Odd Number Possibilities with a Die: 50% (2 Rolls)

  • Context: High School 
  • Thread starter Thread starter Hepic
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Discussion Overview

The discussion revolves around the probabilities of rolling odd numbers with a die, specifically focusing on the outcomes of rolling a die twice and then extending the inquiry to three rolls. Participants explore the calculations and reasoning behind these probabilities.

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant asserts that the probability of rolling an odd number with a die is 50% and questions the possibilities when rolling twice.
  • Another participant claims there is a 25% chance of rolling two odd numbers, a 25% chance of rolling two even numbers, and a 50% chance of rolling one of each, leading to a 75% probability of getting at least one odd number.
  • A participant seeks clarification on how the 75% probability is derived.
  • One response suggests listing all 36 combinations of outcomes or calculating it as 100% minus the probability of rolling two even numbers.
  • A participant attempts to extend the reasoning to three rolls, proposing a probability of 83.33% for at least one odd number.
  • Another participant corrects this, stating that the probability of rolling all even numbers in three rolls is (1/2)^3 = 1/8, leading to a probability of 87.5% for at least one odd number.

Areas of Agreement / Disagreement

Participants express differing views on the calculations of probabilities, with some asserting specific percentages and others challenging those claims. The discussion remains unresolved regarding the correct probabilities for multiple rolls.

Contextual Notes

Some calculations depend on assumptions about the independence of rolls and the definitions of odd and even outcomes. There are unresolved mathematical steps in deriving the probabilities for three rolls.

Hepic
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To bring odd number with a die,is 50%.
If I play with the die two times,how possibilities I have to bring odd number?
 
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There would be a 25% chance of getting two odd numbers, 25% chance of two even numbers and a 50% chance of getting one of each. So the probability of getting at least one odd number would be 75%.
 
Yea,I am talking about bringing at least one time "odds".
How we find the 75%?
 
Well you can write out all 36 combinations and add them up. Or you can think of it as 100% less the probabability of getting both even which is 100% - 50%x50% = 100%-25%=75%
 
So If I undestood good,If I play "3" times,for the same result,I have 83,3333%.

Right?
 
No, apparently you didn't understand at all. The probability of getting all three even is [itex](1/2)^3= 1/8= 0.125[/itex] so the probability of "at least one head in three rolls" is [itex]1- 0.125= 0.875[/itex].
 

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