Tim and Al are playing a game with two dice. They are not using numbers, but instead the die faces are colored. Some of the faces are colored blue and others are red.
Each player throws the dice in turn. Tim wins when the two top faces are the same color. Al wins when the colors are different. Their chances at winning are even.
The first die has 5 red faces and 1 blue face. How many red and how many blue are there on the second die?
I'm posting one puzzle, riddle, math, or statistical problem a day. Try to answer each one and post your answers in the comments section. I'll post the answer the next day. Even if you have the same answer as someone else, feel free to put up your answer, too!
Wednesday, December 16, 2009
Even the Odds
Subscribe to: Post Comments (Atom)
The second die has 3 read and 3 blue faces.ReplyDelete
I agree with Heather.ReplyDelete
Yes Heather is right.ReplyDelete
Tim wins when top two faces are of same colour.
Probablity of both the dice to be Red is
5/6 *3/6 = 5/6*1/2= 5/12
Probablity of both the dice to be Blue is
1/6*3/6=1/6*1/2 = 1/12
Total probablity is 5/12 + 1/12 = 6/12 = 1/2
As long as one Die is always 3 red and 3 blue, you will always have a 50-50 chance.
You can change the colors on the first die as much as you want, but it will only change the color pairs and differences.
And, going the other way around, it's impossible to get a fair game unless at least one of the dice is equally split between red and blue.ReplyDelete
This also works for dice with more (or less) than six sides.
Looks like you have it all figured out, so rather than write out the explanation again, I think I'll leave it as is.ReplyDelete
Sorry I've been posting the questions so late, lately. It seems like life has been busy for me in the mornings. I need to get in the habit of scheduling the posts...