Question

Question: What is the expected sum of the numbers that appear on two dice, each biased so that a 3 comes up twice as often as each other number?

Short answer

Answer

We are expected to lose $0.37 (or 37 cents) per lottery ticket.

Step-by-step solution

Given Information  

$1 lottery ticket

Definition of Integral and Integration  

Probability distribution Types: Discrete Probability distribution and Continuous probability distribution. They each define Discrete and Continuous random variables respectively.

Discrete Probability distribution gives the probability that a discrete random variable will have a specified value. Such a distribution will represent data that has a finite countable number of outcomes. The two conditions discrete probability distributions must satisfy are:

• \[{\bf{0}}{\rm{ }} \le {\rm{ }}{\bf{P}}\left( {{\bf{X}}{\rm{ }} = {\rm{ }}{\bf{x}}} \right){\rm{ }} \le {\rm{ }}{\bf{1}}\]. This implies that the probability of a discrete random variable,\[{\bf{X}}\], taking on an exact value,\[{\bf{x}}\], lies between\[{\bf{0}}\]and\[{\bf{1}}\]

• \[\sum {\bf{P}}\left( {{\bf{X}}{\rm{ }} = {\rm{ }}{\bf{x}}} \right){\rm{ }} = {\bf{1}}\]. The sum of all probabilities must be equal to\[{\bf{1}}\]

Some commonly used Discrete probability distributions are Geometric distributions, binomial distributions and Bernoulli distributions

Calculation of the expected value when a $1 lottery ticket is bought in which the purchaser wins exactly $10 million

We need to select 6 numbers from the 50 numbers\[\left\{ {1,2,3,......,50} \right\}\].

Exactly one of the ways to select these 6 numbers will result in the 6 winning numbers.

The probability is the number of favourable outcomes divided by the number of possible outcomes.

\[\begin{array}{l}P(win) = \frac{{Numberoffavourableoutcomes}}{{Numbeopossibleoutcomes}}\\ = \frac{1}{{C\left( {50,6} \right)}}\\ = \frac{{6!44!}}{{50!}}\\ = \frac{1}{{15,890,700}}\\\end{array}\]

\[\begin{array}{l}P(loss) = 1 - P(win)\\ = 1 - \frac{1}{{15,890,700}}\\ = \frac{{15,890,699}}{{15,890,700}}\end{array}\]

When we win, then we gain $10 million and lose the cost of the lottery ticket

\[\left( {\$ 10,000,000 - 1} \right) = \$ 9,999,999\]

When we lose, then we lose the cost of the lottery ticket ($1).

The expected value is the sum of the product of each possibility\[x\]with its probability\[p(x).\]

\[E(x) = \sum {xp\left( x \right)} \]\[\]

\[\begin{array}{l} = \left( {\$ 9,999,999} \right) \times \frac{1}{{15,890,700}} + \left( { - \$ 1} \right) \times \frac{{15,890,699}}{{15,890,700}}\\ = - \frac{{\$ 5,890,700}}{{15,890,700}}\\ \approx - \$ 0.37\end{array}\]

Therefore, per lottery ticket, we are expected to lose $0.37 (or 37 cents).