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Math Help - Easy probability question.

  1. #1
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    Easy probability question.

    OK this might be too advanced a forum for this question, hence the 'easy' in the title


    I might answer this myself as I go along but anyway here goes.

    Say two people roll a dice, the highest number wins you would expect
    each person to win 50% of the time. But what if one person was luckier
    and won 55% of the time. You could say he was 5% luckier than average.

    How extend that to 3 people, you would expect each person to win 33%
    of the time. Now say one persoon was luckier as in the first example, and
    also the same %age luckier. Would that be 5% luckier, ie 33+5=38%?

    I think this is too high, I think the answer is he would be 3.3% luckier for
    it
    to be the same amount of 'luckierness' than in the first example.

    Also I think I am into problems here with sample size possibly.

    Going back to the first example another way is to say 55% is 10%
    luckier than normal (5/50 times 100).Thats how I get the 3.3% value for
    the second example.

    So for four players it would be 2.5%, 5 players=2%, 10 players=10% etc...

    You see I want to compare how lucky a person is in poker 'showdowns'
    there would be no problem if it was just 2 players but sometimes there are 3
    or maybe 4, the max possible (but rather unlilkely) being 10.

    Problem X
    ------------

    But lets take 10 players as an example, say one player wins 20% of the time,
    which is twice as lucky as expect. Now if he got the same amount of luck
    in 2 player games what would the figures be? It can't be twice because
    twice 50% is 100% and that don't seem right.
    -----------------


    I think I am into sample size problems here too, I never did statistics
    unfortunately.

    Well I pretty much know I am into sample size stuff because I know that
    over say a million game even winning 5% more than average would be
    impossible (at least statistically anyway).
    I guess I need to look at something like standard deviation (yuk) or
    something like that?

    OK back to problem X.
    I can't even guess the right answer so I am going to make it easier and say
    it is on a sample of 100 to see if that helps.

    So in the 10 player games normal is 10 in 100 but he wins 20 in 100

    So in a 2 player game normal is 50 in 100 but he wins ?? in 100 with the
    same amount
    of good luck.

    What is the correct value of ??
    One guess would be 60 but that seems to low, another guess is 100 but that
    seems
    way to high. 75 feels about right but I can't explain why.

    OK I will have a crack at Problem X.
    I think a better way of thinking is how often would he win 20 in 100 which
    would have
    a value 'x' which means it would happen once in every x number of trials so
    then I
    would need to work out the value which happened once in every 'x' trials for
    two
    players, and this would be the answer!??

    Trouble is I am not sure how to do it (yet). I think I could do a simulation
    though.
    On my computer :O)
    I would just get the maximum value which occured in 'x' trials. I better
    still obtain
    an average value over repeated trials??

    I expect there is a (well known?) equation for this, which I might be able
    to work
    out if I spent few years thinking about it!!!

    Any thoughts/answers?
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  2. #2
    GAMMA Mathematics
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    I like the logic of 10% luckier because you are actually using proportions.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by colby2152 View Post
    I like the logic of 10% luckier because you are actually using proportions.
    If one has to do this that does seem the beat way, but I still don't like it

    RonL
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  4. #4
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    Quote Originally Posted by colby2152 View Post
    I like the logic of 10% luckier because you are actually using proportions.
    Yes I kind of wanted to quantify 'luck' so I could aggregate over different situations, maybe I was going about it in to wrong way.
    For example you could say if the chances of winning the lottery is 14 million to one then that is the same as spinning heads on a coin a particular number of times (x) in a row x would be about 24. Anyway his luck would be 14 million
    percent in each case. I was perhaps wrong to think of average luck because the average of the two values would obviously be 14 million percent.
    What I should be doing is multiplying them and getting a result of 196 trillion percent or whatever. So I guess my thinking was wrong from the start and it is simply a case of working the odds out of each event and multiplying them up. Thus in the long run the value should tend to 1.

    I think I got confused in my thinking because I was thinking about a number of things at the same time. Such as how skillful a player is in card games, his skill level might vary across 2, 3,......10 player games. So maybe I could average these values to find his 'skill level'.

    The trouble is poker is a game of skill and luck and it is kind of hard to seperate the too. Having said that, in the long run I would say it is all skill.

    What actually prompted the whole thing was to see if I was getting my fair share of luck at poker, which can be complicated. The simplest situation is when two players go all-in preflop. Your chances of winning depend on the cards you hold and I wanted to see if I was getting my fair share of luck.

    Of course often you fold your hand before the showdown so you never see your opponents cards and thus cannot work anything out.

    So...if you think you are really good at maths, have a go at playing poker online and find out how rubbish you really are
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