Transcendental Dice Rolls

We will find the probability that the product is imaginary, then subtract this value from 1. We could roll i, i, i, or i, x, y. i^0 and i^2 are real, so the mentioned cases are the only viable ones. In addition, x or y cannot be 0, or else the product would be 0. Let us order the dice as A, B, and C. The first case clearly leads to 1 permutation. In the second case, choose one die to have i. Then each other die has 4 possibilities. This case leads to (3)(4)(4)=48 permutations. The number of imaginary permutations is 1+48=49. The total number of permutations is 6^3=216, so the number of real permutations is 216-49=167. The final probability is 167/216, so a+b=167+216=383.

We count the complement. In order for the product to be imaginary, there are only $2$ ways for this to happen: 1) Exactly one dice contains $i$ , and the other $2$ sides contain non-zero real numbers. 2) All $3$ dice contain $i$ .

For the first case, the number of ways of this happening is $3 \times 4 \times 4$ - After you choose a face which to have $i$ (there are 3 faces), there are $2$ other faces, each of which can have only $4$ other possible numbers. For the second case, there is only $1$ way. Total, we have got $49$ cases out of a total of $6*6*6$ different configurations, so the probability that the product is imaginary is given by: $\frac{49}{216}$ , and the probability that the product is real is given by $1 - \frac{49}{216} = \frac{167}{216}$ , so $a + b = 167 + 216 = \boxed{383}$

Let $A,B,C$ be the top face of that three dice. The numbers all possibility of $A,B,C$ is $6^3$ . We will find the numbers possibility of $A,B,C$ such that $A\times B\times C$ is real.

Case 1 : At least one of $A,B,C$ is zero. Then we have $A\times B\times C$ is real ( $A\times B\times C=0$ ). By IE principle, the numbers possibility of $A,B,C$ such that at least one of $A,B,C$ is zero is $6^3-5^3$ .

Case 2 : No one of $A,B,C$ is zero. Then we have, $A\times B\times C$ is real if and only if all of them are real or two of them are equal to $i$ . The numbers possibility of $A,B,C$ such that $A,B,C$ are non-zero real numbers is $4^3$ . The numbers possibilty of $A,B,C$ such that two of them are equal to $i$ and the other one is non-zero real number is $3\times 4$ .

Thus, we have the numbers possibility of $A,B,C$ such that $A\times B\times C$ is real $6^3-5^3+4^3+12=167$ . Therefore the probability that $A\times B\times C$ is real $\frac{167}{216}$ . Since $167,216$ are coprime positive integers, then we have $a=167,b=216$ , and so $a+b=383$ .

Probablity of product to be real=1-probablity of not real

For product real we require one(at least) i to be there

When there are i in all roll product not real -1case

When there are only two i roll (i * i=real )- 0 case

When there is 1 i in any roll And other two are same (could be pi, 1, root (2), e but not zero) -=combinaton of 3 place and two no. Same than it is 3! / 2! × C (4, 1) =12 ways

When there is 1 i and other two different (pi, e, root (2), 1) =3! × C (4,2) =36 ways

Total 49 ways of not real product, Probablity of not real =no. Of ways /total ways

Total ways=6×6×6=216

Probality of not real=49/216

Probality of real=167/216, a+b=383

Consider when the product of the numbers of the 3 top faces is not real. Clearly, all must be non-zero numbers or else the product will be 0 which is real. It also cannot be the case that all 3 numbers are real as clearly the product is real.

Case 1: All 3 numbers are $i$ . The product is $i^3=-i$ which is not real and this happens with probability $(\frac{1}{6})^3=\frac{1}{216}$

Case 2: There is exactly 1 non-zero real number x. The product is $x \times i \times i=-x$ which is real

Case 3: There are exactly 2 non-zero real numbers x,y. The product is $xyi$ which is not real. As there are 4 non-zero real numbers, this happens with probability ${3 \choose 1}(\frac{1}{6})(\frac{4}{6})^2=\frac{48}{216}$

Therefore, the probability that the product is not real is $\frac{1}{216}+\frac{48}{216}=\frac{49}{216}$ and thus the probability that it is real is $1-\frac{49}{216}=\frac{167}{216}$

Lets find all those cases when product is not real i) All three are 'i' - 1 case

ii) 'i' appears just once and other two are different (0 can not be there else product will be 0) On remaining two dices we can have two different numbers in C(4, 2) ways (among 1, e, pi, √2) we can permute them in 3! ways So, C(4, 2)*3! ways

ii) 'i' appears just once and other two dices show same number other than 0 we can choose the same number in C(4, 1) ways Now, we can permute them in 3!/2! = 3 ways So, C(4, 1)*3 ways

Add all the cases and divide by 216 to get the probability.

There are a total of 6 6 6 =216 possibilities out of which 1 would be surely imaginary where all the 3 dices show i. Let the first dice show i, then the result will be imaginary only if the other two are neither 0 nor i. So ,the number of possibilities are 1 4 4=16. The I can roll up in any of the three dices.So,the number of possibilities for imaginary value is 3*16=48. So the total number of possibilities for imaginary value=48+1=49. Probability of imaginary value=49/216. Probability of real value=1-(49/216)=167/216. a=167 & b=216.a+b=167+216=383.

Since the set of real numbers is closed under multiplication, the product of the three numbers shown in the dice is real if the numbers in the dice are all real. All the faces in the dice are real except for i. Hence we count the complement: the cases when i is included that will result to an imaginary product. Note that when exactly 2 i's are included, the product becomes real. Hence we only consider the case when there's is only 1 i and 3 i's. In the first case, there are 3 ways to select the die on which the only i is shown. For the remaining two dice, no 0 must be shown; hence, each can take 4 values. By multiplication principle, there are 3 4 4 = 48 possible rolls. The last case of having 3 i's has only one possible roll. Hence the probability that the product is real is [216 - (48 + 1)]/216 = 167/216, hence the answer is 167 + 216 = 383.

At least one dice roll is 0 with probability $\frac{\text{(number of cases)} - \text{(number of cases without 0)}}{\text{number of cases}} = \frac{216-125}{216} = \frac{91}{216}$

If no dice rolls are 0 then for the product to be real, there must be no $i$ rolled or only a pair of $i$ 's.

Probability the product is non-zero and all numbers rolled are real $=\frac{4}{6} \times \frac{4}{6} \times \frac{4}{6} = \frac{64}{216}$

Probability that a pair of $i$ 's were rolled

$=\frac{\text{number of ways to roll a pair of } i \text{'s}}{216}$ $= \frac{\text{number of ways to choose a die and set it as one of the 4 real numbers}}{216}$ $= \frac{3 \times 4}{216} = \frac{12}{216}$

Probability of product being real $= \frac{167}{216}$

Thus the answer is $167+216=383$

It's easier to count the non-real numbers so I used complementary counting.

To get a real number there must be an odd number of dice with $i$ . One possibility is getting $i$ on all three dice. Another possibility is getting $i$ on one die and $1$ , $e$ , $\pi$ , or $\sqrt{2}$ (multiplying by $0$ would get $0$ which is real) on the other two. There are $16$ combinations of this but you must multiply this by $3$ to get $48$ since the $i$ can be the first, second, or third die. So there are $1+48=49$ ways to obtain an imaginary number making there be $216-49=167$ ways to obtain a real number. The probability is $\frac{167}{216}$ so $167+216=383$

There are 6x6x6 = 216 points in the sample space S Let E be the event: "product of the three outcomes is real" Its complementary event E' will be "product of the three outcomes is imaginary" E' consists of the points Type 1: (i, a nonzero real, a nonzero real), Type 2: (a nonzero real, i, a non-zero real), Type 3: (a nonzero real, a non-zero real, i)) In the given list there are 4 non-zero reals. For each type we have 4x4 = 16 ways of product being imaginary Also we have to consider the possibility (i, i, i) since the product is - i So E' consists of 3x16 + 1 = 49 points of S and thus P(E') = 49/216. SoP(E) = 1 - P(E') = 1 - (49/216) = 167/216 = a/b since 167 is coprime to 216 Hence a + b = 383

Let say we have 4 real non zero numbers called "R", a zero and an imaginary number called "i". The other easier way to find the probability is to find the complement of probability that product of three numbers is an imaginary number. The possible product to create imaginary number is: i i i ---> there are 1 1 1 1 = 1 way i R R ---> there are 3 1 4 4 = 48 ways Total: 49 ways ----> probability complement= 49/216 Thus, the probability is = 1 - probability complement = 1 - 49/216 = 167/216 167 and 216 are already coprime, so the value is 383

Let us count the number of ways of getting a product that is non-real.

Case 1: $i$ appears exactly once. Choose a position for $i$ and choose the remaining two positions among $1, e, \pi,$ and $\sqrt{2}.$ This can happen in $\binom{3}{1} \times 4 \times 4 = 48$ ways.

Case 2: $i$ appears exactly thrice. This can obviously happen in only $1$ way.

Hence, the probability of getting a real number as product is $\dfrac{6^3 - 49}{6^3} = \dfrac{167}{216}.$

Let $A$ be the event that the product is real and $A^c$ be the event that the product is imaginary.

$P(A)=1-P(A^c)$

Since the product of all real numbers is real, the product will only be imaginary if $i$ is one of the factors and 0 is not one of the factors (since 0 times any number, be it real or imaginary, is 0, which is real). Also, there cannot be exactly two $i$ 's because $i^2=-1$ , which is real.

Thus,

$P(A^c)=P(\text{1 } i\text{ appears} \cap \text{no 0 appears})+P(\text{3 } i\text{'s appear} \cap \text{no 0 appears})$ .

$P(A^c)=\displaystyle\binom{3}{1}\left(\frac{1}{6}\right)\left(\frac{4}{6}\right)^2+\displaystyle\binom{3}{3}\left(\frac{1}{6}\right)^3$

$P(A^c)=\frac{49}{216}$

$P(A)=\frac{167}{216}$

$a+b=167+216=383$ .

Let the number on the first dice be A, the second be B and the third be C.

Instead of finding the real value of $A*B*C$ , we will try to find the imaginary value as finding the real value is more tedious.

The only imaginary number on the dice is $i$ . If all three dices show $i$ , then the result is $- i$ which is an imaginary number. The probability is 1/216.

If A is $i$ , B can be take any value on the dice except for 0 and $i$ , C can also take any value on the dice except for 0 and $i$ . The probability is 1 4 4/216.

If B is $i$ , A can be take any value on the dice except for 0 and $i$ , C can also take any value on the dice except for 0 and $i$ . The probability is 1 4 4/216.

If C is $i$ , B can be take any value on the dice except for 0 and $i$ , A can also take any value on the dice except for 0 and $i$ . The probability is 1 4 4/216.

Adding them, we get the probability of getting an imaginary number is 49/216.

Thus the probability of getting a real number is 167/216. Adding them, we get 383.

We must examine several cases in which the product will be real.

Case 1: All 3 dice are real, non-zero numbers (I excluded the 0's initially so as not to double-count them), creating a real product. This probability is $(\frac{4}{6})(\frac{4}{6})(\frac{4}{6})=\frac{64}{216}$

Case 2: 1 of the dice is 0, creating a product of 0, a real number: $(\frac{1}{6})(\frac{5}{6})(\frac{5}{6}) = \frac{25}{216}$ . We then multiply by 3 since the first could be 0 or the 2nd or the 3rd. Probability = $\frac{75}{216}$

Case 3: 2 of the dice are 0: $(\frac{1}{6})(\frac{1}{6})(\frac{5}{6})(3) = \frac{15}{216}$

Case 4: All of the dice are 0: $(\frac{1}{6})(\frac{1}{6})(\frac{1}{6}) = \frac{1}{216}$

Case 5: 2 of the dice come up $i$ and the 3rd is not 0, so as not to double-count: $(\frac{1}{6})(\frac{1}{6})(\frac{4}{6})(3) = \frac{12}{216}$

Add these probabilities together to get $\frac{167}{216}$ .

$167 + 216 = 383$

One can get a complex product in 2 ways only-

Case 1: All three dices show the side $i$ , which has the probability $\frac 16 \cdot \frac 16 \cdot \frac 16 = \frac {1}{216}$

Case 2: Exactly one of the dice shows $i$ , the other two shows any sides other than $0$ or $i$ , which has the probability $3 \cdot \frac 14 \cdot \frac 14 = \frac 3{16}$ (3 ways to choose the dice from which $i$ comes from, and choosing sides other than $0,i$ from the other two dice in $4 \cdot 4$ ways)

So, the probability of getting a real product is $1- \frac 1{216} - \frac 3{16} = \frac {167}{216}$ and thus the answer is $167+216 = 383$

Out of 6 x 6 x 6 = 216 possible results, the only combinations that will yield non-real solutions are:

iii = 1

(11i, eei, ππi, sqrt{2}sqrt{2}i) x 3 permutations = 12

For combinations like 1ei, 1πi, etc.:
[4C2 (1,e,π,sqrt{2})] x 6 permutations = 36

Total non-real combinations: 49
Total real combinations: 216 - 49 = 167

Probability = a/b = 167/216
a + b = 383

There are 2 main ways in which the product can be real

A) any one of the dice is 0

OR

B) You get no i's or 2 i's, but no 0s

We work out the probability of each of these cases and add them up.

Case A

Probability(at least 1 zero)=1-Probability(no zeros)

So, Probability= $1-(5/6)^3$ = $\frac{91}{216}$

Case B We don't allow any zeros in this case (otherwise we end up in case 1)

Probability that you have no $i$ s is $(4/6)^3=\frac{64}{216}$

Probability that you have 2 $i$ s is $3\times \frac{4}{6} \times (\frac{1}{6})^2 =\frac{12}{216}$

So Adding all the cases up we get the probability is $\frac{167}{216}$

We used complementary counting, so we need to find 1- (the probability that the product is nonreal). For the product to be non-real, either 1 i has to come up with no 0s or all 3 must be i. For the first case there are 16*3 possibilities, and the second has 1 possibilities, so our answer is 1-(49/216)=167/216

since the possibilities of the product being real are cryptic, we calculate the probability for the product to be complex. The total different outcomes for the dice: 6x6x6=216

For the product to be complex: 2 cases are present: 1: all 3 die show i on the top case (1 outcome) 2: the first dice show i, the second dice cannot be i or 0, and the same works for the third dice. (total different outcomes = 1x4x4=16) In this case, only one of the die must show i, therefore the outcome can be 'rotated' giving it 2x16 more possibilities. (48 outcomes)

Total possibilities of the product being complex is 1+48=49, therefore the total possibilities for the product being real is 216-49=167. a/b =167/216, therefore a + b = 383.

We first see that there are $6 \cdot 6 \cdot 6=216$ total possibilities. If all of the numbers on the top face are real, then their product will be real. There are a total of 5 real numbers: $0, 1, e, \pi, \sqrt{2}$ . Therefore, there are a total of $5 \cdot 5 \cdot 5=125$ ways to have 3 real numbers facing up.

We see that $i^2=-1$ . Thus, if there are 2 $i$ 's facing up and another real number facing up, the product will be real. There are 5 choices for which real number faces up. For each of these 5 possibilities, there are $\dbinom{3}{2}=3$ ways to arrange the $i$ 's and the real number, for a total of $3 \cdot 5=15$ . Also, if there is one $i$ facing up, and the other two dice have 0's facing up, the product is 0, which is real. There are $\dbinom{3}{2}=3$ ways in which this can happen. Also, if there is an $i$ facing up, and a 0 facing up, it won't matter what number is facing up on the other die - the product will be 0 no matter what. There are 4 choices for which number is facing up (since we've counted the ones for 0 and i already), and there are $3!=6$ ways to arrange each of these, for a total of $4 \cdot 6 = 24$ . Summing all of these gives $125+15+3+24=167$ , for a final answer of $\dfrac{167}{216}$ , or $\boxed{383}$ .

Suppose that the product is imaginary. First, note that no dice are 0. Now:

a) All are i -> 1 way b) The first die is i, the second and third are not i or 0 -> 16 ways But the second can occur for three dice, so there are 49 ways in total.

So there 167 ways of having a real product, so the probability is $frac{167}{216}$ giving a sum of 383.

Out of all the numbers, all are real except iota( i ) which is √-1 ---> An imaginary number.

Since, P(Get a real number)=1-P(Get an unreal number) ,

Therefore:-

Total Number of all possible outcomes= 6 6 6 = 216 . Lets make a list of all possible cases of getting imaginary product when three dices are rolled.

Case 1:- When i comes as output in all the three dices when rolled i.e. i comes thrice. _ total cases here= 1 _ (only)

Case 2:- When i comes only once in one of the three dices, and '0'
should not come on the other two dices. (because 0 \times i equals 0 which is real). So We'll exclude this from our cases.

            a) When _i_ comes on 1st dice, 
                                                    then cases =  __1*4*4 = 16__.

Because We are excluding 0 and i from our list of unreal products, total unreal elements that can give an unreal product reduces to 6-2=4. And, The same over third dice too. Henceforth, 1 4 4.

Similarly, b) When i comes on 2nd dice, then cases = 1 4 4=16.

Similarly, c) When i comes on 3rd dice, then cases = 1 4 4=16. All the possible unreal products in Case 2 are :- (16+16+16) = 48 .

All Cases for getting an unreal product are 1+48=49.

Cases for getting a real product = total cases - All Cases for getting an unreal product = 216 - 49 = 167.

Probability of getting a real product is 167/216 .

Since, 167 and 216 are relatively prime or coprime. So , a = 167, and b=216.

a+b= 383 , and Thats the answer ==> 383 .

We can split this into three cases: (1) All three numbers are nonzero and real, (2) One or more of the numbers are zero, and (3) Exactly two of the numbers are i, and the remaining number is nonzero. Since there are six possible outcomes for each die, right off the bat we can find that there are $6^3 = 216$ total possible outcomes.

(1) If the three numbers are nonzero and real, then each die will be within four different numbers: $1, e, \pi,$ and $\sqrt{2}$ . Thus, there are $4 \cdot 4 \cdot 4 = 64$ total true outcomes for this case.

(2) If one of the three numbers is zero, then we can find the complementary probability by finding the number of cases in which none of the numbers is zero. This occurs when the die on the three cubes are one of the following: $1, e, \pi, i,$ and $\sqrt{2}$ , so there are $5 \cdot 5 \cdot 5 = 125$ cases. Since this is complementary, we use the total possible outcomes and find the difference, which is $216 - 125 = 91$ .

(3) For exactly two of the die to give $i$ and the other to give neither $i$ nor $0$ , we have the number of cases $1 \cdot 1 \cdot 4 \cdot 3$ . The reason why we multiply by $3$ at the end is because we can reorder the numbers in three different ways (if a is the third number, each outcome can be written as $i, i, a; i, a, i; a, i, i$ . Thus, there are $12$ total outcomes for this case.

Adding up the different cases, we get $\frac{64+91+12}{216}=\frac{167}{216}$ , which is irreducible. Adding up the numerator and denominator gives us $\boxed{383}$ .

We first consider all possible cases when the product of three dice rolls are not real.

Out of the 6 sides of the die, only $i$ is not real. The product of real numbers is real, while the product of $i$ and $i$ is real as well. However, the product of $i$ and a non-zero real number is not real. Therefore, for the product of three dice rolls to be not real, either one or three dice rolls must be of the value $i$ .

For three dice rolls of $i$ , there is only 1 possible case.

For the case of exactly 1 roll of $i$ , the product will not be real if the product of the other two numbers is not zero. This means that each roll can take 4 values (all faces except $0$ or $i$ ). The $i$ roll can happen in any of the three dice rolls, so there is a total of $3 \times 4 \times 4 = 48$ possible cases.

Hence there is a total of 49 cases where the product is not real.

There is a total of $6 \times 6 \times 6 = 216$ possible cases, so the probability that the product is real is $1 - \frac{49}{216} = \frac{167}{216}$ .

If we need a real product, we need at least one i in it. If we assume that i is in all rolls, we shall get 1 case. Next, there will be 2 cases. If we take one i and the other 2 same $pi, 1, \sqrt 2,e.=4C1 \times 3!/2!=12$ ways. Also, now we assume that there is 1 i and the other two are different is $3! \times 4C2=36$ ways. We get a total of 49 ways. Total results one can get= 216 . Probability of the product to be real is $(216-49)/216=167/216$ . Therefore a+b=383.

Solution 1: We count the complement. In order for the product to be imaginary, there are only $2$ ways for this to happen: 1) Exactly one die contains $i$ , and the other $2$ dice contain non-zero real numbers. 2) All $3$ dice contain $i$ .

For the first case, the number of ways of this happening is $3 \times 4 \times 4 = 48$ - After you choose a face which to have $i$ (there are 3 faces), there are $2$ other faces, each of which can have only $4$ other possible numbers. For the second case, there is only $1$ way. Total, we have got $49$ cases out of a total of $6\times 6\times 6 = 216$ different configurations, so the probability that the product is imaginary is given by: $\frac{49}{216}$ , and the probability that the product is real is given by $1 - \frac{49}{216} = \frac{167}{216}$ , so $a + b = 167 + 216 = 383$ .

Solution 2: If at least one die shows 0, then the answer will be real. This occurs with probability $1 - \left(\frac{5}{6}\right)^{3} = \frac{91}{216}$ . If no die shows 0, then the number is real if the number of times $i$ occurs is even and is imaginary if the number of times $i$ occurs is odd. The probability of having $i$ occur exactly $k$ times and $0$ not occurring is $\binom{3}{k} \left(\frac{1}{6}\right)^{k}\left(\frac{4}{6}\right)^{3-k}$ . To find the probability of the total being real, we add $\frac{91}{216} + \binom{3}{0} \left(\frac{1}{6}\right)^{0}\left(\frac{4}{6}\right)^{3} + \binom{3}{2} \left(\frac{1}{6}\right)^{2}\left(\frac{4}{6}\right)^{1} = \frac{91 + 64+ 12}{216} = \frac{167}{216}$ . Hence, $a + b = 167+216 = 383$ .

Lets find all those cases when product is not real

i) All three are 'i' - 1 case

ii) 'i' appears just once and other two are different (0 can not be there else product will be 0) On remaining two dices we can have two different numbers in C(4, 2) ways (among 1, e, pi, √2) we can permute them in 3! ways So, C(4, 2)*3! ways

ii) 'i' appears just once and other two dices show same number other than 0 we can choose the same number in C(4, 1) ways Now, we can permute them in 3!/2! = 3 ways So, C(4, 1)*3 ways

Adding up all the cases, we get,

The probability that the product of all the numbers on the top face is real ,

= 1 - (probability that product is not real)

= 1 - (1/216 + 6 * C(4, 2)/216 + 3 * C(4, 1)/216)

= 167/216

Hence the value of a+b=383.