Arjun Shastry

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posted 14 years ago

This was asked during one test:

Two friends A and B are playing with two dices. One dice has 5 red faces and one blue.They define the game:"If same color appears on top faces of dices ,A wins else B wins."Probability of winning of each is same(1/2).

How many faces of second dice are blue/red?

[hint: requires basic probability]

Two friends A and B are playing with two dices. One dice has 5 red faces and one blue.They define the game:"If same color appears on top faces of dices ,A wins else B wins."Probability of winning of each is same(1/2).

How many faces of second dice are blue/red?

[hint: requires basic probability]

MH

Roy Tock

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Posts: 83

posted 14 years ago

Interesting...the answer is 3/3. I wouldn't have expected that!

Let R be the number of red faces on the second die. The probability of rolling two reds, then, is 5R/36. The probability of rolling two blues is 1(6-R)/36. We know the sum of the two probabilites must be 1/2, so

5R/36 + (6-R)/36 = 1/2

=> 5R + (6-R) = 18

=> 4R + 6 = 18

=> 4R = 12

=> R = 3. QED

Let R be the number of red faces on the second die. The probability of rolling two reds, then, is 5R/36. The probability of rolling two blues is 1(6-R)/36. We know the sum of the two probabilites must be 1/2, so

5R/36 + (6-R)/36 = 1/2

=> 5R + (6-R) = 18

=> 4R + 6 = 18

=> 4R = 12

=> R = 3. QED

Jim Yingst

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Arjun Shastry

Ranch Hand

Posts: 1906

1

posted 14 years ago

Perfect

Originally posted by Roy Tock:

Interesting...the answer is 3/3. I wouldn't have expected that!

Let R be the number of red faces on the second die. The probability of rolling two reds, then, is 5R/36. The probability of rolling two blues is 1(6-R)/36. We know the sum of the two probabilites must be 1/2, so

5R/36 + (6-R)/36 = 1/2

=> 5R + (6-R) = 18

=> 4R + 6 = 18

=> 4R = 12

=> R = 3. QED

Perfect

MH

Roy Tock

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Posts: 83

posted 14 years ago

OK, let's generalize.

Say you have two D-sided dice. On the first, S sides are red and (D-S) sides are blue. How many red sides must there be on die 2 for the win probability to be the same for both players?

Again, let R be the number of red faces on die 2. The probability of rolling two reds is S/D * R/D = RS/DD. The probability of rolling two blues is (D-S)/D * (D-R)/D = (DD-SD-RD+RS)/DD. So since the probability of a win for either is 1/2,

RS/DD + (DD-SD-RD+RS)/DD = 1/2

=> RS + DD - SD - RD + RS = DD/2

=> 2RS - RD = DD/2 - DD + SD

=> R(2S-D) = SD - DD/2

=> R(2S-D) = D(S-D/2)

=> R(2S-D) = (D/2)(2S-D)

=> R = D/2. (assuming 2S!=D)

If 2S=D, again we find R = D/2 (proof left up to the reader).

So... Regardless of how many sides are red and blue on die 1, the probability of a win for either is 1/2 if D is even and, on die 2, half the sides are red and half blue.

It makes sense; think of it this way. As long as D is even, it doesn't matter what's rolled on die 1 if die 2 is colored half and half. If die 1 is red, odds are 50/50 that die 2 will be red. If die 1 is blue, odds are 50/50 that die 2 will be blue.

So the result is independent of the number of red sides on die 1. It could be 0, D, or anywhere in between. As long as D is even, the result holds.

Further, if D is odd, there is no correct painting of die 2.

[ July 22, 2003: Message edited by: Roy Tock ]

Say you have two D-sided dice. On the first, S sides are red and (D-S) sides are blue. How many red sides must there be on die 2 for the win probability to be the same for both players?

Again, let R be the number of red faces on die 2. The probability of rolling two reds is S/D * R/D = RS/DD. The probability of rolling two blues is (D-S)/D * (D-R)/D = (DD-SD-RD+RS)/DD. So since the probability of a win for either is 1/2,

RS/DD + (DD-SD-RD+RS)/DD = 1/2

=> RS + DD - SD - RD + RS = DD/2

=> 2RS - RD = DD/2 - DD + SD

=> R(2S-D) = SD - DD/2

=> R(2S-D) = D(S-D/2)

=> R(2S-D) = (D/2)(2S-D)

=> R = D/2. (assuming 2S!=D)

If 2S=D, again we find R = D/2 (proof left up to the reader).

So... Regardless of how many sides are red and blue on die 1, the probability of a win for either is 1/2 if D is even and, on die 2, half the sides are red and half blue.

It makes sense; think of it this way. As long as D is even, it doesn't matter what's rolled on die 1 if die 2 is colored half and half. If die 1 is red, odds are 50/50 that die 2 will be red. If die 1 is blue, odds are 50/50 that die 2 will be blue.

So the result is independent of the number of red sides on die 1. It could be 0, D, or anywhere in between. As long as D is even, the result holds.

Further, if D is odd, there is no correct painting of die 2.

[ July 22, 2003: Message edited by: Roy Tock ]

Jim Yingst

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posted 14 years ago

Aww, I was hoping to get to to do more detailed math on that one. Then follow up with something like a 100-sided die with 29 red and 71 blue.

Indeed, the details of the first die are irrelevant - all we really need to know is that there are only two possible outcomes for the first die, red or blue. Then we can always get a 50% chance of a match using a second die which has a 50% chance of either color. It's like if you have an exam with true/false questions - you've got a 50% chance of answering correctly if you just flip a coin.

Indeed, the details of the first die are irrelevant - all we really need to know is that there are only two possible outcomes for the first die, red or blue. Then we can always get a 50% chance of a match using a second die which has a 50% chance of either color. It's like if you have an exam with true/false questions - you've got a 50% chance of answering correctly if you just flip a coin.

"I'm not back." - Bill Harding, *Twister*

Arjun Shastry

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Posts: 1906

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posted 14 years ago

Will the result be same even if second dice is biased?I think combination of colors on second dice should be irrelevant.

MH

Joe Pluta

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Posts: 1376

posted 14 years ago

It's too late for me to work this out in my head, so I'll post it as a thought exercise: how does the math work if the FIRST die is 50/50? Since the answer should be that ANY combination on the second die works, I'd be interested to see what happens with the math.

Just wondering, before I wander off into snoozeland.

Joe

Just wondering, before I wander off into snoozeland.

Joe

Bhau Mhatre

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Posts: 199

posted 14 years ago

Look at Roy's equation:

RS/DD + (DD-SD-RD+RS)/DD = 1/2

=> RS + DD - SD - RD + RS = DD/2

=> 2RS - RD = DD/2 - DD + SD

=> R(2S-D) = SD - DD/2

=> R(2S-D) = D(S-D/2)

=> R(2S-D) = (D/2)(2S-D)

=> R = D/2. (assuming 2S!=D)

If 2S=D, again we find R = D/2 (proof left up to the reader).

If 2S=D, then R need not be D/2 as he said. It can be anything and its value will be irrelavant.

Originally posted by Joe Pluta:

how does the math work if the FIRST die is 50/50? Since the answer should be that ANY combination on the second die works, I'd be interested to see what happens with the math.

Look at Roy's equation:

RS/DD + (DD-SD-RD+RS)/DD = 1/2

=> RS + DD - SD - RD + RS = DD/2

=> 2RS - RD = DD/2 - DD + SD

=> R(2S-D) = SD - DD/2

=> R(2S-D) = D(S-D/2)

=> R(2S-D) = (D/2)(2S-D)

=> R = D/2. (assuming 2S!=D)

If 2S=D, again we find R = D/2 (proof left up to the reader).

If 2S=D, then R need not be D/2 as he said. It can be anything and its value will be irrelavant.

-Mumbai cha Bhau

Joe Pluta

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Posts: 1376

posted 14 years ago

Here's why (paraphrasing the original calcs):

Say you have two D-sided dice. On the first, D/2 sides are red and D/2 sides are blue. How many red sides must there be on die 2 for the win probability to be the same for both players?

Let R be the number of red faces on die 2. The probability of rolling two reds is (D/2)/D * R/D = R/2D. The probability of rolling two blues is (D/2)/D * (D-R)/D = (D-R)/2D. So since the probability of a win for either is 1/2,

R/2D + (D-R)/2D = 1/2

=> R + D - R = D

=> D = D

R factors out, so all values of R are correct!

And what this says is that if either die is 50/50, then no matter the configuration of the second die (provided it has only faces of either red or blue), the chances are 50/50 of getting two of the same color.

This is certainly not intuitive at first glance, but is very cool!

Joe

Originally posted by Mumbai cha bhau:

If 2S=D, then R need not be D/2 as he said. It can be anything and its value will be irrelavant.

Here's why (paraphrasing the original calcs):

Say you have two D-sided dice. On the first, D/2 sides are red and D/2 sides are blue. How many red sides must there be on die 2 for the win probability to be the same for both players?

Let R be the number of red faces on die 2. The probability of rolling two reds is (D/2)/D * R/D = R/2D. The probability of rolling two blues is (D/2)/D * (D-R)/D = (D-R)/2D. So since the probability of a win for either is 1/2,

R/2D + (D-R)/2D = 1/2

=> R + D - R = D

=> D = D

R factors out, so all values of R are correct!

And what this says is that if either die is 50/50, then no matter the configuration of the second die (provided it has only faces of either red or blue), the chances are 50/50 of getting two of the same color.

This is certainly not intuitive at first glance, but is very cool!

Joe

Roy Tock

Ranch Hand

Posts: 83

posted 14 years ago

Yup, it's clearly correct that if the first die is 50/50, then the second die can be painted in whatever manner we wish.

Kinda reminds me of a few types of proof I learned in grad school:

Proof by intimidation: Because I SAID SO!

Proof by embarrasment: Can't you see it's obvious that...

Kinda reminds me of a few types of proof I learned in grad school:

Proof by intimidation: Because I SAID SO!

Proof by embarrasment: Can't you see it's obvious that...

It is sorta covered in the JavaRanch Style Guide. |