Braingle: 'Red and Blue Dice' Brain Teaser

Red and Blue Dice

Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

Puzzle ID:	#5445
Fun:	(2.67)
Difficulty:	(2.52)
Category:	Probability
Submitted By:	im_para_noid
Corrected By:	Winner4600

Timothy and Urban play a game with two dice. But they do not use the numbers. Some of the faces are painted red and the others blue. Each player throws the dice in turn. Timothy wins when the two top faces are the same color. Urban wins when the colors are different. Their chances 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?

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Comments

cathalmccabe Jul 29, 2002	nice, very good!
bmotts Jul 31, 2002	Ouch
sillywilly Jul 31, 2002	I used an equation and solved for x, the probability of the second die to show red (and 1-x for p that second die shows blue): 5/6(x) + 1/6(1-x) = 1/2 Since I got 1/2, there can only be 3 sides red.
pizzahead Aug 14, 2002	I don't think it adds up. You say that for the chances to be even, there have to be 18 ways to get the same color on top, but you only solve for red. The flaw is that not every combination is a winner for someone. Simply put, in your solution Urban is screwed. He has only 3 winning combinations. I think you'd have to have 5 blue and 1 red on the second die to even the odds. That would give a total of 5 winning combinations to each.
Deucex2 Aug 20, 2002	Actually pizzahead, it does add up. As long as one of the dice is half red and half blue it doesn't matter what the other die has on it. If you had one that was all but one red and one that was all but one blue, the odds would be strongly in favor of the person who wins on two different colors- the guy with same colors will win only about 28 percent of the time.
badolina Aug 20, 2002	There is a solution without probability calculations: No matter what the the first dice shoes, the second should provide us 50:50 chances. That is why it should have equel number of sides: 3 and 3. The question is much more general. 3 sides each is the right answer regardless what the ratio is in the first dice.
peterbradfor Aug 20, 2002	I agree with Badolina. That's the way I solved it, too.
pizzahead Aug 20, 2002	My bad! I read it wrong! I thought one won on red and one won on blue.
cs526columbia Aug 20, 2002	yeah badolina is right, if you plug in any other numbers into the equation from the answer, like 2 & 4, x always = 3
MJs_Kitten Aug 22, 2002	....ok i didnt really like that one...no matter how much u explain it does not make any sense
Bender Nov 08, 2002	As long as one of the dice has 3 red sides and 3 blue, the game has even odds regardless of the configuration of the other die. To see this, imagine you throw them one at a time. First the one with arbitrary configuration: it comes up either red or blue. Whichever color it is, when you throw the second die (the 3/3 one) you have half a chance of matching it.
xpitxbullx Jul 03, 2003	Being dice games are my life, I may enlighten some people on this one... We will call the first die (a) and number the red sides R1, R2, R3, ect. We will call the second die (b) and same numbering system. Consider 5 red and 1 blue side on die (a) and 3 red and 3blue on die (b) 01. (a)R1 + (b)R1 = red wins 02. (a)R1 + (b)R2 = red wins 03. (a)R1 + (b)R3 = red wins 04. (a)R1 + (b)B1 = nothing 05. (a)R1 + (b)B2 = nothing 06. (a)R1 + (b)B3 = nothing 07. (a)R2 + (b)R1 = red wins 08. (a)R2 + (b)R2 = red wins 09. (a)R2 + (b)R3 = red wins 10. (a)R2 + (b)B1 = nothing 11. (a)R2 + (b)B2 = nothing 12. (a)R2 + (b)B3 = nothing 13. (a)R3 + (b)R1 = red wins 14. (a)R3 + (b)R2 = red wins 15. (a)R3 + (b)R3 = red wins 16. (a)R3 + (b)B1 = nothing 17. (a)R3 + (b)B2 = nothing 18. (a)R3 + (b)B3 = nothing 19. (a)R4 + (b)R1 = red wins 20. (a)R4 + (b)R2 = red wins 21. (a)R4 + (b)R3 = red wins 22. (a)R4 + (b)B1 = nothing 23. (a)R4 + (b)B2 = nothing 24. (a)R4 + (b)B3 = nothing 25. (a)R5 + (b)R1 = red wins 26. (a)R5 + (b)R2 = red wins 27. (a)R5 + (b)R3 = red wins 28. (a)R5 + (b)B1 = nothing 29. (a)R5 + (b)B2 = nothing 30. (a)R5 + (b)B3 = nothing 31. (a)B1 + (b)R1 = nothing 32. (a)B1 + (b)R2 = nothing 32. (a)B1 + (b)R3 = nothing 32. (a)B1 + (b)B1 = blue wins 32. (a)B1 + (b)B2 = blue wins 32. (a)B1 + (b)B3 = blue wins Odds are 15/36 per roll that red will win Odds are 3/36 per roll that blue will win If a win/lose decision is made, red is 5 times more likely to win.
xpitxbullx Jul 03, 2003	Therefore, Red has more cracks at the '50/50' chance of winning.
gcriotgurl Sep 01, 2003	WAY too much math for me to enjoy this 1. \
j72883 Sep 01, 2003	I thought this one was pretty fun, as it tested elementary probability. It would be a good brain teaser for the casual math person.
(user deleted) Sep 01, 2003	this is why i simply hated Finite Math... i have a logical, mathimatical mind... but throw probability into the mix and im not even gonna bother.. i just dont get it... enuf said.... great teaser .. and ive enjoyed reading all the comments... smile
marrianna Sep 02, 2003	I agree with xpitxbullx, their chances are NOT even.
JO47 Sep 06, 2003	but pitbull its not a comparison between red and blue, its a comparison between the same color and different colors. so when you said 15/36 for red and 3/36 for blue, that adds up to 18/36 for the same color leaving 18/36 for different colors. which makes the probabilities even.
Varthen Jul 05, 2004	I got it right!!!! although I really dont know any of this math mumbo jumbo...
mr_brainiac Jan 04, 2006	I was confused until I realized that the number of ways to make a pair of bue faces did not have to equal the number of ways to make a red pair. Only the number of ways to match the faces color(whether they are blue or red) needs to equal the number of ways that they mismatch. It's plain to see how easy it is to go off down the wrong path with even one bit of faulty logic.
garul Feb 23, 2006	Nice one
brainjuice Mar 31, 2006	first i think the answer is 1/2. but i actually found it is wrong. the solution is: (i) find the probability red-red will come p(red-red)= 2C2(5/6)^2(1/6)^0= 25/36 (ii) find the probability blu-blu will come p(blu-blu)= 2C2(1/6)^2(5/6)^0= 1/36 (iii) find the probability the same color will appear (either blu-blu or red-red) continued to my next comment..
brainjuice Mar 31, 2006	so sorry.. i've just noticed that my solution at previous comment is wrong. the answer 1/2 is actually right.. my solution is like this: let x be the probability red will come on the 2nd dice then the probability blu will come on the second dice is 1-x. p(red-red)=5/6x p(blu-blu)=1/6(1-x) p(sameColor)=p(red-red)+p(blu-blu)=2/3x+1/6 p(red-blu)=5/6(1-x) p(blu-red)=1/6x p(diffColor)=p(red-blu)+p(blu-red)=5/6-2/3x coz p(sameColor)=p(diffColor) then 2/3x+1/6= -2/3x+5/6 4/3x= 4/6 x= 1/2 coz the probability red will come is 1/2, then there must be 3 red face and 3 blu face on the 2nd dice. the answer is 3 red 3 blu.. so sorry
leftclick Jul 29, 2007	OMG all this over the top maths is doing my head in! Let's simplify... 1. It doesn't matter what the faces of the first die are, that's a red herring. 2. It only matters whether the second die is the same as or different to the first one. 3. Therefore, the second die must have an equal chance of red or blue, i.e. 3 red and 3 blue. To the people who think that there is some chance that nobody wins, please read the question again. There are only two possibilities - the dice are the same or the dice are different.
Bonus66 Oct 29, 2008	How about this.....? b=blue face, r=red face P(bb) or P(rr) = P(rb) + P(br) (1/6 * b/6) + (5/6 * r/6) = (5/6 * b/6) + (1/6 * r/6) b + 5r = 5b + r b=r solution could be (1,1) or (2,2) or (3,3) or (4,4)..... but dice only has six sides, so (3,3) must be answer .
opqpop Jan 27, 2010	leftclick I really like your line. It's at a much higher level than cranking out the math imo. Though the math isn't exactly hard either.