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Thread: Coin and die probability

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    Cool Coin and die probability




    If I repeatedly and alternately toss a coin and roll a 6 sided die beginning with the coin what is probability of getting a head before a 6

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    Re: Coin and die probability

    We can start simple, what is the probability of each independent event?
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    Re: Coin and die probability

    Heads 1/2 and die 1/6

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    Re: Coin and die probability

    Good problem, which I am incapable of solving after having 15 oz. of Riesling on Valentine's Day, but. . .

    2nd trial for getting a head:
    .5^2 = 1/4th
    3rd:
    .5^3 = 1/8th
    and so on. . .

    The strategy of the Non-Ames person seems to be: decompose the problem.
    http://www.google.com/search?q=polya...88.WbnGkpvNlyE

    I'm not a mathematician but it may be that .5^n crosses (1/6)^n at some high n value. One equation, one unknown.

    Dig it out of here
    https://en.wikipedia.org/wiki/List_o...mic_identities
    or use brute force. you may need it later.

    Some person with a Blood Alcohol Content less than mine may be able to solve this within a reasonable time frame.

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    Re: Coin and die probability

    Nope not me right now. I don't drink any more but had a Crawfish Cruiser with dinner. Just imagine a aqua orange drink in a Mason jar, for some reason it was warm with ice? Oh, yeah - and add some dark chocolate with some thing like cayenne in it.


    Yeah, there is probably a fine formula for this, but I can also dig the idea of just thinking of all of the possible combinations of the two independent events.One event, just hit heads first time - game over; second time tails and a 1; third time tails and a 2, etc... So this fit a conditional that the first flip is tails...


    We will see if BGM comes sweeping in and solves this with little effort. Though, I have not seen them around for awhile.
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    Re: Coin and die probability

    I think the problem can be solved by:

    Let X be the number of trials before coin is head including the trial resulting in head
    Let Y be the number of trials before a 6

    The event of getting a head before a 6 is the event that \{X=n,Y\geq n,n=1,2,3,...\}

    Hence one geometric variable higher than another ... theres a solution for similar problem here: https://www.google.dk/search?newwind....0.KaCDiINPaQs

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    Re: Coin and die probability


    As summoned by hlsmith so I make one more comment here.

    JesperHP has nicely setup the problem, and also googled the result. So I do not have to type once again here - just recommend the first search result as a reference:

    http://math.stackexchange.com/questi...ndom-variables

    The derivation is not hard, as what they had shown.

    Side note: One more thing is that geometric distribution has nice analog with exponential distribution in continuous settings, so if someone has found out the minimum of two or more independent exponential random variables, then they should find it to be very similar. In fact in a birth-death process, the process goes upward or downward depends on which of the two exponential arrival time is smaller, and you have a similar form. In discrete settings there are ties so it is a little bit more complicated.

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