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OT maths problem

bonescpbonescp Posts: 181
edited June 2007 in Workshop
what are the odds of tossing a coin and getting 10 heads in row,
this is a friendly arguement at work someone said it was 50 50 but this didnt sound right to me .ive looked on the maths forums but this only got more confussing.

john
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Posts

  • 1 in 1024 by recockoning.

    The chance each time is one in two = 0.5. Then you have ten separate independent trials so the chance of all being heads is 0.5 to the power 10 = 1/1024.
  • natureboyznatureboyz Posts: 82
    50/50
    No other factors come into it.
  • Bones,

    Every time you toss a coin, the probability of getting a 'Head' P(H)=0.5. Similarly the probability of getting a 'Tail' , P(T)=0.5

    The probability of getting 10 Heads in a row is P(H)x P(H)x P(H)x P(H)x P(H)x P(H)x P(H)x P(H)x P(H)x P(H)

    which is 0.5 x 0.5 x 0.5 x 0.5 x 0.5 x 0.5 x 0.5 x 0.5 x 0.5 x 0.5 = 0.0009765635 or 1 in 1024

    Apologies I am a very sad former statistician!

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  • gordycpgordycp Posts: 2,341
    pme is correct.
    You could "prove" it by writing out all 1024 combinations;

    HHHHHHHHHH
    HHHHHHHHHT
    HHHHHHHHTH
    HHHHHHHTHH

    and so on.
    Enjoy!
  • Two shedsTwo sheds Posts: 446
    Just to add to the confusion...your friend might be talking about conditional probability. If you've tossed a coin and got 9 heads in a row, then the odds of making it 10 in a row are 50:50. The coin has no memory.

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  • Mister PaulMister Paul Posts: 719
    natureboy is wrong.

    Look at it with two. You get the following possible combinations-

    HH
    HT
    TH
    TT

    So you've got a 1 in 4 chance of getting two consecutive heads. Keep going like this and you get the 1/1024 with 10 throws.



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  • natureboyznatureboyz Posts: 82
    <blockquote id="quote"><font size="1" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Mister Paul</i>

    natureboy is wrong.

    Look at it with two. You get the following possible combinations-

    HH
    HT
    TH
    TT

    So you've got a 1 in 4 chance of getting two consecutive heads. Keep going like this and you get the 1/1024 with 10 throws.



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    <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
    He asked about one not two so your example is meaningless.
  • Mister PaulMister Paul Posts: 719
    C'mon Mildred. You could at least concede gracefully, rather than making yourself look more dense.

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  • craigwendcraigwend Posts: 321
    http://en.wikipedia.org/wiki/Gambler's_fallacy


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  • natureboyznatureboyz Posts: 82
    <blockquote id="quote"><font size="1" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Mister Paul</i>

    C'mon Mildred. You could at least concede gracefully, rather than making yourself look more dense.

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    <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

    The fact that it is you who is the dense one conveniently escapes you attention DUMBO .
  • mr_hippomr_hippo Posts: 1,051
    I thought it might be one of those delicious problems that would make me think but sadly I am wrong! Answer: 50/50

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  • JadedJaded Posts: 6,663
    It isn't 50:50.

    Not in the way the question was asked by the OP.

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  • nunnun Posts: 434
    <blockquote id="quote"><font size="1" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by pme</i>

    1 in 1024 by recockoning.

    The chance each time is one in two = 0.5. Then you have ten separate independent trials so the chance of all being heads is 0.5 to the power 10 = 1/1024.
    <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

    Give that man a GCSE in maths. He's right.

    A good one is to ask "After a snooker break, what's the probability that the snooker balls come to rest in exactly the same positions as before the break"?
  • <blockquote id="quote"><font size="1" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by bones</i>



    what are the odds of tossing a coin and getting 10 heads in row,
    <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
    Those of you that think that the odds are 50/50: I'd love to meet you in a pub. Tell you what, I'll even give you double the "odds". If you can throw 10 heads in a row, I'll give you œ20. If you can't, you pay me œ10. [:D]

    I think that some of you must have read the OP's question as "If a coin has been tossed 9 times and come down as heads each time, what are the odds of the tenth throw being heads?"
    That would indeed be evens. But that wasn't the question, was it?

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  • Mister PaulMister Paul Posts: 719
    <blockquote id="quote"><font size="1" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by natureboyz</i>

    <blockquote id="quote"><font size="1" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Mister Paul</i>

    C'mon Mildred. You could at least concede gracefully, rather than making yourself look more dense.

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    <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

    The fact that it is you who is the dense one conveniently escapes you attention DUMBO .
    <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

    Toss two coins. Toss the same coin twice. Same probabilities.

    Get it yet?

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  • natureboyznatureboyz Posts: 82
    <blockquote id="quote"><font size="1" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by nun</i>

    <blockquote id="quote"><font size="1" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by pme</i>

    1 in 1024 by recockoning.

    The chance each time is one in two = 0.5. Then you have ten separate independent trials so the chance of all being heads is 0.5 to the power 10 = 1/1024.
    <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

    <b>Give that man a GCSE in maths. He's right.</b>

    A good one is to ask "After a snooker break, what's the probability that the snooker balls come to rest in exactly the same positions as before the break"?


    <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">


    No, he isn't.
    Each of the two possible outcomes has equal probability no matter how many times the coin has been flipped previously and no matter what the result.
  • I have A level maths, oooohh!
  • natureboyznatureboyz Posts: 82
    <blockquote id="quote"><font size="1" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Mister Paul</i>

    <blockquote id="quote"><font size="1" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by natureboyz</i>

    <blockquote id="quote"><font size="1" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Mister Paul</i>

    C'mon Mildred. You could at least concede gracefully, rather than making yourself look more dense.

    __________________________________________________________
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    <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

    The fact that it is you who is the dense one conveniently escapes you attention DUMBO .
    <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

    Toss two coins. Toss the same coin twice. Same probabilities.

    Get it yet?

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    <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

    Don't be stupid over here as well .Save it for Campaign
  • <snigger>
    Natureboyz, please read the original post again.

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  • natureboyznatureboyz Posts: 82
    I have read it thank you sniggerer.
  • OK [:D][:D][:D]

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  • JadedJaded Posts: 6,663
    <snigger> [:D]

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  • natureboyznatureboyz Posts: 82
    How old are you Miss Goody ?
    Five ,six maybe .
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  • andrewgturnbullandrewgturnbull Posts: 3,861
    <blockquote id="quote"><font size="1" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by natureboyz</i>

    I have read it thank you sniggerer.
    <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

    Hi there.

    natureboyz - why don't you test your hypothesis and prove the sniggers wrong?

    Toss a coin 10 times in a row and write down the results. If you don't get ten heads in a row, then try again - using your argument you'll probably crack it the second time.

    I'd be genuinely interested to know how you got on - can you let us know?

    Cheers, Andy

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  • GarybeeGarybee Posts: 815
    Natureboyz, you are correct for any indivudual throw, regardless of the number of throws before. But the question was regarding the result of numerous throws. As a more practical example, have you ever bet on an accumulator at the bookies? If what you are saying was correct, they would lose a lot of money.

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  • andrewgturnbullandrewgturnbull Posts: 3,861
    On a slightly different topic. natureboyz - have you ever met Monty Hall?

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  • JadedJaded Posts: 6,663
    <blockquote id="quote"><font size="1" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by natureboyz</i>

    How old are you Miss Goody ?
    Five ,six maybe .
    <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

    oh, paganini.

    What class. [:D][:D][:D]

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  • Mister PaulMister Paul Posts: 719
    Natureboyz. You're getting mixed up.

    Toss a coin ten times. Each time you have a 50% change of heads. Now, that's where you've stopped thinking.

    Now, toss it twice. The possible outcomes are

    Head head
    head tails
    tails head
    tails tails

    So with 2 throws, you have a 1 in 4 chance of getting both heads. Not a 1 in 2, as you think.

    Now with three throws-

    HHH
    HHT
    HTT
    HTH
    THH
    THT
    TTH
    TTT

    So after three throws there are 8 possible outcomes. Only one is all heads, so that's a one in 8 chance. Not a 1 in 2, as you think.

    And so on, all the way up to ten throws.

    That's about as simple as it can be explained. Do you still not understand?

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  • nunnun Posts: 434
    <i>Originally posted by natureboyz</i>

    <blockquote id="quote"><font size="1" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by nun</i>

    <blockquote id="quote"><font size="1" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by pme</i>

    1 in 1024 by recockoning.

    The chance each time is one in two = 0.5. Then you have ten separate independent trials so the chance of all being heads is 0.5 to the power 10 = 1/1024.
    <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

    <b>Give that man a GCSE in maths. He's right.</b>

    A good one is to ask "After a snooker break, what's the probability that the snooker balls come to rest in exactly the same positions as before the break"?


    <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

    So nobody wants to answer the snooker question?
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