Riddles, brain puzzles and mathematical problems

[Froge]

Ok, it's a bad puzzle. It takes way too long to do without knowledge of statistics, and even then comes down to an extraordinary amount of grindwork, unless you have a graphics calculator.

Spoiler below!

Part 1 is just finding out all the different possible outcomes of one round. The easy way to do this is with binomial probabilities.

ALIAS Probability generator: (0.5 + 0.5)^3
Kman Probability generator: (0.6 + 0.4)^2

A-win: 3-2, 3-1, 3-0, 2-1, 2-0, 1-0

= (0.5)^3 * (0.6)^2 + 0.5^3 * 2 * (0.6)(0.4) + 0.5^3 * (0.4)^2 + 3 * 0.5^3 * 2 * 0.6 * 0.4 + 3 * 0.5^3 * 0.4^2 + 3 * 0.5^3 * 0.4^2
= 0.425

K-win: 1-2, 0-2, 0-1

= 3 * 0.5^3 * 0.6^2 + 0.5^3 * 0.6^2 + 0.5^3 * 2 * 0.6 * 0.4
= 0.24

Tie: 0-0, 1-1, 2-2

= 0.5^3 * 0.4^2 + 3 * 0.5^3 * 2 * 0.4 * 0.6 + 3 * 0.5^3 * 0.6^2
= 0.335

Check: 0.335 + 0.425 + 0.24 = 1, so it's correct.

Alias E(X) = 6 * 0.425 - 10 * 0.24 - 2 * 0.335
= -$2.08

But he gets $5 from Kman so ALIAS is actually at $5 - $2.08 = $2.92

SECOND HALF:

Redo all of the above binomial probabilities with Kman's values fixed to fair coins (I had to resort to a calculator at this point) and you should get:

P(ALIAS wins) = 1/2
P(Kman wins) = 3/16
P(Tie) = 5/16

Probability of ALIAS winning round 1: P(ALIAS wins) = 1/2

Probability of ALIAS winning round 2: P(Tie) * P(ALIAS wins) = (5/16) * (1/2)

Probability of ALIAS winning round 3: 2 * P(Tie) * P(ALIAS wins) = (5/16)^2 * (1/2)

Probability of ALIAS winning round 4: 3 * P(Tie) * P(ALIAS wins) = (5/16)^3 * (1/2)

If this continues you should see that it's a geometric series with first term 1/2 and ratio of 5/16. Using the formula for summation of an infinite geometric series:

Sn = u1 / (1 - r)
= (1/2) / (1 - 5/16)
= 8/11

ALIAS E(X): 8/11 * $11 - 3/11 * $11 = $8 - $3 = $5

Adding $5 to his winnings from round one, $5 + $2.92 = $7.92.

ANSWER: $7.92

I'll give y'all another question to do as replacement for this one. This question doesn't require much knowledge of stats at all.

Horror Games


An interview with citizens of Chronoville gives us the following information:

>45% of the population has played Amnesia: The Dark Descent.

>x% of the population has played Silent Hill.

>If someone has not played Silent Hill, then the probability of them having played Amnesia is 3/7.

>If someone has played Silent Hill, then the probability of them having played Amnesia is 6/13.

One person is picked at random. What is the probability that they have not played Amnesia and Silent Hill? (This means that they can have played one of the games, or none of the games at all. Just as long as it's not both games)