#Define the Catalan Numbers

C:=List([0..300],n->Binomial(2*n,n)/(n+1));

#C[i]=i-1st Catalan Number

#Now we take the Catalan number and divide by 2^2n-1.

P:=List([1..5000],n->(Binomial(2*(n-1),n-1)/(n))/(2^(2*n-1)));

#P[i] is the probability that get equal # of heads and tails for the first time exactly after 2i tosses.

for i in [1..15] do
Print(P[i],"\n");
od;

#What about the partial sums?

L:=P[1];
for i in [2..10] do
L:=L+P[i];
od;
Float(L);

#10th partial sum is 0.828

L:=P[1];
for i in [2..20] do
L:=L+P[i];
od;
Float(L);

#20th partial sum is 0.8746


L:=P[1];
for i in [2..5000] do
L:=L+P[i];
od;
Float(L);
#40th is 0.911
#60th is 0.927
#80th is 0.937
#100th is 0.9436
#200th is 0.96
#300th is 0.96744
#500th is 0.974775
#1000th is 0.9821 (FLOATING POINT FAILS)
#5000th is 0.9919