Probability: Getting Sum of 8 vs Sum of 7 with 2 Dice

  • Thread starter Thread starter semidevil
  • Start date Start date
  • Tags Tags
    Probability
Click For Summary

Homework Help Overview

The discussion revolves around calculating the probability of rolling a sum of 8 before rolling a sum of 7 with two fair dice. Participants are exploring the probabilities associated with these outcomes and the implications of their calculations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the number of combinations that yield sums of 7 and 8, questioning the initial calculations and assumptions about their probabilities. Some suggest breaking down the problem based on the sequence of rolls, while others propose a more straightforward approach of considering only the last roll.

Discussion Status

The discussion is active, with participants providing different perspectives on how to approach the problem. Some guidance has been offered regarding the relevant combinations and the focus on the last roll, but no consensus has been reached on a definitive method or solution.

Contextual Notes

There is a noted discrepancy in the initial probability calculations, with participants questioning the number of combinations leading to sums of 7 and 8. The discussion also highlights the relevance of only considering the outcomes of 7 and 8 in the context of the problem.

semidevil
Messages
156
Reaction score
2
ok, so what is the probability that the sum of 8 will appear before the sum of 7 when rolling 2 fair dices. How do I even start this?

so with 2 fair dice, the probabality of gettng a sum of 8 is 5/36 right(2 6, 3 5, 4 4, 5, 3, 6 2)? and I got the same for the sum of 7(1 6, 2 5, 3 4, 4 3, 5 2, 6 1), 5/36.

so what do I do next?
 
Physics news on Phys.org
I think you meant there are 6 ways to get a sum of 7:

(1 6, 2 5, 3 4, 4 3, 5 2, 6 1) or 6/36 so there is a slightly greater probability of getting a 7 instead of 8 (only 5 combinations, or 5/36 of chance).

If they were equal, the probability of getting a 7 first would be 0.5.
 
Break it up according to when the 8 appears. Can you find the probability that an 8 is thrown on toss n, but neither a 7 nor an 8 was thrown before that? How will this help?
 
That's making the question more complicated than it is. You just imagine rolling the die until the first 7 or 8. So you have a given: the last roll of this sequence is either a 7 or an 8. If it's a 7 then you rolled a 7 before rolling an 8 (imagine for a moment extending the sequence beyond the 7 until the first 8 to be sure of this); if it's an 8 then you rolled an 8 before rolling a 7. The conclusion is easy.
 
There are 6 combinations that yield 7, and 5 combinations of getting 8.

All other combinations are irrelevant to this problem, because you are only interested in the total of 11 combinations that give 7 or 8.

So the probability of getting a combination that gives 7 = 6/(6+5) and the probability of getting 8 is 5/(6+5).

So what's the answer?
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K