A friend offers to play a game, in which you roll 3 standard 6-sided dice. If all

The expected value is $0.32

Step-by-step explanation:

The expected value can be calculated as:

EV = ∑xₙ*pₙ

Where:

xₙ is the n-th event.

pₙ is the probability of the n-th event.

First, let's see the events in this situation and the probability of each event.

Lossing $1.

This happens when you roll all different numbers.

To calculate the number of combinations where you roll all different values, we first need to count the number of possible options for each roll, this is:

First dice: you have 6 options

Second dice: You have 5 options, because we remove the option that was rolled in the first dice.

Third dice: You have 4 options, because we removed the two outcomes showed in the two previous rolls.

Then the number of combinations is:

c = 6*5*4 = 120.

And the total number of combinations for rolling 3 dice is equal to the product between the number of outcomes for each dice, each dice has 6 possible outcomes, then the total number of outcomes is:

C = 6*6*6 = 216

The probability of rolling all different numbers will be equal to the quotient between the number of combinations where you roll all different values and the total number of combinations:

P₁ = 120/216 = 0.56

In the other case, where you roll at least two equal outcomes, you win $2.

The probability for this case will be the complement of the previous one, then the probability can be calculated as:

P₂ = 1 - P₁ = 1 - 0.56 = 0.44

Then the expected value will be:

EV = -$1*0.56 + $2*0.44 =  $0.32

The expected value is $0.32

AC bisects ∠BAD, => ∠BAC=∠CAD (1)

thus in ΔABC and ΔADC,    ∠ABC=∠ADC (given),

                                      ∠BAC=∠CAD      [from (1)],

AC (opposite side side of ∠ABC) = AC (opposite side side of ∠ADC), the common side between ΔABC and ΔADC

Hence,      by AAS axiom, ΔABC ≅ ΔADC,

Therefore,  BC (opposite side side of ∠BAC) = DC (opposite side side of ∠CAD),  since (1)

Hence, BC=DC proved.