribution
+// and be computed as sqrt(np(1-p)). Hence, we have the equation:
 //
-//   kn <= 5.00 * sqrt(np(1-p))
+//   kn <= 4.00 * sqrt(np(1-p))
+//
+// E.g., with p=0.5 k=0.1, n >= 400; with p=0.5 k=0.05, n >= 1600; with p=0.5
+// k=0.01, n >= 40000.
 size_t ComputeIdealNumRpcs(double p, double error_tolerance) {
   GPR_ASSERT(p >= 0 && p <= 1);
   size_t num_rpcs =
-      ceil(p * (1 - p) * 5.00 * 5.00 / error_tolerance / error_tolerance);
+      ceil(p * (1 - p) * 4.00 * 4.00 / error_tolerance / error_tolerance);
   gpr_log(GPR_INFO,
           "Sending %" PRIuPTR " RPCs for percentage=%.3f error_tolerance=%.3f",
           num_rpcs, p, error_tolerance);