Difference between revisions of "2001 AIME I Problems/Problem 6"
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The red path corresponds to the sequence of rolls <math>2, 3, 5, 5</math>. This establishes a one-to-one correspondence between valid dice roll sequences and block walking paths. | The red path corresponds to the sequence of rolls <math>2, 3, 5, 5</math>. This establishes a one-to-one correspondence between valid dice roll sequences and block walking paths. | ||
− | The solution to this problem is therefore <math>\dfrac{\binom{9}{4}}{6^4} = {\dfrac{7}{72}} | + | The solution to this problem is therefore <math>\dfrac{\binom{9}{4}}{6^4} = {\dfrac{7}{72}}</math>. So the answer is <math>\boxed{079}</math>. |
=== Solution 2 === | === Solution 2 === | ||
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*If the first subset ends in a 2, there is <math>2</math> such subsets and there are <math>5 + \ldots + 1 = \frac{5}{2}(5 + 1) = 15</math> ways of making the second subset. | *If the first subset ends in a 2, there is <math>2</math> such subsets and there are <math>5 + \ldots + 1 = \frac{5}{2}(5 + 1) = 15</math> ways of making the second subset. | ||
− | Thus, the number of combinations is <math>\sum_{i=1}^6 i \cdot \left(\frac{(7 - i)(8 - i)}{2}\right) = 21 + 30 + 30 + 24 + 15 + 6 = 126</math>, and the probability again is <math>\frac7{72}</math>. | + | Thus, the number of combinations is <math>\sum_{i=1}^6 i \cdot \left(\frac{(7 - i)(8 - i)}{2}\right) = 21 + 30 + 30 + 24 + 15 + 6 = 126</math>, and the probability again is <math>\frac7{72}</math>, giving <math>m+n=\boxed{079}</math>. |
+ | |||
+ | === Solution 7 Recursion Formula === | ||
+ | If you're too tired to think about any of the above smart transformations of the problem, a recursion formula can be a robust way to the correct answer. | ||
+ | |||
+ | We just need to work out the valid cases for each roll. Denote by <math>N_{k}(n)</math> the number of valid cases in the <math>k+1</math>-th roll when the number <math>n</math> is rolled, for <math>k=1,2,3</math> and <math>1 <= n <= 6</math>. Then we have the following recursion formula: | ||
+ | <cmath> N_{1}(n) = n</cmath> | ||
+ | <cmath> N_{2}(n) = N_{1}(1) + N_{1}(2) + ... + N_{1}(n)=N_{2}(n-1)+N_{1}(n)</cmath> | ||
+ | <cmath> N_{3}(n) = N_{2}(1) + N_{2}(2) + ... + N_{2}(n)=N_{3}(n-1)+N_{2}(n)</cmath> | ||
+ | The logic is that, if <math>n</math> is rolled, then the number of valid cases is the subtotal of valid cases in the preceding roll for the outcome of 1 to <math>n</math>. The recursion can be easily calculated by hand when <math>N_k(n)</math> are put in columns side by side, given the fact that te numbers are smaller than 100. | ||
+ | Finally, the total number of cases is | ||
+ | <cmath> N = \sum_{n=1}^{6} N_{3}(n)</cmath> | ||
+ | and | ||
+ | <cmath> P = \frac{N}{6^4}</cmath> | ||
+ | |||
+ | === Solution 8 === | ||
+ | We use expected value. | ||
+ | |||
+ | The first dice averages to <math>3.5</math>. So, we start off with <math>\frac{7}{2}</math>. Now, we have a <math>\frac{1}{2}</math> chance. | ||
+ | |||
+ | Next, we consider the number that we picked. There is a total of <math>1</math> for the probabilities when we pick <math>1</math>, <math>2</math>, or <math>3</math>. So, we get <math>\frac{1}{3}</math>. Finally, we have <math>1</math> and <math>2</math>. This gives <math>\frac{\frac{1}{3}}{2}=\frac{1}{6}</math>. Multiplying, we get <math>\frac{7}{72} \implies \boxed{079}</math>. | ||
+ | ~asdf334 | ||
+ | (I actually don't know if this is a valid sol, it seems sketchy to me lol) | ||
+ | |||
+ | ===Solution 9: Observation of each case=== | ||
+ | Lets try casework and observe the cases. Notice that if the last roll is a <math>1</math>, then the only dice rolls may be <math>1-1-1-1</math>, which is only <math>1</math> possibility. Observe that if the last roll is <math>2</math>, then there are <math>4 = 1 + 3</math> possibilities. When the last roll is a <math>3</math>, there are <math>10 = 1 + 3 + 6</math> possibilities. Notice when the last roll is <math>n</math>, the number of cases is the sum of the first <math>n</math> positive triangular numbers. This is easily provable when observing the numbers of possibilities after assigning a value to the last and second to last rolls. | ||
+ | So there are a total of <math>1 + 1 + 3 + 1 + 3 + 6 + 1 + 3 + 6 + 10 + 1 + 3 + 6 + 10 + 15 + 1 + 3 + 6 + 10 + 15 + 21 = 126</math> possibilities. So the probability is <math>\frac{126}{6^4} = \frac{7}{72}</math> and <math> 7 + 72 = \boxed{079}</math>. ~skyscraper | ||
+ | |||
+ | ===Solution 10: Distributions=== | ||
+ | |||
+ | This is equivalent to picking a four-element sequence of <math>\{1, 2, 3, 4, 5, 6\}</math> with repetition. Notice that once the sequence is picked, there is one and only one way to order these so that they form a sequence of rolls satisfying our conditions. | ||
+ | |||
+ | Now count the number of such four-element sequences, let <math>a</math> be the number of <math>1</math>s in the sequence, <math>b</math> be the number of <math>2</math>s, <math>c</math> <math>3</math>s, <math>d</math> <math>4</math>s, <math>e</math> <math>5</math>s, and <math>f</math> <math>6</math>s. Now we see that we must have <cmath>a + b + c + d + e + f = 4</cmath> with <math>a, b, c, d, e, f</math> being nonnegative integers since there are a total of <math>4</math> numbers picked. The number of solutions to this is <math>\dbinom{9}{4},</math> so our total number is equal to <math>\dfrac{\binom{9}{4}}{6^4} = \dfrac{7}{72},</math> making our answer <math>\boxed{079}.</math> | ||
+ | |||
+ | ~Ilikeapos | ||
+ | |||
+ | == Solution 11: The proof no one will read == | ||
+ | We can construct a bijection using stars and bars as follows: | ||
+ | Consider the sequence of <math>6</math> balls <math>OOOOOO</math>. Now place <math>4</math> bars in between them. For instance, the configuration <math>O||O|OO|OO</math> represents the rolls <math>1,1,2,2</math>. Now you might think, "Oh now it's easy. <math>\binom{10}{4}</math>." However, what if we have <math>|O|OOO|O|O</math>? Maybe we could remove the last ball. But then we have <math>5</math> balls. How do we rid of this ambiguity? Well we could give it a value. If there is nothing to the right of a bar, give it the value of <math>6</math>. Now the answer is simple. <math>\binom{5+4}{4}=\binom{9}{4}</math>. Then we have <math>6^4</math> different rolls, and now, the answer is simply <math>\frac{\binom{9}{4}}{6^4}</math>, which gives the answer of <math>\boxed{079}</math>. (Remember to always include the <math>0</math> in this competition for <math>1</math> or <math>2</math> digit numbers). | ||
+ | |||
+ | ~th1nq3r | ||
== See also == | == See also == |
Latest revision as of 18:41, 13 July 2021
Problem
A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form where and are relatively prime positive integers. Find .
Contents
Solutions
Solution 1
Recast the problem entirely as a block-walking problem. Call the respective dice . In the diagram below, the lowest -coordinate at each of , , , and corresponds to the value of the roll.
The red path corresponds to the sequence of rolls . This establishes a one-to-one correspondence between valid dice roll sequences and block walking paths.
The solution to this problem is therefore . So the answer is .
Solution 2
If we take any combination of four numbers, there is only one way to order them in a non-decreasing order. It suffices now to find the number of combinations for four numbers from . We can visualize this as taking the four dice and splitting them into 6 slots (each slot representing one of {1,2,3,4,5,6}), or dividing them amongst 5 separators. Thus, there are outcomes of four dice. The solution is therefore , and .
Solution 3
Call the dice rolls . The difference between the and distinguishes the number of possible rolls there are.
- If , then the values of are set, and so there are values for .
- If , then there are ways to arrange for values of , but only values for .
- If , then there are ways to arrange , and there are only values for .
Continuing, we see that the sum is equal to . The requested probability is .
Solution 4
The dice rolls can be in the form
ABCD
AABC
AABB
AAAB
AAAA
where A, B, C, D are some possible value of the dice rolls. (These forms are not keeping track of whether or not the dice are in ascending order, just the possible outcomes.)
- Now, for the first case, there are ways for this. We do not have to consider the order because the combination counts only one of the permutations; we can say that it counts the correct (ascending order) permutation.
- Second case: ways to pick 3 numbers, ways to pick 1 of those 3 to duplicate. A total of 60 for this case.
- Third case: ways to pick 2 numbers. We will duplicate both, so nothing else in this case matters.
- Fourth case: ways to pick 2 numbers. We pick one to duplicate with , so there are a total of 30 in this case.
- Fifth case: ; all get duplicated so nothing else matters.
There are a total of possible dice rolls.
Thus,
Solution 5
Consider the number of possible dice roll combinations which work after roll, after rolls, and so on. There is 6 possible rolls for the first dice. If the number rolled is a 1, then there are 6 further values that are possible for the second dice; if the number rolled is a 2, then there are 5 further values that are possible for the second dice, and so on.
Suppose we generalize this as a function, say return the number of possible combinations after rolls and being the beginning value of the first roll. It becomes clear that from above, ; every value of after that is equal to the sum of the number of combinations of rolls that have a starting value of at least . If we slowly count through and add up all the possible combinations we get possibilities.
Solution 6
In a manner similar to the above solution, instead consider breaking it down into two sets of two dice rolls. The first subset must have a maximum value which is the minimum value of the second subset.
- If the first subset ends in a 1, there is such subset and there are ways of making the second subset.
- If the first subset ends in a 2, there is such subsets and there are ways of making the second subset.
Thus, the number of combinations is , and the probability again is , giving .
Solution 7 Recursion Formula
If you're too tired to think about any of the above smart transformations of the problem, a recursion formula can be a robust way to the correct answer.
We just need to work out the valid cases for each roll. Denote by the number of valid cases in the -th roll when the number is rolled, for and . Then we have the following recursion formula: The logic is that, if is rolled, then the number of valid cases is the subtotal of valid cases in the preceding roll for the outcome of 1 to . The recursion can be easily calculated by hand when are put in columns side by side, given the fact that te numbers are smaller than 100. Finally, the total number of cases is and
Solution 8
We use expected value.
The first dice averages to . So, we start off with . Now, we have a chance.
Next, we consider the number that we picked. There is a total of for the probabilities when we pick , , or . So, we get . Finally, we have and . This gives . Multiplying, we get . ~asdf334 (I actually don't know if this is a valid sol, it seems sketchy to me lol)
Solution 9: Observation of each case
Lets try casework and observe the cases. Notice that if the last roll is a , then the only dice rolls may be , which is only possibility. Observe that if the last roll is , then there are possibilities. When the last roll is a , there are possibilities. Notice when the last roll is , the number of cases is the sum of the first positive triangular numbers. This is easily provable when observing the numbers of possibilities after assigning a value to the last and second to last rolls. So there are a total of possibilities. So the probability is and . ~skyscraper
Solution 10: Distributions
This is equivalent to picking a four-element sequence of with repetition. Notice that once the sequence is picked, there is one and only one way to order these so that they form a sequence of rolls satisfying our conditions.
Now count the number of such four-element sequences, let be the number of s in the sequence, be the number of s, s, s, s, and s. Now we see that we must have with being nonnegative integers since there are a total of numbers picked. The number of solutions to this is so our total number is equal to making our answer
~Ilikeapos
Solution 11: The proof no one will read
We can construct a bijection using stars and bars as follows: Consider the sequence of balls . Now place bars in between them. For instance, the configuration represents the rolls . Now you might think, "Oh now it's easy. ." However, what if we have ? Maybe we could remove the last ball. But then we have balls. How do we rid of this ambiguity? Well we could give it a value. If there is nothing to the right of a bar, give it the value of . Now the answer is simple. . Then we have different rolls, and now, the answer is simply , which gives the answer of . (Remember to always include the in this competition for or digit numbers).
~th1nq3r
See also
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.