Question

A game has 3 possible outcomes, with probabilities p 1, p 2, and p 3. The amount of money that you will win or lose for each outcome is v 1, v 2, and v 3, respectively. What is the expression p 1 v 1 + p 2 v 2 + p 3 v 3 equal to?

Answer 1

The total amount you will win (or lose) in the long run.
The average amount you will win (or lose) per game in the long run.
The exact amount you will win (or lose) per game.
The amount that you will win (or lose) for 3 games.

Answer 2

Do you mean "p^1 , p^2"?

Answer 3

@AaronAndyson

Answer 4

I so hate this kind of question, at least on this site :D
No real way to guide you to it... it's more of a question of vocabulary... ugh

Anyway, short of outright telling you the answer (which I'm not even totally sure of) the value in question is called the "expected value"

Which of those four choices best looks like an elaboration of the phrase "expected value"?
Realistically, of course...

Answer 5

The average amount you will win (or lose) per game in the long run.

Answer 6

We're in agreement :)

Answer 7

Okay. Thank you.

Answer 8

No problem :)

Answer 9

Could you help me with another question please?

Answer 10

I do what I can.

Answer 11

A lottery ticket costs $5. You pick 3 two-digit numbers (anything from 00 to 99) in order.

If you match the first number only, you win $200.
If you match the first two numbers and not the third, you win $5000.
If you match all three numbers, you win $100,000.

Answer 12

It wants to know if the game is fair or not. I do not know how to set up the work to solve it.

Answer 13

This is a real brainbuster.
Well, not really...it might actually be enjoyable ^^

Let me ask you a question first... how do YOU define a fair game?

Answer 14

One in which the expected value is 0

Answer 15

Ugh, you're so boring... did you look up the textbook definition? -_-


I'm kidding, of course ^^
You're absolutely right, though I'd have hoped you'd phrase it in more "fun" words since we are talking of a game, after all XD

Nonetheless, you ARE right, so can you go ahead and find the expected value?

Or... is that where we begin? ^_^

Answer 16

Okay. That is where we begin the "game" i am unsure how because my example is not clear.

Answer 17

Let me think for a minute

Answer 18

Alright.

Answer 19

And I'm done.

Answer 20

Okay, step right up, go ahead and pick three two digit numbers out of a hundred and line them up in order, anything from 00 to 99...

Or, we stop and think for a moment, if we line up three two-digit numbers, what are the chances that we win (in the first category)?
That is to say, what are the odds that the first number we pick is the first number drawn?

Answer 21

24,53, 64.
So, the chance that I would win the first number is \[\frac{ 24 }{ 99 }\] Would that be correct?

Answer 22

Or wait, no it should be \[\frac{ 1 }{ 99}\]

Answer 23

Unfortunately, no, the numbers that the you (the player) choose do not figure into the calculations at all, they are not even treated as numbers, but just objects.

double unfortunately, it seems LaTeX is not rendering on my browser.
Hold for a moment while I refresh, and see if that solves it...

Answer 24

Okay

Answer 25

Done, and done.
Now, the problem, no matter how tempting, is that it's a little more complex than that.
Ever heard of permutations?

Answer 26

By the way, it's \(\Large \frac1{100}\)
there is a 1% chance of the first number drawn being the same as the first number on your ticket, HOWEVER, that also takes into account the possibility that the second number drawn is the second number on your ticket, which is a win of the SECOND category, and should not be counted in the first...

Catch me so far?

Answer 27

Kind of. So, that would make the next fraction \[\frac{ 2 }{ 100 }\]?

Answer 28

I'm sorry

Answer 29

The way I began this problem was unnecessarily confusing, so can we start over? I think I have my thoughts better organised now ^^

Answer 30

Alright. That works.

Answer 31

Expected value. To get this, we need the list of ALL possible outcomes. And they are...

-You LOSE
-1st Category Win (ONLY the first number is correct; pays 200)
-2nd Category win (ONLY the first two numbers are correct; pays 5000)
-Jackpot (ALL numbers correct; pays 100,000)

So far, so good, yes?

Answer 32

Yes.

Answer 33

That's actually, not very accurate ugh... It's so hard to phrase XD
But I'll try...

Better phrased:
X = wrong
O = right
? = doesn't matter

-Loss: X ? ?
-Win1: O X ?
-Win2: O O X
-Jackpot: O O O

In other words, in the first category win, the third drawn number could literally be anything, right, or wrong, it doesn't matter, as long as the second number is wrong.

Does that make sense? I feel like I'm being convoluted T.T

Answer 34

It makes since.

Answer 35

Well, good. Now, permutations, if you haven't heard of them, then l'll make do.

Question: How many ways are there, total, to draw three numbers from a hundred?

Answer 36

Draw three numbers in ORDER, that is.
There is a simple way to figure this out:

For the first number, there are 100 possible outcomes
For the second number, there will only be 99 left, since one is already picked
For the third, there will be only 98.

So the number of total possible ways to draw three numbers in order from a hundred is
100 x 99 x 98
which I would rather not work out, it's simpler this way :>

Catching me still?

Answer 37

Yes, perfectly.

Answer 38

Good. This will be the basis of our probabilities.
Now, how many ways are there to draw a LOSING set of three numbers?
Can you work that out? ^^

Answer 39

Just to remind you, for a draw to be a total loss, the first number has to be incorrect. Nothing else matters.
X ? ?

So... how do we get this?

Answer 40

So, if losing and only the first number has to be wrong then it would still be 100, correct?

Answer 41

Not exactly. Put it this way, of the 100 numbers you could draw, how many of them are NOT the first number in your ticket?

Answer 42

oh 99.

Answer 43

Yes. 99 possibilities for the first number of the draw.
And the second?

Answer 44

98

Answer 45

Nope :)
There were 100 numbers in total, but for the first number, we refrained from one of them because that particular one would have made this draw, well, not a loss.
So only 99 for the first number

But we still have only taken away one number, so for the second number of the draw, upon which there are NO restrictions, there are still 99 possible picks!

Get me? ^^

Answer 46

Okay, which means the third on would also be 99.

Answer 47

No.
Because by the time we get to the third, we'd have taken away two numbers ^^
The first, and the second. So 98 for the third ^^

Are you feeling the brainbusting yet? :D

Answer 48

Hmm. Okay i think i follow what you are saying there. Oh yeah. It's exploding.

Answer 49

Let me run it down.
To get a losing draw:
For the first number, only 99, since we can't pick the correct first number
For the second number, still 99, since the only number we can't pick is whatever we picked for the first one.

For the third number, 98, since we can't pick whatever was the first or second number

Got it now?