// Original C++ solution (slower):
#include <iostream>

#include <cstdio>

#include <vector>

#include <list>

#include <complex>

#include <algorithm>

#include <set>

#include <map>

#include <queue>

#include <string>

#include <cstring>

#include <stack>

#include <cmath>

#include <iomanip>

#include <sstream>

#include <fstream>

#include <cassert>

#include <numeric>

using namespace std;

typedef long long ll;

typedef pair<int,int> PI;

#define EPS (1e-6)

#define rep(i,n) for(int i=0;i<(int)(n);++i)

#define REP(i, n) rep (i, n)

#define F first

#define S second

#define mp(a,b) make_pair(a,b)

#define pb(a) push_back(a)

#define min3(a,b,c) min((a),min((b),(c)))

#define min4(a,b,c,d) min((a),min3((b),(c),(d)))

#define SZ(a) (int)((a).size())

#define ALL(a) a.begin(),a.end()

#define FLL(a,b) memset((a),b,sizeof(a))

#define CLR(a) memset((a),0,sizeof(a))

#define FOR(it,a) for(__typeof(a.begin())it=a.begin();it!=a.end();++it)

template<typename T,typename U> ostream& operator<< (ostream& out, const pair<T,U>& val){return out << "(" << val.F << ", " << val.S << ")";}

template<class T> ostream& operator<< (ostream& out, const vector<T>& val){out << "{";rep(i,SZ(val)) out << (i?", ":"") << val[i];return out << "}";}

typedef double FP;

typedef complex<FP> pt;

typedef pt P;

typedef pair<pt,pt> line;

FP dot(P a,P b){return real(conj(a)*b);}

FP crs(P a,P b){return imag(conj(a)*b);}

P ortho(P a){return conj(P(imag(a),real(a)));}

P ortho(line a){return ortho(a.S-a.F);}

P crspt(P a,P b,P c,P d){b-=a,d-=c;return a+b*crs(d,c-a)/crs(d,b);}

P crspt(line a,line b){return crspt(a.F,a.S,b.F,b.S);}

bool onl(P a1,P a2,P b){return abs(b-a1)+abs(b-a2)<abs(a1-a2)+EPS;}

bool onl(line a,P b){return onl(a.F,a.S,b);}

bool iscrs(line a,line b){P c=crspt(a,b);return onl(a,c)&&onl(b,c);}

void pkuassert(bool t){t=1/t;};

int dx[]={0,1,0,-1,1,1,-1,-1};

int dy[]={1,0,-1,0,-1,1,1,-1};

enum{TOP,BTM,LFT,RGT,FRT,BCK};

int dxdy2ce[]={RGT,FRT,LFT,BCK};

template<class T> T shift(T a,int b,int c,int d,int e){

 __typeof(a[0])t=a[b];a[b]=a[c];a[c]=a[d];a[d]=a[e];a[e]=t;

 return a;}

template<class T> T rgt(T a){return shift(a,TOP,LFT,BTM,RGT);}

template<class T> T lft(T a){return shift(a,TOP,RGT,BTM,LFT);}

template<class T> T frt(T a){return shift(a,TOP,BCK,BTM,FRT);}

template<class T> T bck(T a){return shift(a,TOP,FRT,BTM,BCK);}

line mkl(P a,P v){return line(a,a+v);}

FP lpdist(line a,P b){return abs(b-crspt(a,mkl(b,ortho(a))));}

FP spdist(line a,P b){

  P c(crspt(a,mkl(b,ortho(a))));

  return onl(a,c)?abs(b-c):min(abs(a.F-b),abs(a.S-b));

}

FP ssdist(line a,line b){

 return

  iscrs(a,b)?0.:

   min4(spdist(a,b.F),spdist(a,b.S),

        spdist(b,a.F),spdist(b,a.S));

}

namespace std{

  bool operator<(const P&a,const P&b){

    return mp(real(a),imag(a))<mp(real(b),imag(b));

  }

}



int n,m,k;

ll allsum;

void dfs(int sum,int depth){

  if(depth==n){

    allsum+=max(sum-k,1);

    return;

  }

  for(int i=1;i<=m;++i)

    dfs(sum+i,depth+1);

}



void solve(){

  allsum=0;

  dfs(0,0);

  printf("%.8f\n",allsum/pow(m,n));

}



int main(int argc, char *argv[])

{

  while(cin >> n >> m >> k && n) solve();

  return 0;

}

// Optimized solution to the same problem (faster):
//#include <bits/stdc++.h>

#include <stack>

#include <cassert>

#include <set>

#include <queue>

#include <iostream>

#include <vector>

#include <complex>

#include <algorithm>

#include <cstring>

#include <cstdio>

#include <map>

#include <cmath>

#include <numeric>



using namespace std;

typedef long long ll;

typedef pair<int,int> PI;

#define EPS (1e-6)

#define rep(i,n) for(int i=0;i<(int)(n);++i)

#define REP(i, n) rep (i, n)

#define F first

#define S second

#define mp(a,b) make_pair(a,b)

#define pb(a) push_back(a)

#define min3(a,b,c) min((a),min((b),(c)))

#define min4(a,b,c,d) min((a),min3((b),(c),(d)))

#define SZ(a) (int)((a).size())

#define ALL(a) (a).begin(),(a).end()

#define RALL(a) a.rbegin(),a.rend()

#define FLL(a,b) memset((a),b,sizeof(a))

#define CLR(a) memset((a),0,sizeof(a))

#define declare(a,it) __typeof(a) it=(a)

#define FOR(it,a) for(declare(a.begin(),it);it!=a.end();++it)

#define FORR(it,a) for(declare(a.rbegin(),it);it!=a.rend();++it)



template<typename T,typename U> ostream& operator<< (ostream& out, const pair<T,U>& val){return out << "(" << val.F << ", " << val.S << ")";}

template<class T> ostream& operator<< (ostream& out, const vector<T>& val){out << "{";rep(i,SZ(val)) out << (i?", ":"") << val[i];return out << "}";}

typedef double FP;

typedef complex<FP> pt;

typedef pt P;

typedef pair<pt,pt> line;

namespace std{

  bool operator<(const P&a,const P&b){

    if(abs(a.real()-b.real())>EPS)

      return a.real()<b.real();

    return a.imag()<b.imag();

  }

}

FP dot(P a,P b){return real(conj(a)*b);}

FP crs(P a,P b){return imag(conj(a)*b);}

P ortho(P a){return P(imag(a),-real(a));}

P ortho(line a){return ortho(a.S-a.F);}

P crspt(P a,P b,P c,P d){b-=a,d-=c;return a+b*crs(d,c-a)/crs(d,b);}

P crspt(line a,line b){return crspt(a.F,a.S,b.F,b.S);}

bool onl(P a1,P a2,P b){return abs(b-a1)+abs(b-a2)<abs(a1-a2)+EPS;}

bool onl(line a,P b){return onl(a.F,a.S,b);}

bool iscrs(line a,line b){P c=crspt(a,b);return onl(a,c)&&onl(b,c);}

void pkuassert(bool t){t=1/t;};

int dx[]={0,1,0,-1,1,1,-1,-1};

int dy[]={1,0,-1,0,-1,1,1,-1};

enum{TOP,BTM,LFT,RGT,FRT,BCK};

int dxdy2ce[]={RGT,FRT,LFT,BCK};

int s2i(string& a){stringstream ss(a);int r;ss>>r;return r;}

template<class T> T shift(T a,int b,int c,int d,int e){

  __typeof(a[0])t=a[b];

  a[b]=a[c];a[c]=a[d];a[d]=a[e];a[e]=t;return a;}

template<class T> T rgt(T a){return shift(a,TOP,LFT,BTM,RGT);}

template<class T> T lft(T a){return shift(a,TOP,RGT,BTM,LFT);}

template<class T> T frt(T a){return shift(a,TOP,BCK,BTM,FRT);}

template<class T> T bck(T a){return shift(a,TOP,FRT,BTM,BCK);}

line mkl(P a,P v){return line(a,a+v);}

FP lpdist(line a,P b){return abs(b-crspt(a,mkl(b,ortho(a))));}

FP spdist(line a,P b){

  P c(crspt(a,mkl(b,ortho(a))));

  return onl(a,c)?abs(b-c):min(abs(a.F-b),abs(a.S-b));

}

FP ssdist(line a,line b){

  return

    iscrs(a,b)?0.:

    min4(spdist(a,b.F),spdist(a,b.S),

         spdist(b,a.F),spdist(b,a.S));

}



int n,m,k;



void solve(){

  double ans = 0;

  if(n==1){

    rep(j,m) ans += max(1,j+1-k)*1./m;

  }else{

    double dp[n*m+1];

    rep(i,n*m+1) dp[i]=0;

    dp[0] = 1;

    rep(jj,n){

      double ndp[n*m+1];

      rep(i,n*m+1) ndp[i]=0;

      

      rep(i,n*m+1) if(dp[i]> 1e-9)

        rep(j,m) ndp[i+j+1]+=dp[i]/m;

      rep(i,n*m+1) dp[i]=ndp[i];

    }

    for(int i =1;i<=n*m;++i)

      ans += max(1,i-k)*dp[i];

  }

  printf("%.13f\n",ans);

}



int main(int argc, char *argv[])

{

  while(cin >> n >> m >> k,n) solve();

  return 0;

}

// Fastest known solution for this problem: