SOLUTION: If a player rolls doubles when she uses two dice, she wins $5. If the person rolls a 3 or 12, she wins $15. The cost to play the game is $3. Find the expectation of the game. 2(

Algebra ->  Probability-and-statistics -> SOLUTION: If a player rolls doubles when she uses two dice, she wins $5. If the person rolls a 3 or 12, she wins $15. The cost to play the game is $3. Find the expectation of the game. 2(      Log On


   



Question 806336: If a player rolls doubles when she uses two dice, she wins $5. If the person rolls a 3 or 12, she wins $15. The cost to play the game is $3. Find the expectation of the game.
2(5)+3+12(15)*3=

Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
If a player rolls doubles when she uses two dice, she wins $5. If the person rolls a 3 or 12, she wins $15. The cost to play the game is $3. Find the expectation of the game.
------------------------------
Random "winnings":: 5.........15........-3
Probabilities:::::: 6/36......3/36......|..27/36
===================================================
E(x) = (6*5 + 3*15 - 27*3)/36 = -6/36 = -16 2/3 cents
=====================
Cheers,
Stan H.
================

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!

Refer to the below table of dice results for 2 six-sided dice:
Sum Ways # of Ways
21,11
31,2; 2,12
41,3; 2,2; 3,13
51,4; 2,3; 3,2; 4,14
61,5; 2,4; 3,3; 4,2; 5,15
71,6; 2,5; 3,4; 4,3; 5,2; 6,16
82,6; 3,5; 4,4; 5,3; 6,25
93,6; 4,5; 5,4; 6,34
104,6; 5,5; 6,43
115,6; 6,52
126,61


Note that there are 36 different possible results, 6 of which are doubles, 2 of which are 3, and one of which is 12. However, the 12 result is also doubles. Because you don't specify what happens when a 12 is rolled (score 15 for the 12 which overrides the doubles or score 5 for the doubles which overrides the 12, or score 20 for meeting both criteria), I'm going to go with a literal interpretation of the given rules: The player gets 5 for rolling doubles AND 15 for rolling 12.

So the probability of rolling doubles OTHER than 2 sixes = 12 is , the probability of rolling 12 is , the probability of rolling 3 is , and the probability of rolling anything else, which we must presume represents a loss of the initial $3 wager, is (28 ways to lose out of 36 outcomes).

And then the expected payout is:



You can do your own arithmetic, but in the long run you lose a quarter every time you play.

John

Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
The Out Campaign: Scarlet Letter of Atheism