A friend asks you to play a game using a fair 6 sided # cube. If you roll a 2 you win 4 points, and if you roll a 5 you win 7 points. Rolling any other # causes

Question

Please help me solve this. 

Answer 1

Sounds like the problem description got cut off.  Rolling any other # causes ???

 

1) To summarize:

 

• Roll a 2, you win 4 points
• Roll a 5, you win 7 points
• Roll a 1, 3, 4 or 6, you lose 3 points
• This covers all the possibilities.  That is, there are only 6 possible outcomes when you roll
• The fact that the dice is fair means every number has a 1/6 chance of appearing on a given roll

2)  This is an expected value problem.  Expected Value = (P[2] * 4) + (P[5] * 7) + (P[1,3,4 or 6] * -3) where P[x] means the probability of x occurring

3)  Plug in the numbers to above equation:  Expected Value = (1/6 * 4) + (1/6 * 7) + (4/6 * -3)

4)  And do the math :  Expected Value = 4/6 + 7/6 + -12/6 =  -1/6

5)  This means that, over the long term, you can expect to lose -1/6 point on every roll.  So, for example, if you roll 6000 times, you can expect to lose around 1000 points (i.e., -1/6 * 6000 = -1000)

6)  If the game is not fair, then this implies the probability is not 1/6 for each roll of the cube.  The advantage depends on how the probabilities are distributed.  For example, in the extreme case, if the cube is skewed so that the number 5 will never be rolled and the other number each has a 1/5 probability, then your friend has the advantage.  But if it is skewed so that number 5 always appears, then you have the advantage. 

 

Just for kicks, I ran an excel program using the RANDBETWEEN function to see what happens if a cube is rolled 6000 times with the above scoring system.  The average score per roll was -.172, which is close to the -.167 predicted by probability theory.  Of course, the average score will change each time I re-run the excel spreadsheet because the random numbers are regenerated. 

 

So, one take-away from Expected Value is that the number (-1/6 in this case) is actually not a possible outcome if you roll the cube once.  Expected value is sort of a long-term average.

Answer 2

P(roll a 2) = 1/6

P(roll a 5) = 1/6)

P(roll 1,3,4,6) = 4/6 = 2/3

 

The expected value is the probability of an outcome * the value of that outcome

 

(1/6)(4) + (1/6)(7) * (2/3)(-3)

[2/3 + 7/6 - 2]

[4/6 + 7/6 - 12/6]

-1/6

 

So on each roll you would be expected to lose 1/6 point.