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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).

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@runaros: Would you please explain the game, what does it mean "to get yahtzee"? –  Tom Stephens Aug 10 '10 at 12:25
Throwing five or six dice and trying to get various combinations of those dice. Three throws every round (but some rule variations allow saving throws to later rounds), and you can choose how many dice to rethrow between throws. en.wikipedia.org/wiki/Yahtzee –  runaros Aug 10 '10 at 12:34
@runaros: I was hoping you would edit the question itself. Only you and our moderators (and high-rep users) have that option. –  Tom Stephens Aug 10 '10 at 13:05
you start by rolling N dice, you get Y rolls, on every roll but the first you can choose to save the results of some of the dice and only roll the rest of them. A yahtzee is when at the aftermath of a roll, all dice (those rolled and those saved from previous rolls) show the same number. –  yohay kaplan Aug 10 '10 at 13:16
@runaros do you mean in a single round(with rerolls allowed in a round) or over the course of the entire game? In the second case we need to talk about strategy. :/ –  BBischof Aug 10 '10 at 17:16

5 Answers 5

up vote 5 down vote accepted

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}]]].)

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Nice answer. I'll upvote it once I get enough rep, and probably accept it after some time as well :) –  runaros Aug 11 '10 at 8:59
Very nice solution. –  JohnD Dec 16 '12 at 21:01

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."

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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 $\frac{347897}{7558272}$.

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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.


#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> );
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++){

    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
            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;
                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;
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++){
    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;
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The odds are 1 in 1,296 with five six-sided dice because of this math sentence: 1*1/6*1/6*1/6*1/6. Also, to get a Yahtzee, the first number on one die has to math the other numbers on the other dice.

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This is only valid for $Y=1$ (using the OP's notation). Also note that the question is over three years old. How does your answer add to the ones already given? –  mrf Nov 26 '13 at 8:36

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