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Percentage question (probability , game)

  1. Mar 18, 2013 #1

    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 :)
     
  2. jcsd
  3. Mar 18, 2013 #2

    mathman

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    Calculation looks OK. Style question - why did you use comma for decimal points?
     
  4. Mar 18, 2013 #3

    reenmachine

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    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 ;)
     
  5. Mar 18, 2013 #4

    reenmachine

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    (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
     
  6. Mar 19, 2013 #5

    SteamKing

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    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.
     
  7. Mar 19, 2013 #6

    reenmachine

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    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 knowledgable 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!!!
     
  8. Mar 20, 2013 #7
    No, that's correct.

    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: Mar 20, 2013
  9. Mar 20, 2013 #8

    reenmachine

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    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
     
  10. Mar 20, 2013 #9

    mathman

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