# Samuel Pepys Wrote Isaac Newton Asking Which Event is More Likely to Occur

**Problem**

Samuel Pepys wrote Isaac Newton to ask which of three events is more likely: that a person get (*a*) at least 1 six when 6 dice are rolled (*b*) at least two sixes when 12 dice are rolled, or (*c*) at least 3 sixes when 18 dice are rolled. What is the answer?

*a*) is more likely than (

*b*) and (

*c*)

B. (

*b*) is more likely than (

*a*) and (

*c*)

C. (

*c*) is more likely than (

*a*) and (

*b*)

D. (

*a*), (

*b*), and (

*c*) are equally likely

**Answer Key**

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**Solution**

## Click here to expand or collapse this section

*a*) In rolling 6 dice:

$Q_0 = (5/6)^6 = 0.335$

Probability that at least 1 six will come out

$P_a = 1 - Q_0$

$P_a = 1 - 0.335 = 0.665$

Event (*b*) In rolling 12 dice:

$Q_0 = (5/6)^{12} = 0.112$

Probability that exactly 1 six will come out

$Q_1 = (1/6)^1(5/6)^{11} \times 12 = 0.269$

Probability that at least 2 sixes will come out

$P_b = 1 - Q_0 - Q_1$

$P_b = 1 - 0.112 - 0.269 = 0.619$

Event (*c*) In rolling 18 dice:

$Q_0 = (5/6)^{18} = 0.038$

Probability that exactly 1 six will come out

$Q_1 = (1/6)^1 (5/6)^{17} \times 18 = 0.135$

Probability that exactly 2 sixes will come out

$Q_2 = (1/6)^2 (5/6)^{16} \times {_{18}C_2} = 0.230$

Probability that at least 3 sixes will come out

$P_c = 1 - Q_0 - Q_1 - Q_2$

$P_c = 1 - 0.038 - 0.135 - 0.230 = 0.597$

*P _{a}* >

*P*>

_{b}*P*

_{c}*a*) ← Answer: [ A ]

**Another Solution**

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$= (_6C_1)(1/6)^1(5/6)^5$

The probability of getting exactly *x* sixes when 6 dice are rolled

$= (_6C_x)(1/6)^x (5/6)^{6-x} ~ ~ ~ \text{where} ~ x = 0, \, 1, \, 2, \, 3, \, 4, \, 5, \, 6$

The probability of getting 1 or more sixes when 6 dice are rolled

$= 1 - (_6C_0)(1/6)^0 (5/6)^{6}$

The probability of getting exactly *x* sixes when *n* dice are rolled

$= (_nC_x)(1/6)^x (5/6)^{n - x} ~ ~ ~ \text{where} ~ x = 0, \, 1, \, 2, \, \ldots , \, n$

The probability of getting *n* or more sixes when 6*n* dice are rolled is

P & = \sum_{x = n}^{6n} ({_{6n}C_x})(1/6)^x (5/6)^{6n-x} \\

& = 1 - \sum_{x = 0}^{n - 1} ({_{6n}C_x})(1/6)^x (5/6)^{6n - x}

\end{align}$

For 6 dice, *n* = 1. The probability of getting 1 or more sixes is

$\begin{align} \displaystyle

P_a & = \sum_{x = 1}^{6} ({_6C_x})(1/6)^x (5/6)^{6-x} \\

& = 0.665

\end{align}$

For 12 dice, *n* = 2. The probability of getting 2 or more sixes is

$\begin{align} \displaystyle

P_b & = 1 - \sum_{x = 0}^{1} ({_{12}C_x})(1/6)^x (5/6)^{12 - x} \\

& = 0.619

\end{align}$

For 18 dice, *n* = 3. The probability of getting 3 or more sixes is

$\begin{align}

P_c & = 1 - \sum_{x = 0}^{2} ({_{18}C_x})(1/6)^x (5/6)^{18 - x} \\

& = 0.597

\end{align}$

*P _{a}* >

*P*>

_{b}*P*

_{c}*a*) ← Answer: [ A ]

See also this problem at Wolfram post

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