Compute the experimental probability of a particular score on a die given a sample of throws, and compare it with the theoretical probability.

\nThe last part asks what you expect to happen to the experimental probability as the sample size increases.

"}, "advice": "There are two ways of assigning probability:

\n- \n
**Experimental Probability**is where you do an experiment, observe the outcome and record the relative frequency results. \n**Theoretical Probability**is where the probability is calculated based on fairness and symmetrical properties of the results. \n

To calculate the experimental probability (relative frequency) of an outcome we divide the frequency of the outcome in the experiment by the number of trials.

\nWe are given that the experiment was repeated $\\var{no_rolls}$ times.

\nWe then need the number of times that the sum of the faces of the dice was equal to $\\var{sum[0]}$. From the frequency table, we can see that the frequency of rolling a $\\var{sum[0]}$ in the experiment was $\\var{Freq[sum[0]-2]}$.

\nTherefore, the experimental probability of rolling a $\\var{sum[0]}$ is

\n\\[ \\begin{align} P(\\text{total}=\\var{sum[0]}) &= \\displaystyle\\frac{\\text{number of times a total of $\\var{sum[0]}$ was rolled}}{\\text{total number of rolls}}\\\\&= \\displaystyle\\frac{\\var{Freq[sum[0]-2]}}{\\var{no_rolls}}.\\end{align}\\]

\n\\[\\begin{align} P(\\text{total}=\\var{sum[0]}) &= \\displaystyle\\frac{\\text{number of times a total of $\\var{sum[0]}$ was rolled}}{\\text{total number of rolls}}\\\\&= \\displaystyle\\frac{\\var{Freq[sum[0]-2]}}{\\var{no_rolls}}\\\\&= \\displaystyle\\var[fractionNumbers, simplifyFractions]{{Freq[sum[0]-2]/no_rolls}}.\\end{align}\\]

\nWhen two unbiased 6-sided dice are rolled and their scores are added, there are $11$ possible outcomes: the total must be between $2$ and $12$ inclusive.

\nTo work out the probabilities of each outcome occurring, we must be very careful about how the experiment is performed.

\nThere are a total of $36$ different outcomes when rolling two dice one after the other; these are shown in Table $1$.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n1 | 2 | 3 | 4 | 5 | 6 | |
---|---|---|---|---|---|---|

1 | (1,1) | (1,2) | (1,3) | (1,4) | (1,5) | (1,6) |

2 | (2,1) | (2,2) | (2,3) | (2,4) | (2,5) | (2,6) |

3 | (3,1) | (3,2) | (3,3) | (3,4) | (3,5) | (3,6) |

4 | (4,1) | (4,2) | (4,3) | (4,4) | (4,5) | (4,6) |

5 | (5,1) | (5,2) | (5,3) | (5,4) | (5,5) | (5,6) |

6 | (6,1) | (6,2) | (6,3) | (6,4) | (6,5) | (6,6) |

Note that these outcomes are all different. For example, the outcomes (2,1) and (1,2) are not the same because we know the order in which the dice were thrown so we can distinguish between these two outcomes; when the first die lands on $2$ and the second die lands on $1$ the outcome is (2,1), however when the first die lands on $1$ and the second die lands on $2$ the outcome is (1,2).

\nThe sum of the numbers in these outcomes is the same, we just count them as different outcomes.

\nAny one of these $36$ outcomes is equally likely to occur, so the probability of each of these outcomes is $\\displaystyle\\frac{1}{36}$.

\nHowever, if you roll two indistinguishable dice at the same time, you can't differentiate (a 1 and a 2) from (a 2 and a 1). From your point of view, they are the same outcome. The probabilities of the underlying events (each die's score) haven't changed, but the outcomes you observe have.

\nIn this case you have a total of $21$ outcomes; these outcomes are *not* all equally likely. There's only one way of obtaining two 1s, while there are two ways of obtaining a 1 and a 2, corresponding to two squares in Table $1$. Hence the probability of obtaining two 1s would be $\\displaystyle\\frac{1}{36}$, whereas the probability of obtaining a 1 and a 2 would be

\\[\\displaystyle\\frac{2}{36}=\\displaystyle\\frac{1}{18}.\\]

\nYou *can* do the experiment this way, but it's easier when you can tell the dice apart.

Table $2$ shows the corresponding total of the faces of the dice for each of the outcomes in Table $1$.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n+ | 1 | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|---|

1 | 2 | 3 | 4 | 5 | 6 | 7 |

2 | 3 | 4 | 5 | 6 | 7 | 8 |

3 | 4 | 5 | 6 | 7 | 8 | 9 |

4 | 5 | 6 | 7 | 8 | 9 | 10 |

5 | 6 | 7 | 8 | 9 | 10 | 11 |

6 | 7 | 8 | 9 | 10 | 11 | 12 |

For equally likely outcomes you can calculate the probability using the formula

\n\\[\\text{Probability of an event} = \\displaystyle\\frac{\\text{number of favourable outcomes}}{\\text{total number of outcomes}}.\\]

\nTable $3$ gives the theoretical probabilities of each of the possible totals of the two dice occurring:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nTotal | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

Probability | $\\displaystyle\\frac{1}{36}$ | $\\displaystyle\\frac{2}{36}$ | $\\displaystyle\\frac{3}{36}$ | $\\displaystyle\\frac{4}{36}$ | $\\displaystyle\\frac{5}{36}$ | $\\displaystyle\\frac{6}{36}$ | $\\displaystyle\\frac{5}{36}$ | $\\displaystyle\\frac{4}{36}$ | $\\displaystyle\\frac{3}{36}$ | $\\displaystyle\\frac{2}{36}$ | $\\displaystyle\\frac{1}{36}$ |

\\[P(\\text{total}=\\var{sum[0]}) = \\displaystyle\\frac{\\var{Freq2[x[0]]}}{36}.\\]

\\[\\begin{align} P(\\text{total}=\\var{sum[0]}) &= \\displaystyle\\frac{\\var{Freq2[x[0]]}}{36}\\\\ &= \\displaystyle\\var[fractionNumbers,simplifyFractions]{Freq2[x[0]]/36}. \\end{align}\\]

\nFor any experiment, as we make the number of trials very large the experimental probability tends towards the theoretical probability.

\nFor this experiment, for example, the theoretical probability of the sum of the faces of the two dice being $3$ is

\n\\[\\displaystyle\\frac{2}{36} = \\var{sigformat(1/18,3)} \\text{ (rounded to $3$ significant figures)}.\\]

\nIf we were to roll the two dice a very large number of times, for example $10000$ times, the frequencies of each outcome would be different.

\nThe table below was produced by the recording the frequency of each outcome after rolling the two dice 10,000 times.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nTotal | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

\n Frequency \n | $\\var{ceil(10000/36)+{add}}$ | $\\var{ceil(10000/18)+{add}}$ | $\\var{ceil(2500/3)+{add}}$ | $\\var{ceil(10000/9)+{add}}$ | $\\var{ceil(12500/9)+{add}}$ | $\\var{10000-{remainder}}$ | $\\var{ceil(12500/9)-{add}}$ | $\\var{ceil(10000/9)-{add}}$ | $\\var{ceil(2500/3)-{add}}$ | $\\var{ceil(5000/9)-{add}}$ | $\\var{ceil(2500/9)-{add}}$ |

This means that the new experimental probability of the sum of the faces of the two dice being $3$ is

\n\\[\\displaystyle\\frac{\\var{ceil(10000/18) + {add}}}{10000} = \\var{sigformat((ceil(10000/18)+{add})/10000,3)} \\text{ (rounded to $3$ significant figures)}.\\]

\nThis is very close to the theoretical probability, so we can see how the experimental probability changes as we increases the number of trials.

", "extensions": [], "type": "question", "statement": "Two unbiased, 6-sided dice were rolled together and the total of the numbers shown on the faces was recorded.

\nThe experiment was repeated $\\var{no_rolls}$ times.

\nThis table gives the frequency of each outcome.

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\nTotal | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
---|---|---|---|---|---|---|---|---|---|---|---|

Frequency | $\\var{Freq[0]}$ | $\\var{Freq[1]}$ | $\\var{Freq[2]}$ | $\\var{Freq[3]}$ | $\\var{Freq[4]}$ | $\\var{Freq[5]}$ | $\\var{Freq[6]}$ | $\\var{Freq[7]}$ | $\\var{Freq[8]}$ | $\\var{Freq[9]}$ | $\\var{Freq[10]}$ |

Sums obtained from no_rolls of two dice part a)

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\nEnter your answer as a fraction.

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