Phillip's Phun Phactory: Probability Problem

[SprucerMoose]

Homework Statement

Phillip's Phun Phactory operates a number of different lotto-type games.
In "Choose 2", you are required to choose two numbers from 1, 2, 3, 4, and 5, and if you win you receive double your money.
In "Choose 3", you are requied to choose three numbers from 1, 2, 3, 4, 5 and 6, and if you win you receive triple your money.
Each game costs $1 to play.
(a) Which game gives you the better chance of making a profit?PART 2
Second prizes are introduced as follows:
In "Choose 2" you receive 1.5 times your money if you correctly select one of the two numbers.
In "Choose 3" you receive double your money if you correctly select two of the three numbers.

(b) Which would you choose if there was a second prize in each game.

The Attempt at a Solution

This problem relates to combinations.

I calculated the amount of outcomes possible using the combinations below.

$$\binom{5}{2} = 10$$ and $$\binom{6}{3} = 20$$

For the remainder of this problem, I am just going to focus on "choose 2", as the methods are the same.

Thus the probability of selecting the correct combination for "Choose 2" is 1/10. It would then be expected that after 10 games, $10 will have been spent and $2 won back ($1 spent on that game and $1 profit), resulting in a net loss of $8 and an average loss of 80 cents per game. However, the book is telling the answer is a 70 cent loss per game. I imagine that they are calculating this based on a winner receiving their money back PLUS double their money or have I done something wrong?

Using the same method, I obtained an average loss of 85 cents for the "Choose 3" game, which the book says is correct.Onto part (b) - where I am struggling. For "Choose 2" I calculated the amount of possible combinations where only a single number of the 2 numbers chosen could appear in the two randomly selected numbers as follows.

$$\binom{3}{1} + \binom{3}{1} = 6$$

Calculated this way because I am calcuating the amount of combinations with each selected number and the other excluded, leaving 1 more to be selected from 3.

This result was confirmed with the ten possible combinations below.

(12)
(13) (23)
(14) (24) (34)
(15) (25) (35) (45)

Say I choose (12)

(13) (14) (15) (23) (24) (25) are the 6 that give one or the other, but not both.

Thus probability of getting a second chance prize is 6/10. Over 10 games I would expect to lose the $8 from the calcuation above and gain 6 x 1.5 = $9.

$9 - $8 = $1 (expected net profit over 10 games)

1/10 = 0.1

Thus a person would expect to win an average of 10 cents per game. The book says a loss of 20 cents would be incurred.

I also came to the solution of an average profit of 5 cents per game in the case of "Choose 3" with a second chance prize, whereas the book says it should be a loss of 55 cents

Can someone please show me where i have gone wrong? Is the book incorrect?Thanks

 

[verty]

I get the same answers for the Choose 2 case.

Chance of getting both numbers = 1/10
Chance of getting neither = 3/10
Chance of getting one only = 6/10

Return: -1 + (1/10)(2) + (6/10)(3/2)
= -1 + 2/10 + 9/10
= 1/10

 

[SprucerMoose]

So the book is wrong. Thanks for the help.