Solving a Challenging Probability Problem - Detailed Steps

  • Context: Undergrad 
  • Thread starter Thread starter Brad_Ad23
  • Start date Start date
  • Tags Tags
    Probability
Click For Summary

Discussion Overview

The discussion revolves around a probability problem involving the outcome of rolling 20 six-sided dice and calculating the chances that the total sum is equal to or greater than 100. Participants explore various methods to approach the problem, including combinatorial calculations, polynomial expressions, and the Central Limit Theorem.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests a combinatorial approach by calculating the probabilities of achieving specific sums (e.g., 120, 119, 118) and summing them to find the probability of exceeding 100.
  • Another participant proposes using a polynomial expression to find the coefficients corresponding to the desired sums, indicating that this method can yield the probabilities for sums of 100 or higher.
  • A different approach involves applying the Central Limit Theorem, suggesting that the sum of the dice outcomes can be approximated by a normal distribution, with specific mean and variance calculations provided.
  • One participant mentions a specific probability result, although the context and derivation of this result are not fully detailed.
  • There is a suggestion that the methods discussed may simplify to recognizable patterns when varying the number of dice.

Areas of Agreement / Disagreement

Participants present multiple competing views and methods for solving the problem, with no consensus on a single approach or solution. The discussion remains unresolved as different methods are explored.

Contextual Notes

Some methods involve complex calculations and assumptions about distributions, while others rely on combinatorial reasoning. The discussion does not resolve the accuracy or efficiency of the proposed methods.

Brad_Ad23
Messages
497
Reaction score
1
Ok, a friend at another forum posted this challenging probability problem.

Mike throws 20 6-sided die on the table. He then adds up the results of all the dice, resulting in a number between 20 and 120. What are the chances the total number is equal to or greater than 100?

How, in detailed steps, would one go about solving this?[?]
 
Physics news on Phys.org
I'll be honest, I have no clue in hell... However by the sounds of the question I think this is better suited for the homework section. You sure it was your friend, and not a mathematics textbook that ponders the answer?
 
The "detailed steps" are long and tedious. I don't think you are paying us enough!

One way to get "above 100" is to get 120. To get that, all dice must land on 6. The probability of that is (1/6)20.
Another way is to get 119. To get that all dice except one must come up 6 and the one die must be 5. The probability of that is
20C1(1/6)20= 20(1/6)20.
Another way is to get 118. One way to get that is for all dice except one to come up 6 and that one die to come up 4. The probability of that is 20C1(1/6)20=
20(1/6)20. Another way to get 18 is for 18 of the dice to come up 6 and the other two to come up 5. The probability of that is
20C[/sub]2[/sub](1/6)2= 190(1/6)20.
Adding those gives a probability of 210(1/6)20 of getting 118.

Now continue like that down to 101. Believe me, they get really messy really fast!
 
I found another way. If one takes the polynomial

(x+x2+x3+x4+x5+x6) and raises it to the power of however many dice there are, the coefficient of the sum you wish to get will be the numerator in the probability.

So if we want to know the probability of getting the sum of 120, we look for the coefficient of x120 when that is all factored out. It turns out to be 1. So then we divide by 620 and we have the probability. Do find the probability of a sum 100 or higher, simply sum up the coefficients and then divide it by 620. It seems to work out pretty good!
 
What about using the Central Limit theorem?

X(1) = The outcome from dice number 1, it can be 1,2...6

We want Y = X(1)+X(2)+... X(20). This should be approximately normal in distributoin. Mean = 20 *3.5, variance = 20*2.92.

We want P(Y>100)

Sam
 
The odds are:

\frac{137 846 528 820}{3 656 158 440 062 976}
 
So, the method I posted works.
 
Originally posted by Brad_Ad23
So, the method I posted works.

Yes.

In this case it simplifies out to
2d choose d

It's also easy to spot a pattern if you work your way up throug varying numbers of dice.
 
Last edited:

Similar threads

  • · Replies 66 ·
3
Replies
66
Views
8K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 41 ·
2
Replies
41
Views
6K
  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 29 ·
Replies
29
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K