Percentage question (probability , game)

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Discussion Overview

The discussion revolves around calculating probabilities in a dice game with specific win/loss conditions. Participants explore the odds of winning based on different scenarios and rules, including multiple rounds of throws and conditional probabilities.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant outlines the rules of a dice game and attempts to calculate the odds of winning, arriving at a probability of 42% based on their reasoning.
  • Other participants confirm the calculations but question the use of commas for decimal points, highlighting regional differences in numerical representation.
  • A separate participant presents a different probability scenario involving multiple chances to win and attempts to calculate the overall odds, arriving at 44.5%.
  • Another participant asks about calculating overall winning odds given specific probabilities for winning and losing, expressing uncertainty about how to sum probabilities across multiple rounds.
  • One participant suggests that the method of summing probabilities resembles a geometric series, providing a formula for calculating the total probability of winning.
  • There is a discussion about the terminology of "odds" versus "probability," with some participants indicating that they are often used interchangeably, while others seek clarification on their differences.

Areas of Agreement / Disagreement

Participants generally agree on the calculations presented, but there is some confusion regarding the terminology of odds versus probability. The discussion remains unresolved regarding the best approach to summing probabilities in the context of the game.

Contextual Notes

Some participants express uncertainty about the concept of conditional probabilities and the implications of summing probabilities across multiple rounds, indicating a need for further clarification on these topics.

Who May Find This Useful

Individuals interested in probability theory, game theory, or those seeking to understand the nuances of calculating odds in games of chance may find this discussion beneficial.

reenmachine
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Suppose you throw a dice and on the first throw you win if you score 1 , you lose if you score 2 & 3 and have to throw the dice again if you hit 4 , 5 or 6.If you have to throw the dice again the rules changes and you win if you hit 1 but lose if you hit 2.If you hit any other number , you continue to throw it until you win or lose (with the same rules as the second throw).How do I find out what are my odds to the nearest percentage of winning in this game before the first throw?

I tried this , please help me understand what I did wrong or what I did right:

First throw:

1/6 = win (16,67%)
2/6 = lose (33,33%)
3/6 = throw again (50%)

Second throw and beyond:

1/6 = win (16.67%)
1/6 = lose (16.67%)
4/6 = throw again (66.67%)

Since the odds are the same for winning or losing beyond the first throw , once I get passed it my odds becomes 50% to win overall.

Calculations(?):

50(0,50) + 16,67(1) + 33,33(0)
25 + 16,67 + 0 = 41,67 = 42%

any thoughts would be appreciated , I'm pretty insecure everytime I encounter a probability problem so go easy on me please :)
 
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Calculation looks OK. Style question - why did you use comma for decimal points?
 
mathman said:
Calculation looks OK. Style question - why did you use comma for decimal points?

Thanks a lot , I had no idea how to attack the problem , I have almost no background in probability so I just went with my intuition.

As for the style question , I guess I always used commas since my youth.Is this considered wrong or is it just a computer problem? I rarely use computers to do or communicate mathematics.Do the vast majority of programs don't count the commas as decimal numbers? If that's the case I guess the more math I'll do on computers the more I'll have to change that habit ;)
 
(Out of context , but if on a random first throw (ignoring the previous problem and rules)):

If I have a 25% chance to have a 25% chance to win on my next throw , a 60% chance to have a 60% chance on my next throw , and a 15% chance to have a 15% chance on my next throw , how do I calculate my odds of winning on my next throw?

I tried:

25(0.25) + 60(0.60) + 15(0.15)
6.25 + 36 + 2.25 = 44.5%
So I would have a 44.5% chance to win on my next throw.

does that make any sense to you guys?

thanks
 
N.B.: In most English speaking countries, the decimal point is a dot (.), whereas on the Continent it is a comma (,). The separator for thousands is the opposite.

A similar situation occurs with dates: in the US, it is month-day-year; elsewhere day-month-year.
 
Ok I'll ask it here also even though I asked it in the homework subforum , but if I have a 5.7% chance to win and a 8.33% to lose a game (with the additionnal 85.97% meaning I have to roll the dice (or whatever) again with the same odds until I win or I lose , how do I calculate my odds of winning the game overall before the game even begins?

In the other thread they talked about conditionnal probabilities , a concept I'm not knowledgeable about.

I thought my odds of winning were 40.6% because of 5.7+8.33=14.03. 5.7/14.03*100 = 40.6 , but it seems I was wrong.

If I have to add up the probability of winning round 1 + P of winning round 2 etc... how do I do it?

My try:

round 1: 0.057
round 2: 0.8597 (0.057)
round 3: 0.8597 (0.8597)(0.057)
etc...

I have a very strong feeling that I'm wrong , and even if I'm right how do I know when to stop adding up? (can't add up forever)

thanks!
 
reenmachine said:
I thought my odds of winning were 40.6% because of 5.7+8.33=14.03. 5.7/14.03*100 = 40.6 , but it seems I was wrong.

No, that's correct.

reenmachine said:
round 1: 0.057
round 2: 0.8597 (0.057)
round 3: 0.8597 (0.8597)(0.057)
etc...

I have a very strong feeling that I'm wrong , and even if I'm right how do I know when to stop adding up? (can't add up forever)

You are right again! The trick is that you can add up forever, this is called the sum of a geometric series
 whose properties are well known. In this case the sum is ## \frac{0.057}{1-0.8597} = \frac{0.057}{0.1403}##, just the same as your intuitive result.
 
Last edited:
wow I'm actually impressed that I was right :D

Last question , very basic , but 40,6% is the odds of winning right? If I used the word probability instead of odds , would the result be the same?

Thank you for your input :X

cheers
 
reenmachine said:
wow I'm actually impressed that I was right :D

Last question , very basic , but 40,6% is the odds of winning right? If I used the word probability instead of odds , would the result be the same?

Thank you for your input :X

cheers

Don't get caught up in language quibble. Odds of winning and probability of winning are the same.
 

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