Math  /  Data & Statistics

QuestionA dice game involves rolling 2 dice. If you roll a 2,3,4,10,112,3,4,10,11, or 12 you win $5\$ 5. If you roll a 5,6,7,85,6,7,8, or 9 you lose $5\$ 5. Find the expected value you win (or lose) per game.
What is the expected value of this game? \ \square$ type your answer.(include the negative if needed here)
Is this a fair game? \square type your answer...

Studdy Solution

STEP 1

What is this asking? If we play this dice game many times, how much money can we expect to win or lose on average per game? Watch out! Don't forget there are different ways to roll the same total with two dice!

STEP 2

1. Calculate winning probabilities
2. Calculate losing probabilities
3. Calculate expected value

STEP 3

Let's **figure out** the probability of rolling each **winning number**.
Remember, there are 3636 possible outcomes when rolling two dice (6 outcomes for the first die times 6 outcomes for the second die).

STEP 4

There's only **one way** to roll a 22 (1+11+1), so the probability is 136\frac{1}{36}.
Similarly, there's only **one way** to roll a 1212 (6+66+6), so the probability is 136\frac{1}{36}.

STEP 5

There are **two ways** to roll a 33 (1+21+2 and 2+12+1), so the probability is 236\frac{2}{36}.
Same goes for an 1111 (5+65+6 and 6+56+5), giving us 236\frac{2}{36}.

STEP 6

There are **three ways** to roll a 44 (1+31+3, 2+22+2, 3+13+1), making the probability 336\frac{3}{36}.
A 1010 is similar (4+64+6, 5+55+5, 6+46+4), also 336\frac{3}{36}.

STEP 7

Now, let's **add up** all the winning probabilities: 136+136+236+236+336+336=1236=13\frac{1}{36} + \frac{1}{36} + \frac{2}{36} + \frac{2}{36} + \frac{3}{36} + \frac{3}{36} = \frac{12}{36} = \frac{1}{3}.
So, we have a 13\frac{1}{3} chance of **winning**!

STEP 8

We could calculate the probability of each losing number like we did for the winning numbers, but there's a **faster way**!
Since the total probability of all outcomes must add to one, and we know the winning probability is 13\frac{1}{3}, the probability of **losing** is simply 113=231 - \frac{1}{3} = \frac{2}{3}.
Slick, right?

STEP 9

The **expected value** is calculated by multiplying each outcome by its probability and then adding everything up.

STEP 10

If we **win**, we get $5\$5, and the probability of winning is 13\frac{1}{3}.
If we **lose**, we lose $5\$5 (which is the same as winning $5-\$5), and the probability of losing is 23\frac{2}{3}.

STEP 11

So, the **expected value** is: ($5)13+($5)23=$53$103=$53$1.67(\$5) \cdot \frac{1}{3} + (-\$5) \cdot \frac{2}{3} = \frac{\$5}{3} - \frac{\$10}{3} = -\frac{\$5}{3} \approx -\$1.67.

STEP 12

The expected value of this game is approximately $1.67-\$1.67.
Since the expected value is not zero, this is *not* a fair game.
On average, you'll lose about $1.67\$1.67 each time you play!

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