### Brain Teasers

# Red and Blue Dice

Probability
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?

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.

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## Comments

nice, very good!

Ouch

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.

5/6(x) + 1/6(1-x) = 1/2

Since I got 1/2, there can only be 3 sides red.

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.

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.

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.

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.

I agree with Badolina. That's the way I solved it, too.

My bad! I read it wrong! I thought one won on red and one won on blue.

yeah badolina is right, if you plug in any other numbers into the equation from the answer, like 2 & 4, x always = 3

....ok i didnt really like that one...no matter how much u explain it does not make any sense

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.

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.

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.

Therefore, Red has more cracks at the '50/50' chance of winning.

WAY too much math for me to enjoy this 1. \

I thought this one was pretty fun, as it tested elementary probability. It would be a good brain teaser for the casual math person.

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*

I agree with xpitxbullx, their chances are NOT even.

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.

I got it right!!!!

although I really dont know any of this math mumbo jumbo...

although I really dont know any of this math mumbo jumbo...

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.

Nice one

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..

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..

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

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

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.

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.

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 .

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 .

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.

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