Red and Blue Dice
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
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?
Answer
Each die has 6 faces. When two dice are thrown, there are 36 equally possible results. For chances to be even, there must be 18 ways of getting the same color on top. Let X be the number of red faces on the second die. We have: 18 = 5X + 1(6  X)
X = 3
The second die must have 3 red faces and 3 blue faces.
Hide
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 1x for p that second die shows blue):
5/6(x) + 1/6(1x) = 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 redred will come
p(redred)= 2C2(5/6)^2(1/6)^0= 25/36
(ii) find the probability blublu will come
p(blublu)= 2C2(1/6)^2(5/6)^0= 1/36
(iii) find the probability the same color will appear (either blublu or redred)
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 1x.
p(redred)=5/6x
p(blublu)=1/6(1x)
p(sameColor)=p(redred)+p(blublu)=2/3x+1/6
p(redblu)=5/6(1x)
p(blured)=1/6x
p(diffColor)=p(redblu)+p(blured)=5/62/3x
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. 
Back to Top
 
