Math Games

Archive for August, 2006

Question: The impossible quiz at school?

I've been trying to find a website with the game but I haven't found one that passes through. I found one game website called cool Math Games which has bloxorz but that one makes me mad. Well what game sites can I use to pass the school internet thing?

Answer: www.addictinggames.com or www.miniclip.com

These are two of the largest free games websites and they have the sequels to the quiz as well.

cool Math Games bloxorz 1-10

Question: Probability distributions maths problem help. 10 pts easy?

Hi, could someone please help solve and explain this maths problem on probability distributions. Thanks! The problem is:

A game involves 2 balls being released, one after the other, rolling down a slope and each ending up in one of 5 numbered slots, as shown below. (it's in the link)

http://www.flickr.com/photos/andromeda-4-eva/4313927070/

The random release mechanism is such that each slot has an equal chance of receiving each ball. A player pays $5 for one release of the two balls. The total of the scores achieved by the two balls is the player's score for the game and the player receives that number of dollars as a prize.
(i.e. if you look in the diagram the $6 mechanism has two balls in slot 3 because 3+3 = $6 etc)

Hence,:
a) what total score would we expect 40% of the plays to exceed?
b) Approximately how much should the organizers expect to be "up" after 100 plays of the game?

Answer: First, we must determine the expected return of one play.

This is 2[(1/5)(0) + (1/5)(1) + (1/5)(1) + (1/5)(3) +(1/5)(5)] = 20/5 = 4.
The two is because each play drops 2 balls.
The 1/5 is because each slot has equal probability of being dropped into.
The 0, 1, 3, and 5 are the returns for landing in each slot.

This expected value is the mean of the probability distribution. We also need to know the standard deviation. This is [sum(x - mu)^2 P(x)]^(1/2)

[(1/5)(0 - 4)^2 + (1/5)(1 - 4)^2 + (1/5)(1 - 4)^2 + (1/5)(3 - 4)^2 + (1/5)(5 - 4)^2]^(1/2) = 2.68

The formula for the z-score is z = (x - mu)/sigma. We need to manipulate this to solve for x and we get z(sigma) + mu = x.

We need to consult the standard normal table to determine how many standard deviations above the mean represents 60% (we want 40% to score more than this, so 60% score less). I get this to be about 0.255 standard deviations. (0.255)*(2.68) = 0.6834. This means 40% time, the player's score will be higher than 4.68.

Since the expected return to the player is $4 and the player pays $5, the organizer can "expect" to make $1 every time someone plays. After 100 plays, he expects to be "up" $100.

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