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Probability Rolling Sums BEFORE another sum

  1. May 11, 2008 #1
    The question is:
    Rolling a sum of 8 and a sum of 6 BEFORE rolling two sums of 7
    Experimental Probability: 55%
    Theoretical Probability: 54.5%

    How did they do this?
    I understand that to get a sum of
    8: 13.888%
    6: 13.888%
    7: 16.666%
    6/8: 27.777%

    but I dont understand how they figured out those values for theoretical probability...
    thanks so much
     
  2. jcsd
  3. May 11, 2008 #2

    DaveC426913

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    Perhaps a little more detail?

    What were they rolling? Cigars?
     
  4. May 11, 2008 #3
    two fair dice
     
  5. May 15, 2008 #4

    CRGreathouse

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    So there are 5 + 5 + 6 = 16 outcomes you care about; all others are rerolled. On the first roll*, there is a 10/16 chance of getting either a 6 or 8, and a 6/16 chance of rolling a 7.

    In the first case (discarding rolls with the original 6 or 8), you have a 5/11 chance of rolling the number you need and a 6/11 chance of rolling a 7. To 'win' you need to avoid rolling a 7 twice (chance 6/11 * 6/11 = 36/121); thus the odds of winning at this point are 85/121. Multiplying by the chance of getting to this case we get 85/121 * 10/16 = 425/968.

    In the second case you need to roll a 6 or 8 (chance 10/16), immediately followed by the other (chance 5/11) to 'win', thus chance 25/88. Overall, this is 75/704 since there's only a 6/16 chance of getting here.

    425/968 + 75/704 = 4225/7744 = 54.55...%. Now you'll have to check all of this, of course, since I do make arithmetic mistakes. The method's good, though.

    * Remember, this is really a sequence of rolls terminated when you roll one of {6, 7, 8}.
     
  6. May 16, 2008 #5

    mathwonk

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    i learned this stuff reading scarne on dice when i was 13 years old, a nice introduction to probability.
     
  7. May 18, 2008 #6

    uart

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    I got the same answer as CRGreathouse but I'm not sure how he got his answer without considering an infinite series. Normally I approach this sort of problem by looking at the probability of winning (exactly) on the first throw, then on the second, then on the third, and so on until a pattern emerges and you get a series you can sum.
     
  8. May 18, 2008 #7

    CRGreathouse

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    I have implicit series in my rolls: each "roll" I consider is actually rolling the dice until some relevant number appears. With that simplification I need only consider a finite number of "rolls", though they represent a potentially infinite number of rolls.
     
  9. May 18, 2008 #8

    uart

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    Yes I followed that. I also first simplified the problem by using the marginal probabilities (eg the probability of getting a seven given that either six, seven or eight was actually rolled). Even with this simplification you can still get an arbitrarily large sequence of events (each event being one or more rolls eventually resulting in a 6, 7 or 8) before you actually win. For example you could roll repeated sixes all day before eventually rolling an eight to “win”. I’ll post my working so you can see what I mean.

    As already mentioned start with the marginal probabilities and consider the subsequences of rolls eventually terminating in either a 6, 7 or 8 to be single events.

    There are now only three events with (marginal) probabilities s follows,

    Event A : Roll 6, probability a = 5/16

    Event B : Roll an 8, probably b = a = 5/16

    Event C : Roll a 7, probability c = 6/16

    Now consider the probability of winning (that is, getting both 6 and 8 before two 7’s) on exactly 1st then 2nd then 3rd …. events.

    On first event obviously P1 = 0
    On second event P2 = 2ab = 2a^2
    On third event P3 = a^2 b + b^2 a + 4abc = 2a^3 + 4a^2 c

    In general for n>=3 we have P = 2 a^n + 2(n-1) a^(n-1) c

    Since each of the above P’s are mutually exclusive we can add them to find the overall probability of winning. The series is easy enough to sum (I split it into a GP and a derivative of a GP) and I got the final result of

    [tex] P = \frac{2a^2}{1-a} \left( 1 + \frac{(2-a)c}{1-a} \right) [/tex]
     
    Last edited: May 19, 2008
  10. May 19, 2008 #9

    CRGreathouse

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    Yes, but once you've rolled a 6 or an 8 that number is dead (just like 2, 3, 4, 5, 9, 10, 11, 12) and can be ignored, putting you back to a finite number of 'rolls'.
     
  11. May 19, 2008 #10

    uart

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    Ok I see how you did it without the need for infinite series now. Thanks :)
     
  12. Jun 15, 2008 #11
    I made an error on this problem before; thanks for helping me find it and for showing a remarkably simpler solution.
     
  13. Mar 28, 2011 #12
    Ah, I don't think the answer posted really addresses the question. I'm not questioning the math, just the postulates.
    It's pretty obvious that the posted wants to know the odds of breaking even on craps place bets. WizardOfOdds answers this partially, but again doesn't fully examine the situation.

    Here's the reason for the question and the answer that is sought.
    In craps, if place-bets placed on 6 or 8, each for $6, those bets can pay out $7 when those numbers roll out. The bets are lost if 7 rolls at any time.
    It is pretty popular for gamblers to place $6 on the 6 and 8 once a point has been established, and to 'press' those bets once they have paid for themselves. That means doubling the bet for an even bigger payout.
    So, the answer sought of course is what is the probability of rolling at least 6,6 or 6,8 or 8,8 before rolling a 7. The initial bet has a value of $12 at risk which is lost of 7 rolls.
    Here are the situations:
    roll 7 -> $0
    roll 6 -> $12 risk + $7 bank, roll 7 -> $7
    roll 8 -> $12 risk + $7 bank, roll 7 -> $7
    roll 6 -> $12 risk + $7 bank, roll 6 -> $12 risk + $14 bank, roll 7 -> $14
    roll 6 -> $12 risk + $7 bank, roll 8 -> $12 risk + $14 bank, roll 7 -> $14
    roll 8 -> $12 risk + $7 bank, roll 6 -> $12 risk + $14 bank, roll 7 -> $14
    roll 8 -> $12 risk + $7 bank, roll 8 -> $12 risk + $14 bank, roll 7 -> $14
    etc...

    The overall odds of throwing 7 is 6/36 and the odds of 6 and 8 are 5/36, but because only 7,6,8 are interesting here, and there are 6 ways to make 7 and only 5 ways to make 6 or 8 there are a total of 16 combinations that are interesting with odds of
    6/16 for 7
    5/16 for 6
    5/16 for 8

    This means if we're not going to change the original $12 bet the odds of rolling either 6 or 8 on each roll is 10/16 against the 6/10 odds of rolling 7.
    So, we have a 6/10 chance of losing all $12.
    Then, we have a 10/16 chance of earning $7 on the first roll.
    (Notice I didn't say at least $7 because that scenario includes winning any amount above $7, and corresponds to any sequence of rolls of 6's and 8's.)
    The odds of rolling either 6 or 8 twice in a row is 10/16^2=25/64 or 39%. Of course that's the only way you can make more than $7 on the second roll and so the probability of making $14 is 39%.
    The expected amount you win is for the whole set of possibilities:

    0 x 6/10 +
    $7 x 10/16 +
    $7 x 10/16^2 +
    $7 x 10/16^3 etc...

    We're not counting the first $7 you win because that's taken care of by the 10/16 probability.
    The sum of 0 + q + q^2 + q^3 ... is (1/1-q)) so the sum of this is: $7 * (1/(1-10/16))
    which comes to: $2.66666

    Over time, counting the fact you will lose $12 every time (because we keep rolling until we lose them) you will lose $12-2.66 = $9.33

    Now that we've done the math for just the 6 & 8 place bets and leaving them on until we seven-out, what's the odds for place bets 5,6,8,9 all together ?

    Remember that 5 & 9 pay out $7, with a bet of $5, while 6 and 8 pay out $7 for a bet of $6. The answer uses the same method, since the payout is the same, but this time, the odds are
    6/24 for 7
    5/24 for 6
    5/24 for 8
    4/24 for 5
    4/24 for 9

    You can get the probability of winning $7 simply by 'not rolling 7, which is 18/24. ;-)

    Now, my question to people on this forum is this: What happens if we press bets after they're paid ?
    Pressing, means that once a place bet has paid (say the six rolled, and paid $7), we double the bet to $12 (2x$6) and pocket only $1, for the chance that next time it rolls we'll pocket $14 instead of $7.
    I'll only ask what happens if we press the bet once and leave it alone, but we have to consider doing this on any bet that pays, independently of others.
    If you feel like tackling the strategy where two winnings (say 6 and 8) get used to double a bet on the 6, knock yourself out. I can't figure it out.
     
  14. Nov 18, 2011 #13
    Correction:

    The sum of 0 + q + q^2 + q^3 ... is (1/1-q)) so the sum of this is: $7 * (1/(1-10/16))
    which comes to: $7*2.6666 = $18.666

    Over time, counting the fact you will lose $12 every time (because we keep rolling until we lose them) you will $18.66-$12 = $6.66

    Of course, this doesn't take into account the fact that when these place bets are available, you are waiting on someone to make the point. If the point is 6 or 8, the game ends prematurely.
     
    Last edited: Nov 18, 2011
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