Winning on a Crooked Roll

For both dice to show the same number n, if there is probability p(n) of each die showing n, there is probability $p(n)^2$ of both showing n. The probability of both dice showing the same number is therefore $(1/6)^2+(1/6)^2+(1/9)^2+x^2+(2/9)^2+y^2$ , where $x$ and $y$ are the probabilities of rolling a 4 and and a 6 respectively. After simplifying this and setting it equal to $(2/3)^4=16/81$ , we find that $x^2+y^2=13/162$ . Additionally, as the probabilities of landing on all 6 numbers adds to 1, $x+y=1/3$ .

At this point, we could substitute and solve the resulting quadratic, but there is a much better way of finding our answer. The probaility of rolling a 10 on the two dice equals the probability of rolling two 5's, which we have calculated to be $4/81$ , plus $2*xy$ (Either the first die shows a 4 and the second a 6, or vice versa, hence the 2.) We have $2*xy=(x+y)^2-(x^2+y^2)=1/9-13/162=5/162$ , so $a/b=4/81+5/162=13/162$ , and our answer is $13+162=175$ .

[Latex edits - Calvin]

Let the probability of the number 4 showing up be $x$ , and let number 6's probability be $y$ . Then the probability of a total of 10 is $x*y+\frac{2}{9}*\frac{2}{9}+y*x=2xy+\frac{4}{81}$ . The probability that the two dice show the same number is the sum of the squares of the probabilities (for example, in order for both to have 4, we would need to have 4 on the first with probability $x$ , and 4 on the second with probability $x$ , for a combined probability of $x*x=x^2$ ). So we know that $(\frac{1}{6})^2+(\frac{1}{6})^2+(\frac{1}{9})^2+x^2+(\frac{2}{9})^2+y^2=(\frac{2}{3})^4$ . We also know that the total probability must be 1, so $\frac{1}{6}+\frac{1}{6}+\frac{1}{9}+x+\frac{2}{9}+y=\frac{2}{6}+\frac{3}{9}+x+y=\frac{1}{3}+\frac{1}{3}+x+y=$ $\frac{2}{3}+x+y=1$ , and $x+y=\frac{1}{3}$ . Solving our other equation, we get $\frac{1}{36}+\frac{1}{36}+\frac{1}{81}+x^2+\frac{4}{81}+y^2=\frac{16}{81}$ , $\frac{2}{36}+\frac{5}{81}+x^2+y^2=\frac{16}{81}$ , $\frac{1}{18}+x^2+y^2=\frac{11}{81}$ , $x^2+y^2=\frac{22}{162}-\frac{9}{162}=\frac{13}{162}$ . So we now have: $x+y=\frac{1}{3}$ , and $x^2+y^2=\frac{13}{162}$ , and we seek $2xy+\frac{4}{81}$ . Squaring the first equation, we get $x^2+2xy+y^2=\frac{13}{162}+2xy=\frac{1}{9}$ , and $2xy=\frac{1}{9}-\frac{13}{162}=\frac{18}{162}-\frac{13}{162}=\frac{5}{162}$ . Thus our probability is $\frac{5}{162}+\frac{4}{81}=\frac{5}{162}+\frac{8}{162}=\frac{13}{162}$ , and the answer is $13+162=175$

Let the probability of rolling 4 and 6 on one dice be x and y respectively. We $x+y=1-\frac{1}{6}-\frac{1}{6}-\frac{1}{9}-\frac{2}{9}=\frac{1}{3}$ . The probability of rolling two dice that show the same number is clearly $\frac{1}{36}+\frac{1}{36}+\frac{1}{81}+\frac{4}{81}+x^2+y^2=\frac{19}{162}+x^2+y^2$ . We have $\frac{16}{81}-\frac{19}{162}=x^2+y^2 \Rightarrow x^2+y^2=\frac{13}{162}$ . Subtracting this from $(x+y)^2=x^2+y^2+2xy=\frac{1}{9}$ gives $2xy=\frac{5}{162}$ . The probability of rolling 10 on two dice is either rolling 4 and 6, 5 and 5, or 6 and 4. The probability is then $\frac{4}{81}+2xy=\frac{4}{81}+\frac{5}{162}=\frac{13}{162}$ . Therefore $a+b=175.$

Let $a$ be the probability of rolling a 4 on either die. Note that the probability of rolling a 6 on either die is thus $\frac{1}{3}-a$ . Then the probability that the two dice both show the same numbers is

$\frac{1}{36}+\frac{1}{36}+\frac{1}{81}+a^2+\frac{4}{81}+(\frac{1}{3}-a)^2=\frac{16}{81}$

Simplifying this yields

$2a^2-\frac{2}{3}a+\frac{1}{9}=\frac{13}{162}$

Now note that the probability of rolling a 10 is

$2a(\frac{1}{3}-a)+\frac{4}{81}=-2a^2+\frac{2}{3}a+\frac{4}{81}=(\frac{1}{9}-\frac{13}{162})+\frac{4}{81}=\frac{13}{162}$

Thus $a+b=\boxed{175}$ .

Let the probability of rolling 4 be A. Let the probability of rolling 6 be B.

We get that $\frac {1}{6} + \frac {1}{6} + \frac {1}{9} + A + \frac {2}{9} + B = 1$ from the table. So $A + B = \frac {1}{3}$ . Call this equation 1.

By the probability of two dice both show the same, we get: $(\frac {1}{6})^2 +(\frac {1}{6})^2 + (\frac {1}{9})^2 + A^2 + (\frac {2}{9})^2 + B^2 = (\frac {2}{3})^4$ After simplifying, we get $A^2 + B^2 = (\frac {13}{162})^2$ . Call this equation 2.

If we subtract (equation 1) $^2$ from equation 2, we get $2AB=\frac {5}{162}$ .

The probability of getting a 10 is precisely rolling 4,6 or 5,5, or 6,4: which is $2AB + (\frac {2}{9})^2 = \frac {13}{162}$ . Thus, the solution is $13 + 162 = 175$

Let x = P(4). Then P(6) = 1/3 - x. We have that 1/36+1/36+1/81+x^2 + 4/81+(1/3-x)^2 = 16/81. We want to find 2x(1/3-x) + 4/81. Solving, we get that 2x^2 - 2/3 x + 5/162 = 0, so 2x(1/3-x) + 4/81 = 5/162 + 4/81 = 13/162, so 175

As we all know, the sum of probability of all events in a random experiment is 1. therefore , p(1)+p(2)+p(3)+p(4)+p(5)+p(6) = 1 [NOTE: p(1) donates probability of occurrence of 1 on dice] => p(4) + p(6) = 1/3

let, p(4) = x , then p(6) = 1/3 - x.

Now according to the question , probability of two dice showing the same numbers is (2/3)^4. so, 1/36 + 1/36 + 1/81 + x^2 + 4/81 + (1/3 - x) ^ 2 = 16/81

on solving we get x = 1/18 or x = 5/18

case I : x = 1/18 therefore p(4) = 1/18 and p(6) = 5/18 [as p(6) = 1/3 - p(4)]

case II : x=5/18 therefore p(4) = 5/18 and p(6) = 1/18 [ as p(6) = 1/3 - p(4)]

now, we can get 10 in three ways - FIRST WAY : 4 on first dice and 6 on second SECOND WAY: 5 on both dice THIRD WAY: 6 on first dice and 4 on second.

so considering all cases, probability of getting 10 on these two dice is :- 5/324 + 4/81 + 5/324 = 10/324 + 4/81 =26/324 =13/162 = a/b [ according to the question]

so a+ b = 162 + 13 = 175

Let the probability of rolling a 4 be $x$ . Thus the probability of rolling a 6 is $\frac{1}{3}-x$ .

The probability of rolling doubles is $\frac{1}{36}+\frac{1}{36}+\frac{1}{81}+x^2+\frac{4}{81}+(\frac{1}{3}-x)^2 = \frac{16}{81}$ This reduces to $2x^2-\frac{2}{3}x=-\frac{5}{162}$

The probability of rolling 10 is $2(x)(\frac{1}{3}-x)+\frac{4}{9}=-(2x^2-\frac{2}{3}x)+\frac{4}{9}=\frac{5}{162}+\frac{4}{81}=\frac{13}{162}$ Thus our answer is $13+162=\boxed{175}$ .

Let the probability of 4 falling on a dice be p, and the probability of 6 falling on a dice be q.

Now the possible outcomes of one dice is are {1, 2, 3, 4, 5, 6}. So the sample space of rolling one dice is {1, 2, 3, 4, 5, 6}. Thus, probability of any one event of {1, 2, 3, 4, 5, 6}= 1. Since these are mutually exclusive events, i.e. one of these occurs implies any other event cannot occur the following can be derived.

1= Probability(dice showing 1) + Probability(dice showing 2) + Probability(dice showing 3) + Probability(dice showing 4) + Probability(dice showing 5) + Probability(dice showing 6).

Plugging the values gives 1= 1/6 + 1/6 + 1/9 + p + 2/9 + q. This implies p+q= 1/3......(1) . Now, it is given that the probability of both dice showing same result is (2/3)^4. If one dice shows 1 (probability 1/6), the other dice also must show 1 (probability 1/6). So probability of both dice showing 1 is (1/6)^2. Similarly probability of both dice showing 2 is (1/6)^2. Probability of both dice showing 3 is (1/9)^2. Probability of both dice showing 4 is p^2. Probability of both dice showing 5 is (2/9)^2. Probability of both dice showing 6 is q^2. Here the sample space is {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}. Probability of any one of these 6 events occurring is 1. Again these events being mutually exclusive the following can be concluded.

(1/6)^2 + (1/6)^2 + (1/9)^2 + p^2 +(2/9)^2 + q^2= (2/3)^4, which gives p^2 + q^2= 13/162........(2).

Now, pq= {(p+q)^2 - (p^2 + q^2)}/2. Plugging values from equation (1) and (2), we get pq= 5/324......(3)

Now, the sum of the results in the dice can be 10 in 3 ways:- (5, 5), (4, 6), (6, 4). The probability of (5, 5) occurring is (2/9)^2= 4/81. Probability of (4, 6) occurring is pq. Probability of (6, 4) occurring is pq. So, total probability= pq+pq+4/81= 2pq +4/81 = 2*5/324 + 4/81 (from equation (3))= 13/162. Since g.c.d(13, 162)= 1, it can b seen that a=13 and b= 162. Thus, a+b= 13+162= 175.

$P\left(Both \: 1\right) = \frac{1}{36}$ $P\left(Both \: 2\right) = \frac{1}{36}$ $P \left(Both \: 3 \right) = \frac{1}{81}$ $P\left(Both \: 4\right) = x^2$ $P\left(Both \: 5 \right) = \frac{4}{81}$ $P\left(Both \: 6 \right) = y^2$

$\frac{1}{36} + \frac{1}{36} + \frac {1}{81} + \frac{4}{81} + x^2 + y^2 = \frac{16}{81}$

$x^2 + y^2 = \frac{26}{324}$

$\frac{1}{6} + \frac{1}{6} + \frac{1}{9} + \frac{2}{9} + x + y = 1$

$x+y = \frac{12}{36}$

$P\left(Total \: of \: 10\right) = P\left(Both \: 5\right) + P\left(4 \: and \:6 \right) = \frac{4}{81} + 2xy$

$2xy = \left(x+y\right)^2 - \left(x^2 + y^2 \right)$

$2xy = \frac{144}{1296} - \frac{26}{324}$ $= \frac{40}{1296}$

$P\left(Total \: of \: 10 \right) = \frac{40}{1296} + \frac{4}{81}$ $= \frac{104}{1296}$ $= \frac{13}{162}$

Therefore $a + b = 13+ 162 = 175$

Let $P(x)$ denote the probability of rolling $x$ . Let x be P(4) and y be P(6). We are given that $x+y=1-\left(\frac16+\frac16+\frac19+\frac29\right)=\frac13$ . We are also given that $x^2+y^2=\frac{13}{162}$ . Solving, we get $(x,y)=\left(\frac{1}{18},\frac{5}{18}\right)$ or $(x,y)=\left(\frac{5}{18}, \frac{1}{18}\right)$ . Either way, now we just need to list the ways to get 10. That is, we can roll two fives or a six and a four. We have the equation $\left(\frac{2}{9}\right)^2+2\left(\frac{5}{18}\right)\left(\frac{1}{18}\right)=\frac{13}{162}$ To AWIA (Answer What is Asked), we have $a=13$ and $b=162$ . And $a+b=\boxed{175}$

Let a and b be the probability of getting 4 and 6 respectively on either dice.

We have: 1/6 + 1/6 + 1/9 + a + 2/9 + b = 1 (since the total probability of getting 1 through 6 on one dice must equal 1)

=> a+b = 1/3 (1)

We also have (1/6)^2 + (1/6)^2 + (1/9)^2 + a^2 + (2/9)^2 + b^2 = (2/3)^4 (since the probability of getting the same numbers in two dice is (2/3)^4)

=> a^2 + b^2 = 13/162 (2)

From (1) and (2) we have (a,b) = (1/18,5/18) or (5/18,1/18)

In order for the sum of numbers on 2 dices equals 10, we must roll out either 5 on both dices or 4 on one dice and 6 on the other dice.

Let S be the probability that we roll out 2 numbers with the total sum of 10.

Then, we have: S = P(5,5) + P(4,6) + P(6,4) = (2/9)^2 + 2 (1/18) (5/18) = 13/162

Therefore, a+b, according to the problems, will equal 13 + 162 = 175

Let probability for 4 is p and for 6 is q. 1/6+1/6+1/9+p+2/9+q = 1 So, p+q = 1/3, or q=1/3-p From given condition, (1/6)^2+(1/6)^2+(1/9)^2+p^2+(2/9)^2+(1/3… 16/81 Simplifying, 2 p (1/3-p)=5/162 Rolling 10 can occur through 5+5, 4+6 or 6+4 Probability = (2/9)^2 + p (1/3-p)+(1/3-p) p = 4/81+5/162= 13/162 Hence a+b = 175

Or (5/18,1/18)

let probabilty of getting 4 & 6 be x & y respectively. den 1/6+1/6+1/9+x+2/9+y=1 or x+y=1/3 and (1/6)^2+ (1/6)^2 +(1/9)^2+x^2 +(2/9)^2 +y^2=(2/3)^4 or x^2 + y^2 =13/162 subtracting 2nd eqn fron 1st eqn's square we get xy=5/324 now probability of getting 10 = xy +(2/9)^2 +xy = 5/324 + 4/81 + 5/324 = 13/162 hence 13+162 =175 is required ans.

$P(4) = a, P(6) = b$

$a^2 + b^2 = (\frac{2}{3})^4 - ((\frac{1}{6})^2 + (\frac{1}{6})^2 + (\frac{1}{9})^2 + (\frac{2}{9})^2) = \frac{13}{162}$

$a + b = 1 - (\frac{1}{6} + \frac{1}{6})^2 + \frac{1}{9} + \frac{2}{9}) = \frac{1}{3}$

$2ab = (a + b)^2 - (a^2 + b^2) = (\frac{1}{3})^2 - \frac{13}{162} = \frac{5}{162}$

$P(Rolling a Ten) = P(4) \times P(6) + P(5) \times P(5) + P(6) \times P(4)$ $= 2 \times P(4) \times P(6) + P(5) \times P(5) = 2ab + (\frac{2}{9})^2 = \frac{13}{162}$

$13 + 162 = \fbox{175}$

Let the probability of rolling a $4$ be $x$ , so that the probability of rolling a $6$ is $\frac{1}{3}-x$ The probability of rolling $2$ dice that show the same number, $P(1 \text{ then } 1)+P(2 \text{ then } 2) + P(3 \text{ then } 3) + P(4 \text{ then } 4) + P(5 \text{ then } 5) +P(6 \text{ then } 6)$ , is thus $(\frac{1}{6})^2+(\frac{1}{6})^2+(\frac{1}{9})^2+x^2+(\frac{2}{9})^2+(\frac{1}{3}-x)^2$ Setting this equal to $(\frac{2}{3})^4$ and simplifying, we find $0=2x^2-\frac{2}{3}x+\frac{5}{162}$ By the quadratic formula, the two roots of this equation are $\{\frac{5}{18},\frac{1}{18}\}$ Note that either of these could be $x$ , so that that the other is $\frac{1}{3}-x$ . This comes from the fact that our definition of $x$ could be rolling a $6$ instead of a $4$ To roll a $10$ , we need either a $4$ and a $6$ or two $5$ s. The probability of this is $P(4 \text{ then } 6)+P(5 \text{ then } 5) + P(6 \text{ then } 4)=\frac{5}{18^2}+\frac{4}{36}+\frac{5}{18^2}=\frac{23}{162}$ This is equal to $\frac{a}{b}$ , so $a=23, b=162 \rightarrow a+b=\boxed{175}$

There are 2 ways to roll 10: by rolling 2 5's (p=(2/9)^2) or by rolling a 6 and a 4.

We can calculate the odds for this event by calculating for the event of Die 1 being a 4 OR a 6 and Die 2 being a 4 and a 6. The chance of a 4 or a 6 is 1-(1/6)-(1/6)-(1/9)-(2/9) is 1/3. So the chance for a 4-6 combination is 1/9.

Problem is, we have to subtract the chance for a 4-4 OR a 6-6 from this total.

This chance is equal to (2/3)^4 - 1/36 - 1/36 - 1/81 - 4/81 = 13/162

This leaves 5/162 for a 10 through a 4-6 combination. Plus 4/81 (the odds for a 5-5 roll), this leaves 13/162, so a=13, b=162, a+b=175

Let $P(4) = x$ and $P(6) = y$ , $x + y = \frac{1}{3}$ , P(getting same number on both die) = $(P(1))^2 +(P(2))^2 +(P(3))^2 +(P(4))^2 + (P(5))^2 +(P(6))^2 = \frac{2}{3})^4$ $\Rightarrow (\frac{1}{6})^2 +(\frac{1}{6})^2 +(\frac{1}{9})^2 +x^2 +(\frac{2}{9})^2 +y^2 = (\frac{2}{3})^4$ Solving the two equations, we get $x = \frac{5}{18} , y =\frac{1}{18} or x =\frac{1}{18} , y = \frac{5}{18}$ . Both will yield same answer. P(sum of numbers on dice = 10) = P(4 , 6) +P(5 , 5) + P(6 , 4) = $(\frac{5}{18})(\frac{1}{18}) + (\frac{2}{9})^2 + (\frac{1}{18})(\frac{5}{18})$ = $\frac{13}{162}$ . $\Rightarrow a + b = 13 + 162 = 175$