OT maths problem

bonescp
bonescp 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
«13

Comments

  • 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.
  • natureboyz
    natureboyz 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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  • gordycp
    gordycp 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 sheds
    Two 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 Paul
    Mister 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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  • natureboyz
    natureboyz 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 Paul
    Mister Paul Posts: 719
    C'mon Mildred. You could at least concede gracefully, rather than making yourself look more dense.

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


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  • natureboyz
    natureboyz 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_hippo
    mr_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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  • Jaded
    Jaded Posts: 6,663
    It isn't 50:50.

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

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  • nun
    nun 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 Paul
    Mister 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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  • natureboyz
    natureboyz 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!
  • natureboyz
    natureboyz 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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  • <snigger>
    Natureboyz, please read the original post again.

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

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

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  • natureboyz
    natureboyz Posts: 82
    How old are you Miss Goody ?
    Five ,six maybe .
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  • andrewgturnbull
    andrewgturnbull 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?

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

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  • Jaded
    Jaded 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 Paul
    Mister 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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  • nun
    nun 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?