Question

We roll a pair of dice. If the sum of the dice is 7, you pay me $32. If the sum is not 7, I pay you the number of dollars indicated by the sum of the dice. What is your expected value for the game?

Answer 1

how many ways can you roll to 7?

Answer 2

(die1, die2)
1,6
2,5
3,4
4,3
5,2
6,1

Answer 3

EV = x1*P(x1) + x2*P(x2) + .. xn*P(xn)

Answer 4

if i roll a sum of 7, i lose 32 dollars.
if i dont roll a sum of 7 , i gain the sum of the dice

Answer 5

if i roll a sum of 7, i lose 32 dollars.
if i dont roll a sum of 7 , i gain the dollar amount of the sum of the dice

Answer 6

mmkay do what @perl says XD

Answer 7

now you have to break down the cases

Answer 8

i think there might be a nice symmetry im ust gonna guess before seeing that full form pan out

Answer 9

there is no trick or fancy math required here, its purely mechanical

Answer 10

7= 1/6th of the time
5/6th of the time im still gonna get expected sum = 7 so
7*5=35m and u are gaining 32, so on ag u lose 3 bucks

Answer 11

there are sums 2,3,4,5,6,8,9,10,11,12 (note there is no 7)

Answer 12

im thinkinngg expected is still 7, not sure though :P

Answer 13

im going to upload an image

Answer 14

damn, attach file button is disabled

Answer 15

Here are the probability of P(X) , where x = sum of two die
P(2)=1/36
P(3)=2/36
P(4)=3/36
P(5)=4/36
P(6)=5/36
P(7)=6/36
P(8)=5/36
P(9)=4/36
P(10)=3/36
P(11)=2/36
P(12)=1/36

Answer 16

EV = (-7)P(7) + 2*P(2) + 3*P(3) + 4*P(4) + 5*P(5) + 6*P(6) + 8*P(8) + 9*P(9) + 10*P(10) + 11*P(11) + 12*P(12)

Answer 17

and chucky's gone

Answer 18

EV = 2*1/36 + 3*2/36 + 4*3/36 + 5*4/36 + 6*5/36 + (-7)*6/36 + 8*5/36 + 9*4/36 + 10*3/36 + 11*2/36 +12*1/36

Answer 19

EV = 14/3 = 4.16666666666666666666666666666666666666666666666666

Answer 20

hmm

Answer 21

feels weird

Answer 22

i feel like he shud be gaining 3 bucks on avg

Answer 23

why did my symmetry fail

Answer 24

4.67 dollars gain on average

Answer 25

the dollar amounts you win are weighted by the probability of getting that sum

Answer 26

right look at this argument

Answer 27

they are not uniform,

Answer 28

1/36 times its a 2
1/36 times in a 12
on avg 2/36 times u are paying a 12 or 2 so its avg of 7 bucks

Answer 29

ok

Answer 30

similarliy
2/36 times u pay 3
2/36 times u have 11
again on avg 4/36 u pay an avg of 7

Answer 31

that shud continue till
8 and 6

Answer 32

then -7*6/36

Answer 33

interesting :)

Answer 34

i mean 32*

Answer 35

5/6th of the time u shud be getting 7 and
1/6th u lose 32

Answer 36

something must be wrong with my arguments

Answer 37

ill see what i can find on paper, sometimes things pop out on paper

Answer 38

you dont mind if i get back to you. or if anyone else wants to try to solve dan's symmetry argument @everyone

Answer 39

lol

Answer 40

iheartmath @ Dan

Answer 41

dan, youre pretty creative. you did that without knowing about EV?

Answer 42

you have an intuition for probability

Answer 43

i knew the expected value to work but just thought this was interesting lol

Answer 44

oh ok

Answer 45

kinda lazy to write out all the expected values

Answer 46

yeah it was a chore

Answer 47

and thanks <3

Answer 48

i had to put it in notepad

Answer 49

maybe i made a mistake in arithmetic?

Answer 50

no its probably righ

Answer 51

this is cool question

Answer 52

@ganeshie8 ganeshie likes this kinda stuff too

Answer 53

oh you stopped going there now?

Answer 54

alright bye baby

Answer 55

im not sure if this helps in anyway to think about it like this

Answer 56

if u imagine a binomial distribution 2 to 12, and 7 being the mean, removing the mean still has same ammount left and right of the mean , so 7 should still be the ev

Answer 57

ganeshie can u tell me why my way of thinking fails

Answer 58

http://prntscr.com/50uz55 this part

Answer 59

i like problems that have 'mechanical' solutions, though

Answer 60

im not so great at the creativity part

Answer 61

http://www.wolframalpha.com/input/?i=%282*1%2B3*2%2B4*3%2B5*4%2B6*5-7*6%2B8*5%2B9*4%2B10*3%2B11*2%2B12*1%29%2F36

Answer 62

oh i see it makes sense now!!

Answer 63

it shud be skewed more to the right, since the weight is shifting to the right ofc!

Answer 64

ugh nvm

Answer 65

is that your final answer?

Answer 66

nope, we ganeshie will tell me why im wrong :)

Answer 67

http://www.wolframalpha.com/input/?i=%282*1%2B3*2%2B4*3%2B5*4%2B6*5%2B8*5%2B9*4%2B10*3%2B11*2%2B12*1%29%2F30

Answer 68

look it does make sense if u eliminate!! the -7 every 1/6th time

Answer 69

out of the 30 rolls, u do get a 7 on avg

Answer 70

7*30/36 - 32*6/26 =[ why this fails!!

Answer 71

the probability for each sum is not same rihgt ?

Answer 72

yeah

Answer 73

the values to the right are greater, so will make it seesaw to the right

Answer 74

7 has the highest probabilty because a+b=7
number of ways of choosing two positive integers for a, b = 6

since a,b can only be chosen from {1,2,3,4,5,6}, we are getting some symmetry:
P(n) = P(14-n)

hmm

Answer 75

i wonder if we can do think about it like stick and placement problem to see how much of each sum shud exist