Mabel has a standard 4-sided die with sides numbered 1 through 4. Each time Mabel rolls the die she writes down the number she rolls. After some number of rolls, the total of the numbers Mabel has written down is 5. We know that she must have rolled the dice at least twice. The expected number of times Mabel rolled the die can be expressed as 2 + b a where a and b are positive coprime integers. What is a + b ?
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Considering Mabel to always make 5 throws is a good simplifying assumption, but what does it simplify? It simplifies the solution because it gives us a set of outcomes that each have the same probability of occurring. This allows us to simply divide by the total number of outcomes at the end of the question.
Why are you allowed to assume 5 throws every time?
The minimum number of rolls needed will be 2 and the maximum number of rolls needed will be 5,in order to get a sum total of 5.
Probability of getting 5 with -
2 rolls is 1/4
3 rolls is 3/32
4 rolls is 1/64
5 rolls is 1/1024
So,total probability of rolling a 5 is = 1/4+ 3/32+ 1/64 + 1/1024 = 369/1024
Now,probability of needing -
2 rolls to roll a 5 = (1/4)/(369/1024)= 256/369 = P(2)
3 rolls to roll a 5= (3/32)/(369/1024) = 32/123 = P(3)
4 rolls to roll a 5 = (1/64)/(369/1024) = 16/369 = P(4)
5 rolls to roll a 5 = (1/1024)/(369/1024)= 1/369 = P(5)
Expected number of rolls = E(X) = x1p1 + x2p2 + x3p3 + x4p4 + x5p5
= 2 P(1) + 3 P(3) + 4 P(4) + 5 P(5) = 869/369
E(X)-2 = 869/369 - 2 = 131/369
a=131,b=369 a+b = 500
Consider 5 dice rolls.
The number of possibilities for two dice rolls is 4 ⋅ 4 3 = 2 5 6 . For three dice rolls, there are 2 ⋅ 3 ⋅ 4 2 = 9 6 possibilities. For four dice rolls, there are 4 ⋅ 4 possibilities, while for five dice rolls there is only 1 possibility. Note that 2 5 6 + 9 6 + 1 6 + 1 = 3 6 9 . So, the expected value is
3 6 9 2 × 2 5 6 + 3 × 9 6 + 4 × 1 6 + 5 × 1 = 3 6 9 8 6 9 . So, b a = 3 6 9 1 3 1 , and thus a + b = 5 0 0 .
This solution has the problem that if you don't already understand how to solve the question, it doesn't help.
The first line says to consider 5 rolls, but then we start considering possibilities for different numbers of rolls.
What are these possibilities we are counting?
Where do the numbers in counting them come from?
We're doing an expectation and you're dividing by the total number of possibilities. To do that, we need that the possibilities are equally likely. Are these possibilities all equally likely?
Denote W ( n ) as the set of all possible ways that the total of the numbers Mabel has written down after n rolls is 5 .
W ( 2 ) W ( 3 ) W ( 4 ) W ( 5 ) = { ( 1 , 4 ) , ( 2 , 3 ) , ( 3 , 2 ) , ( 4 , 1 ) } , = { ( 1 , 1 , 3 ) , ( 1 , 2 , 2 ) , ( 1 , 3 , 1 ) , ( 2 , 1 , 2 ) , ( 2 , 2 , 1 ) , ( 3 , 1 , 1 ) } , = { ( 1 , 1 , 1 , 2 ) , ( 1 , 1 , 2 , 1 ) , ( 1 , 2 , 1 , 1 ) , ( 2 , 1 , 1 , 1 ) } , = { ( 1 , 1 , 1 , 1 , 1 ) } .
The chance for any sequence of n rolls with a 4-sided die to come up, is 4 n 1 . Denote P 1 ( n ) as the probability that after n rolls, the total adds up to 5 .
P 1 ( 2 ) P 1 ( 3 ) P 1 ( 4 ) P 1 ( 5 ) = ∣ W ( 2 ) ∣ ⋅ 4 2 1 = 1 6 4 = 4 1 , = ∣ W ( 3 ) ∣ ⋅ 4 3 1 = 6 4 6 = 3 2 3 , = ∣ W ( 4 ) ∣ ⋅ 4 4 1 = 2 5 6 4 = 6 4 1 , = ∣ W ( 5 ) ∣ ⋅ 4 5 1 = 1 0 2 4 1 .
Note that these probabilities do not add up to 1 , but to 1 0 2 4 3 6 9 . This is because if you were to roll a 4-sided dice several times, you might just skip over 5 . However, we're already told that the numbers Mabel wrote down add up to 5 , so:
P 2 ( 2 ) P 2 ( 3 ) P 2 ( 4 ) P 2 ( 5 ) = 3 6 9 / 1 0 2 4 P 1 ( 2 ) = 3 6 9 2 5 6 , = 3 6 9 / 1 0 2 4 P 1 ( 3 ) = 3 6 9 9 6 , = 3 6 9 / 1 0 2 4 P 1 ( 4 ) = 3 6 9 1 6 , = 3 6 9 / 1 0 2 4 P 1 ( 5 ) = 3 6 9 1 ,
where P 2 ( n ) denotes the probability that, when given that the total adds up to 5 , then that total is reached after n rolls. Now, all that's left is calculating the expected value E :
E = 2 ⋅ 3 6 9 2 5 6 + 3 ⋅ 3 6 9 9 6 + 4 ⋅ 3 6 9 1 6 + 5 ⋅ 3 6 9 1 = 2 + 3 6 9 1 3 1 .
So, a + b = 1 3 1 + 3 6 9 = 5 0 0 .
With i is the number of rolls, the probability of a sequence of number is n i = 4 i 1 .
As the composition of an integer theory, the number of ways to write s as the sum of a sequence of k elements is ( k − 1 s − 1 ) . In this problem, s is 5 and k is the number of rolls.
Therefore, if Mabel rolls twice, the number of sequences is n 2 = ( 2 − 1 5 − 1 ) = ( 1 4 ) = 4 and the probability of each sequence is p 2 = 4 2 1 .
If Mabel rolls 3 times, n 3 = ( 2 4 ) = 6 and p 3 = 4 3 1 .
If Mabel rolls 4 times, n 4 = ( 3 4 ) = 4 and p 4 = 4 4 1 .
If Mabel rolls 5 times, n 5 = ( 4 4 ) = 1 and p 5 = 4 5 1 .
Thus, the expected value is E = ∑ i = 2 5 n i p i ∑ i = 2 5 i n i p i = 4 / 4 2 + 6 / 4 3 + 4 / 4 4 + 1 / 4 5 2 ⋅ 4 / 4 2 + 3 ⋅ 6 / 4 3 + 4 ⋅ 4 / 4 4 + 5 ⋅ 1 / 4 5 = 3 6 9 8 6 9
Finally, b a = 3 6 9 1 3 1 and the final solution is 1 3 1 + 3 6 9 = 5 0 0
We look at the partitions of 5. There are 4 partitions of 5 with 2 ordered terms, namely (1,4),(2,3),(3,2),(4,1). The total amount of ordered pairs of 2 is 4 × 4 = 1 6 . Therefore the probability is 1 6 4 = 4 1 . We do similarly for 3, 4, and 5 ordered terms. For 3, there are 6 different ordered triplets. The probability is therefore 4 3 6 = 3 2 3 . For 4, there are 4 ordered quadruplets. The probability is 4 4 4 = 6 4 1 . For 5, there is only one ordered quintuplet. Therefore the probability is 4 5 1 = 1 0 2 4 1 . The expected value of the number of rolls is therefore E ( x ) = 4 1 + 3 2 3 + 6 4 1 + 1 0 2 4 1 2 ⋅ 4 1 + 3 ⋅ 3 2 3 + 4 ⋅ 6 4 1 + 5 ⋅ 1 0 2 4 1 = 3 6 9 8 6 9 = 2 + 3 6 9 1 3 1 so the answer is 1 3 1 + 3 6 9 = 5 0 0
Clearly, the maximum number of rolls is 5 . And the minimum was given to us, 2 .
When the dice is rolled two times, the number of cases that give us sum five is 4 , and the probability is 4 2 4 . (The probability here is the number of cases that give us sum five divided by the total number of cases.)
When the dice is rolled three times, the number of cases that give us sum five is 6 , and the probability is 4 3 6 .
When the dice is rolled four times, the number of cases that give us sum five is 4 , and the probability is 4 4 4
Finally, when the dice is rolled five times, the number of cases that give us sum five is 1 , and the probability is 4 5 1 .
At this point, we just need to calculate our expected value:
E V = 4 2 4 + 4 3 6 + 4 4 4 + 4 5 1 2 ( 4 2 4 ) + 3 ( 4 3 6 ) + 4 ( 4 4 4 ) + 5 ( 4 5 1 )
E V = 3 6 9 8 6 9
But our reached value for EV can be expressed as 2 + 3 6 9 1 3 1 .
Therefore, our answer is 1 3 1 + 3 6 9 = 5 0 0 .
First, we count how many ways she can roll to 5. To calculate this, for n rolls, we must find the number of ways to split 5 into n parts that are at least 1. This means we must find the number of ways to split 5 − n into n nonnegative parts. This is simply ( n − 1 5 − n + ( n − 1 ) ) = ( n − 1 4 ) .
In two rolls, there are 4 ways. In three rolls, there are 6 ways. In four rolls, there are 4 ways. In five rolls, there is 1 way.
Now, we are given that she rolls a total of 5. We can calculate the probability for each of our above sequences of rolls, but then we must normalize them, since not every roll is possible.
The probability of rolling in two rolls is 4 ⋅ 1 6 1 .
The probability of rolling in three rolls is 6 ⋅ 6 4 1 .
The probability of rolling in four rolls is 4 ⋅ 2 5 6 1 .
The probability of rolling in five rolls is 1 ⋅ 1 0 2 4 1 .
These add to 1 0 2 4 3 6 9 , so our normalized probabilities are 3 6 9 2 5 6 , 3 6 9 9 6 , 3 6 9 1 6 , and 3 6 9 1 . Now, our expected value is 2 ⋅ 3 6 9 2 5 6 + 3 ⋅ 3 6 9 9 6 + 4 ⋅ 3 6 9 1 6 + 5 ⋅ 3 6 9 1 = 3 6 9 8 6 9 = 2 3 6 9 1 3 1 Therefore, the answer is 1 3 1 + 3 6 9 = 5 0 0 .
There are 6 non-ordered partitions of the integer 5 into two or more terms: 5 = 1 + 1 + 1 + 1 + 1 = 1 + 1 + 1 + 2 = 1 + 1 + 3 = 1 + 2 + 2 = 1 + 4 = 2 + 3 . The probability of observing the first case is 4 5 1 ; the second ( 1 4 ) 4 4 1 = 4 3 1 = 6 4 1 ; the third and fourth ( 1 3 ) 4 3 1 = 6 4 3 ; and the fifth and sixth ( 1 2 ) 4 2 1 = 8 1 , since we must account for the order of the terms. Let N be the number of dice rolls given the conditions. Then E [ N ] = ∑ n = 2 5 Pr [ N = n ] ∑ n = 2 5 n Pr [ N = n ] = 1 / 8 + 1 / 8 + 3 / 6 4 + 3 / 6 4 + 1 / 1 0 2 4 2 ( 1 / 8 + 1 / 8 ) + 3 ( 3 / 6 4 + 3 / 6 4 ) + 4 ( 1 / 6 4 ) + 5 ( 1 / 1 0 2 4 ) = 3 6 9 8 6 9 = 2 + 3 6 9 1 3 1 , from which we find a + b = 1 3 1 + 3 6 9 = 5 0 0 .
Nice solution, Hero P.
Recursion is another way of doing it: Let E n be the expected number of rolls if the total of the numbers is n (instead of 5). Then E 1 = 1 , E n = ( ∑ m = min ( 1 , n − 4 ) n − 1 E m P m ) / ( 4 P n ) + 1 where P n is the overall probability of the number n being "hit".
Then P 1 = 1 / 4
P 2 = 5 / 1 6
P 3 = 2 5 / 6 4
P 4 = 1 2 5 / 2 5 6
P 5 = 3 6 9 / 1 0 2 4
and E 2 = 6 / 5 ,
E 3 = 7 / 5 ,
E 4 = 8 / 5 ,
E 5 = 8 6 9 / 3 6 9 .
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P.S. The given P n values can be computed by P n = ( ∑ m = min ( 0 , n − 4 ) n − 1 P m ) / 4 where we define P 0 = 1 .
I think you missed one 6 4 1 in denominator on your calculation. Hero P.
when dice is trown two time so total number of cases of sum five (1 4)(4 1)(2 3)(3 2) or 4 cases the probability 4 / 4^2 (as 16 is total number of cases) when dice thrown 3 times total cases are 6 and probablilty 6 / 4^3 again when thrown 4 time stotal number of cases=4 then probablity=4 / 4^4 or five rolls there is only one way (1 1 1 1 1) and probability=1 / 4^5 now calcuate expected value E[x]=[2*(4/4^2)+3(6/4^3)+4(4/4^4)+5(1/4^5)]/[4/(4^2)+6/4^3+4/4^4+1/4^5] =869/369 or E[x]= 2+131/369 so answer is 131+369=500
This solution matches word to word with that submitted by Evan C. , Debjit M. , and Sanjay M. . Just asking out of curiosity, how is this possible?
I did a little digging and I found this: "reflection009 @brilliantperson123 when dice is trown two time so total number of cases of sum five (1 4)(4 1)(2 3)(3 2) or 4 cases the probability 4/4^2 (as 16 is total number of cases) lly when dice thrown 3 times total cases are 6( chk urself) and probablilty 6/4^3 again when thrown 4 time stotal number of cases=4 then probablity=4/4^4 or five rolls there is only one way (1 1 1 1 1) and probability=1/4^5 now calcuate ecxpected value E[x]=[2⋅(4/4^2)+3(6/4^3)+4(4/4^4)+5(1/4^5)]/[4/(4^2)+6/4^3+4/4^4+1/4^5] =869/369 orE[x]= 2+131/369 so answer is 131+369=500" This I found on PaGaLGuy. Here is the link: http://www.pagalguy.com/forums/quantitative/brilliant-org-problems-solutions-t-101694/p-17045293/r-17047415 Why do people cheat like this?
When dice is thrown 2 time so total number of cases of sum five ( 1 , 4 )( 4 , 1 )( 2 , 3 )( 3 , 2 ) i.e. 4 cases. So, the probability is 4 2 4 (as 16 is total number of cases). Similarly, when dice thrown 3 times total cases are 6 and probability is 4 3 6 . Again when thrown 4 times total number of cases is 4 then the probability is 4 4 4 . At last, when 5 rolls there is only one way (1,1,1,1,1) and probability is 4 5 1 . Now by calculating expected value, E [ x ]= 4 2 4 + 4 3 6 + 4 4 4 + 4 5 1 2 . 4 2 4 + 3 . 4 3 6 + 4 . 4 4 4 + 5 . 4 5 1 ⇒ E [ x ]= 3 6 9 8 6 9 ⇒ E [ x ]= 2 + 3 6 9 1 3 1 so answer is 1 3 1 + 3 6 9 = 5 0 0
when dice is thrown two time so total number of cases of sum five (1 4)(4 1)(2 3)(3 2) or 4 cases the probability 4/4^2 (as 16 is total number of cases) similarly when dice thrown 3 times total cases are 6( check yourself) and probablilty 6/4^3 again when thrown 4 times total number of cases= 4 then probablity=4/4^4 or five rolls there is only one way (1 1 1 1 1) and probability=1/4^5 now calculate expected value E[x]=[2⋅(4/4^2)+3(6/4^3)+4(4/4^4)+5(1/4^5)]/[4/(4^2)+6/4^3+4/4^4+1/4^5] =869/369
or E[x]= 2+131/369
so answer is 131+369=500
If we were to construct a series of tables to represent the sum totals of the dice after a certain number of rolls, then we would see that in two rolls there are 1 6 4 ways to get a total of 5 . For three rolls there are 6 4 6 ways, for four rolls there are 2 5 6 4 ways, and for five rolls there are 1 0 2 4 1 ways.
If we multiplied up by 1 0 2 4 , we would expect 2 5 6 , 9 6 , 1 6 , 1 fives from 2 , 3 , 4 , 5 rolls respectively.
Our expected total can be calculated as 2 5 6 + 9 6 + 1 6 + 1 2 ∗ 2 5 6 + 3 ∗ 9 6 + 4 ∗ 1 6 + 5 ∗ 1 = 3 6 9 8 6 9 = 2 3 6 9 1 3 1
So a + b = 1 3 1 + 3 6 9 = 5 0 0
when dice is trown two time so total number of cases of sum five (1 4)(4 1)(2 3)(3 2) or 4 cases the probability 4/4^2 (as 16 is total number of cases) lly when dice thrown 3 times total cases are 6( chk urself) and probablilty 6/4^3 again when thrown 4 time stotal number of cases=4 then probablity=4/4^4 or five rolls there is only one way (1 1 1 1 1) and probability=1/4^5 now calcuate ecxpected value E[x]=[2⋅(4/4^2)+3(6/4^3)+4(4/4^4)+5(1/4^5)]/[4/(4^2)+6/4^3+4/4^4+1/4^5] =869/369 orE[x]= 2+131/369 so answer is 131+369=500
We split cases: 1,4 1/4^2x2! chance 2,3 1/4^2x2! chance
1,1,3 1/4^3x3!/2! chance 1,2,2 1/4^3x3!/2! chance
1,1,1,2 1/4^4x4!/3! chance
1,1,1,1,1 1/4^5 chance
We want to evaluate: (1/4^2x2!x2+1/4^2x2!x2+1/4^3x3!/2!x3+1/4^3x3!/2!x3+1/4^4x4!/3!x4+1/4^5x5)/(1/4^2x2!+1/4^2x2!+1/4^3x3!/2!+1/4^3x3!/2!+1/4^4x4!/3!+1/4^5) which is the sum of possibility of each case x number of rolls/sum of possibility of each case. The fraction is equals to 2 and 131/369
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For simplicity, assume that Mabel always made 5 throws. The statement tells us that her sequence of throws was one of the following ones:
In each pattern the non- ⋆ numbers can be permuted arbitrarily, and the ⋆ s represent throws that follow after Mabel had sum=5. Thus, each star represents an arbitrary outcome of a throw: i.e., 4 possibilities. Taking both of these into account, we can now count:
In total, we have: 1 sequence where Mabel needed 5 throws to get sum=5, 16 sequences with 4 throws each, 96 sequences with 3 throws each, and 256 sequences with 2 throws each.
Thus, the expected number of throws is 1 + 1 6 + 9 6 + 2 5 6 1 ⋅ 5 + 1 6 ⋅ 4 + 9 6 ⋅ 3 + 2 5 6 ⋅ 2 = 3 6 9 8 6 9 = 2 + 3 6 9 1 3 1 and the answer is 1 3 1 + 3 6 9 = 5 0 0 .