What is the probability of getting yahtzee?

Question

What is the probability of getting yahtzee using N dices with X sides in Y throws in a single round?

Which side of the dice one gets yahtzee with doesn't matter (e.g. it doesn't matter if it is ones or twos etc.). I also assume perfect strategy is used, that is, after each throw one saves the number it is the most of (and if equal, just pick one at random).

Answer 1

What follows is a solution to a smaller problem; I don't have a good way to determine the probabilities of exactly k of some kind being the most of any kind on the first roll, but the work below could be used from that point to get an answer.

Let $f(N,X,Y)$ be the probability of getting all 1s with N dice, each with X sides, in at most Y rolls, where after each roll, 1s are retained and only non-1 dice are re-rolled. Working 1 roll at a time, the first roll could have exactly $k=0,\dots,N$ 1s. The probability of exactly k 1s in a single roll is ${N\choose k}\left(\frac{1}{X}\right)^k\left(\frac{x-1}{x}\right)^{N-k}$ and having k 1s in the first roll, we then need $N-k$ 1s in the remaining $Y-1$ rolls. So, $$f(N,X,Y)=\sum_{k=0}^{N}{N\choose k}\left(\frac{1}{X}\right)^k\left(\frac{x-1}{x}\right)^{N-k}f(N-k,X,Y-1).$$ Also, the probability of all 1s on 0 dice is 1 (0 of the 0 dice are guaranteed to be 1), so $f(0,X,Y)=1$, and the probability of all 1s on $N\ge 0$ dice in 0 rolls is 0 (can't get any 1s without rolling some dice), so $f(N,X,0)=0$ for $N\ge 0$. This is a complete definition of $f(N,X,Y)$, though it does not lend itself to easy computation. (However, software like Mathematica may be able to compute from this definition. In Mathematica: f[n_, x_, y_] := If[n == 0, 1, If[y == 0, 0, Sum[ Binomial[n, k]*(1/x)^k*(1 - 1/x)^(n - k)*f[n - k, x, y - 1], {k, 0, n}]]].)

Answer 2

According to Wikipedia "the probability of a Yahtzee for any three-roll turn is about 0.04603 (or $\frac{347897}{7558272}$), or roughly 1 in 22 attempts."

Answer 3

This isn't a complete answer to your question, but at least it is a closed form. For 5 normal (6-sided), dice, the exact probability of achieving a yahtzee, following the optimum strategy, given $n$ rolls is

$1+\frac{53}{13} \left(\frac{5}{6}\right)^{2n+1}+\frac{11 \cdot 5^n}{13 \cdot 2^{n+5} \cdot 3^{3n+1}} -\frac{5^n}{8 \cdot 3^{2n-2}} -\frac{7 \cdot 5^{n+1}}{ 2^{n+5} 3^{n-1}} $

For example, setting $n=3$ gives $\frac{347897}{7558272}$.

Answer 4

Well I am not able to give you an analytical answer, but this question just screamed to be programmed. Now I am sorry that I haven't used Matlab ( which would be much less code ) but I implemented it in c++. The code is simply your rules implemented in my not so good c++. I used g++ 4.6.3 on Ubuntu 12.04. If you run the code, it will print the results and also write the pairs (number of dices | avg. number of throws) in out.txt. The result is plotted attached, plotted with gnuplot for number of dices $\leq$ 40.

See the results here as I dont know much about gnuplot and couldn't upload a .ps to MathSE I had to upload it there. For every number of dices, 10000 experiments have been player. The code is probably terrible slow so everyone feel free to improve it ;)

This here does the same computation for number of dices $\leq 400$ but only 1000 experiments each.

Maybe you can use this to varify any of the formulas.

yahtzee.cpp:

#include <iostream>
#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <vector>
#include <fstream>

using namespace std;


int hasYourNumber(vector<int>,int );
int sum(int* ,int );
int mostFrequent ( vector<int> );
vector<int>throwDice(int);
int role();
void playYahtzee(int,int,bool);

int main ( void ) {

    int REP = 10000; // Number of repeting the experiment
    bool echo = 0;    // print information (yes 1,no 0)
    for (int numberOfDices=1;numberOfDices<40;numberOfDices++){
        playYahtzee(numberOfDices,REP,echo);
    }


    return 0;
}


void playYahtzee(int N,int REP,bool echo){// N dices with 6 faces each;


    srand( time(NULL) );        // initialize random generator  
    int* throws = new int[REP]; // save number of throws for each experiment
    vector <int> dices;     // save dices that came up in one throw inside this, vector just for size(), my way...

    for (int k = 0;k<REP;k++){
        bool first = true;      // is it the first throw?
        int AmountOfYourNumber = 0;     // how often as our number occured
        int yourNumber = -1;        // whats your number
        int numberOfThrows = 0;     // how often have we been throwing in this round
        while (AmountOfYourNumber<N){   // finish if our number has occured N times
            numberOfThrows++;
            dices.clear();
            dices = throwDice(N-AmountOfYourNumber); // place random integers between 1 and 6 inside
            if (first){
                yourNumber = mostFrequent(dices);// choose the most frequent number
                AmountOfYourNumber += hasYourNumber(dices,yourNumber); // how often has it your number
                first = false;
            }else{
                AmountOfYourNumber += hasYourNumber(dices,yourNumber);
            }
            if (echo){ // some output mostly for debugging
                printf("Your number is %d and it has occured %d times in throw 1 to %d.\n", 
                        yourNumber, AmountOfYourNumber,numberOfThrows);
                printf("\tNumbers in this throw:\n\t");
                for (int i=0;i<dices.size();i++) printf("%d\t",dices[i]);
                printf("\n \n");
            }
        }

        throws[k] = numberOfThrows; // save number of throws
    }   

    double avg = double(sum(throws,REP))/double(REP);   // calculate average
    printf("Average number (over %d repetitions) of attempts, when playing with %d dices equals  %g\n",REP,N,avg);
    ofstream output;
    output.open ("out.txt",fstream::app);
    output << N << " " << avg<< endl;
    output.close();
}
int sum(int* throws,int N){ // = sum_i throws[i]
    int s = 0;
    for (int i=0;i<N;i++) s+= throws[i];
    return s;
}
int hasYourNumber(vector<int> dices,int yourNumber){ // calculates how often yournumber is in dices
    int N = 0;

    for (int i=0;i<dices.size();i++){
        if ( dices[i] == yourNumber) N++;
    }
    return N;
}

int mostFrequent ( vector<int> dices){ // which is the most frequent number, takes first best
    int freq [6] = {0,0,0,0,0,0};
    for (int i=0;i<dices.size();i++){
        freq[dices[i]-1]++;
    }
    int yourNumber = -1;
    int max = 0;
    for (int i=0;i<6;i++){
        if (freq[i]>max) {yourNumber = i+1 ; max = freq[i];}
    }
    return yourNumber;
}

vector <int> throwDice(int M){ // throws M dices
    vector <int> dices;
    for (int i=0;i<M;i++) dices.push_back(role());
    return dices;
}
int role(){ // throws one dice
    return rand() % 6 +1;
}