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Thread: tricky algebra: With a fair coin, my opponent flips 50 times, with fair odds, ie,...

  1. #1

    tricky algebra: With a fair coin, my opponent flips 50 times, with fair odds, ie,...

    Hi, I have searched and cannot find a similar question to this that has been answered...


    Q. With a fair coin, my opponent flips 50 times, with fair odds, ie I double my bet on a win.
    The average outcome is 25 wins, 25 losses.
    However, if I make the assumption that I will win at least 15 times, how do I proportion my bets
    to be ahead at the 15th win, whilst minimising bets?


    Further, write an equation to illustrate this with e= expected wins, n= total flips.




    I know e=15, n=50.
    I think if c=expected wins at any point in the series, bet=1+(losses/c) should be right, but
    it does not take into account n.


    I have asked an engineering graduate who is stuck. Any help, please?

  2. #2
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by daveylibra View Post
    Q. With a fair coin, my opponent flips 50 times, with fair odds, ie I double my bet on a win.
    Where does this "doubling my bet" thing come in? I see nothing in the preceding words which says anything about your bet, winning, rules, etc...?

    Quote Originally Posted by daveylibra View Post
    The average outcome is 25 wins, 25 losses.
    "The average outcome" being "of the coin tosses", rather than of "winning" or "bets"...?

    Quote Originally Posted by daveylibra View Post
    However, if I make the assumption that I will win at least 15 times, how do I proportion my bets to be ahead at the 15th win, whilst minimising bets?
    What other assumptions are you supposed to make? For instance, what happens when you "lose"?

    When you reply, please include the full and exact text of the exercise, the complete instructions, and how you feel that this involves only "tricky algebra" rather than probability, etc. Thank you!

  3. #3
    Hi! Apparently...

    1. for example, if I bet 1 and win, I get back 2. If I lose, I get back 0.

    2. Lets assume I bet on heads every time. 25 heads, 25 tails = 25 wins, 25 losses (this is the average.)

    3. I don't think there are any more assumptions. As before, when I lose I get back nothing.
    I can bet any amount I choose on each flip.

    Cheers!

  4. #4
    PS - This is the complete text of the exercise, and I wasn't sure whether to post in probability/statistics or algebra, or maybe calculus ??

  5. #5

    Exclamation Unanswerable?

    I think this problem could be unanswerable.
    Unfortunately, I cannot PROVE its unanswerable....!

  6. #6
    Can anyone help with the above question?

    I didn't think it was very complex... ? I think it must be answerable... ??

  7. #7
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    Quote Originally Posted by daveylibra View Post
    Q. With a fair coin, my opponent flips 50 times, with fair odds, ie I double my bet on a win.
    The average outcome is 25 wins, 25 losses.
    However, if I make the assumption that I will win at least 15 times, how do I proportion my bets
    to be ahead at the 15th win, whilst minimising bets?

    Further, write an equation to illustrate this with e= expected wins, n= total flips.

    I know e=15, n=50.
    I think if c=expected wins at any point in the series, bet=1+(losses/c) should be right, but
    it does not take into account n.

    I have asked an engineering graduate who is stuck. Any help, please?
    The problem is that your problem is not stated clearly enough to be able to answer it. That may be partly from lack of context.

    Being stated in the first person, it does not look like a textbook problem; yet you eventually said, "This is the complete text of the exercise". If it is really an exercise from a book, you must state it exactly as given, and tell us what is being taught in that section, along with any assumptions you are told to make.

    Context is everything when you ask a question.

  8. #8
    Its a question I am trying to answer for a friend...
    How much more information could possibly be required?

  9. #9
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    There are only 32,768 cases for 15 tosses. Spreadsheets work.

    1) Are you betting on each flip?
    2) Do you reset back to the initial amount on a loss? Maybe just 1/2 and back to where you were on the previous bet.
    3) What do you mean by "proportion".

    The direct answer to you question, "How much more information could possibly be required?" is simply, "enough information for a well-defined problem". We're not there, yet.

    Also, we're not really a consulting service. If you wan to LEARN, then we can help you AFTER you show your efforts.

    Note: If there were an easy way to hedge a bet, why don't we have more rich gamblers?
    Last edited by tkhunny; 11-21-2017 at 01:05 PM.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  10. #10
    Elite Member
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    Quote Originally Posted by daveylibra View Post
    I have asked an engineering graduate who is stuck.
    That's enough to show that your problem is not
    sufficiently clear...as everybody has been telling you...
    I'm just an imagination of your figment !

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