How To Solve This Equation?
To be honest, I have no clue how to get X by itself and solve for this... please help me. The question is:
You pick a card at random from an ordinary deck of 52 cards. If the card is an ace, you get 9 points; if not, you lose 1 point.
I somehow found out that the game is not fair because the expected value is -3/13.
What value for the aces would make the game fair? To find this, solve the equation:
0 = 4/52(x) + 48/52(−1)
I have virtually no idea how this equation works or what to start with. I need help! Thank you.
- PuzzlingLv 73 years agoFavorite Answer
Here's the logic.
If you draw an ace (probability = 4/52 = 1/13), you get 9 points.
If you draw any other card (probability = 48/52 = 12/13), you get -1 points.
If you multiply that out, you get the expected number of points per play:
E(X) = (1/13) * 9 + (12/13) * -1
= 9/13 - 12/13
So in the long run you are expected to lose with this game an average of 3/13 points per play.
If it is to be fair, the amount should be 0 (an average of nothing lost or won).
So instead of 9 points, let's use 'x' points. I'm also going to use 1/13 and 12/13 rather than the equivalent fractions of 4/52 and 48/52, but I think you see why they used that.
You want the expected winnings to be 0 in the long run to be fair.
(1/13)x + (12/13)(-1) = 0
x/13 - 12/13 = 0
Multiply everything by 13:
x - 12 = 0
x = 12
So you should be getting 12 points for drawing an ace for the game to be fair.
- Some BodyLv 73 years ago
0 = 4/52(x) + 48/52(-1)
0 = 4/52(x) - 48/52
48/52 = 4/52(x)
48 = 4x
x = 12
- 3 years ago
0=4/52(x)+48/52(-1) -> 0=4/52(x)-48/52 ->
0=4-48x -> 48x=4 -> x= .0833