If a person rolls two dice and obtain a sum of 2 or 12, he wins $15. Find the expectation of the game if the person pays $5 to play.

yuhki  Nov 21, 2014

Best Answer 


Thanks Chris, yes these odds are dreadful!

Melody  Nov 23, 2014

3+0 Answers


I don't know how to do this 




Well the probablility of getting double 2 or double 6 is 2/36 = 1/18

So he has 1/18 chance of winning $15

and a 17/18 chance of losing $5


So maybe

$$\left({\frac{{\mathtt{1}}}{{\mathtt{18}}}}{\mathtt{\,\times\,}}{\mathtt{15}}\right){\mathtt{\,-\,}}\left({\frac{{\mathtt{17}}}{{\mathtt{18}}}}{\mathtt{\,\times\,}}{\mathtt{5}}\right) = {\mathtt{\,-\,}}{\frac{{\mathtt{35}}}{{\mathtt{9}}}} = -{\mathtt{3.888\: \!888\: \!888\: \!888\: \!888\: \!9}}$$

So the person should expect to lose $3.89


This probably is not correct.  I am totally guessing.  Sorry.


If you find out how it is done please let us know.  If I find out I will post it on this thread.  

Someone else probably knows how to do it. I shall ask around.  :))

Melody  Nov 22, 2014

This is correct, Melody.......!!!!

Another way to look at this is.......call the amount of money the person bets = B...........and call the amount of money he wins  = 3B

So the expected value of any bet is

3B(1/18) - B(17/18) =

3B/18 - (17B)/18  =

-(14B)/18 =

-.7778 B 

In other words......the player stands to lose about 78 cents on each dollar he bets........

(Contrast this to the game of roulette where, on the "American Wheel," the player stands to lose a little more than 5 cents per dollar bet on any spin !!! )


CPhill  Nov 22, 2014
Best Answer

Thanks Chris, yes these odds are dreadful!

Melody  Nov 23, 2014

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