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Got drilled playing craps

  1. May 24, 2006 #1
    I was playing dice at a bar tonight, and after a while one of the guys I was playing asks "What do you think are the odds against rolling all the place numbers (4, 5, 6, 8, 9, 10) before rolling a seven?" So I'm thinking, just rolling a 4 before a 7 is a 2-to-1 shot, so rolling all six numbers first has to be something like a 50-to-1 shot. Well, this guy says I don't have a clue about figuring odds, and we go back and forth about it a little bit, and pretty soon I end up agreeing to lay 40-to-1 against on this very proposition. We were betting a buck a turn, and after about three hours I'm stuck $200.:eek: I guess my figuring was off, but by how much? How do you figure the correct odds?
     
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  3. May 24, 2006 #2

    mathwonk

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    as damon runyan wrote in guys and dolls: someday a guy will come up with a brand new deck of cards on which the seal is never broken and offer to bet you $50 that the jack of hearts will jump out of the deck and squirt cider in your ear, but do not bet him my friend, because if you do, you are going to get an ear full of cider.
     
  4. May 24, 2006 #3

    mathwonk

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    i do not know how to figure the exact odds but i estimate you were overpaying by at least 3 to one at 40/1 odds. Think about it: although the odds of throwing a 4 before 7 are only 1/2, he did not have to throw a 4 at first to win, he only had to throw one of the 6 numbers 4,5,6,8,9,10, and the odds of dsoijng that were 24/30 or 4/5.

    Then suppose he threw a 6. After that he only had to throw anyone of the numbers 4,5,8,9,10 before a 7, and the odds of that were still 19/25, or almost 4/5 again.....so even if he had to throw the numbers in a spoecific order, like say 6,5,4,8,9,10, the oidds against doing it are only about 14/1 against. and you were paying 40/1.

    notice he asked you what you thoiught the odds were before agreeing to the bet. if you had gotten closer to the correct odds he would not have played. drinking and gambling is a mistake. in fact gambling at all is a mistake.
     
    Last edited: May 24, 2006
  5. May 24, 2006 #4

    mathwonk

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    or just note that you lost say 7 bets in 3 hours and won say 80 bets. so your advantage was really about 11-12 to one, not 40 to 1.

    and at that rate, the fact it took you 3 hours to stop tells you again that, like the rest of us, you have no business gambling.

    next time please make a donation to physics forums instead for the valuable advice available free here.
     
    Last edited: May 24, 2006
  6. May 24, 2006 #5
    I can't tell you how to calculate the exact odds, but I think that you screwed yourself even if the guy was using honest dice.

    I figure it this way. On any given roll the probability of rolling a 7 is 1/6, the probability of any one of {4,5,6,8,9,10} is 4/6, and the probability of any one of {2,3,11,12} is 1/6. So the probability of 6 rolls in a row without a 7 is [itex](frac{5}{6})^6[/itex] which is approx .335, or 2 to 1 against. If this occurs, 4/5 of those 6 rolls are numbers you want. Of course, you won't get your 6 place numbers all in a row -- there may be duplications and there may be some of {2,3,11,12} -- but if that happens the game goes on.

    So what's the probability of
    12 rolls in a row without a 7: [itex](\frac 5 6)^{12} \sim .112[/itex] or about 8 to 1 against
    18 rolls in a row without a 7: [itex](\frac{5}{6})^{18} \sim .037[/itex] or about 26 to 1 against

    It seems pretty likely that in 12 to 18 rolls you would get your 6 place numbers, considering that their individual probabilities range from 1/6 to 5/36 each.

    You gave the guy 40 to 1, which, if I'm correct is roughly the odds against 20 rolls without a 7.
     
    Last edited by a moderator: May 27, 2006
  7. May 24, 2006 #6
    Well, apparently I didn't do the latex the way the board wants it, and the board isn't letting me edit it either.

    Those numbers were supposed to be:
    (5/6)^6
    (5/6)^12
    (5/6)^18

    Moderators: why can't I edit my posts? "Preview post" doesn't seem to generate latex graphicsl, and "edit post" doesn't work at all, so if we don't guess correctly the first time we're sunk?
     
  8. May 26, 2006 #7
    Thanks for those calculations, mathwonk and gnome. Wow, 14 to 1 huh? I would have thought the odds against throwing the numbers in a specific order would have been lot higher, like 10000 to 1 or something. I just thought the guy was having an amazing streak of luck. Maybe I don't have any business gambling. But some people apparently do. The guy who hustled me at craps seems to do okay at it:biggrin: Even I've done okay at gambling small on games of skill, like pool and chess. If it weren't for peoples fascination with gambling, mathematicians might not have come up with the theory of probability! I mean seriously, how many people do you know that find drawing different colored marbles out of a bag interesting--something worth studying? I sure don't. But finding dice odds or roulette odds, now that's applied math. Of course it would be nice if I were better at it....
     
    Last edited: May 26, 2006
  9. May 26, 2006 #8

    Hurkyl

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    Odds are (ha ha) that they guy already knew the "right" odds. It was kind of him to talk you down from a 50:1 payoff, though!

    He not only had to roll them before 7, but in the right order too?

    My mental calculation went like this:

    The odds are 6:3 against that he gets a 4 before a 7.
    The odds are 6:4 against that he gets a 5 before a 7.
    The odds are 6:5 against that he gets a 6 before a 7.
    The odds are 6:5 against that he gets a 8 before a 7.
    The odds are 6:4 against that he gets a 9 before a 7.
    The odds are 6:3 against that he gets a 10 before a 7.

    So, I just need to multiply all of those to get the odds that he wins. But to estimate: the product of three is
    6*6*6 : 3*4*5 = 216 : 60
    That's about 7:2. The other three is also 7:2, which works out to 49:4, or about 12:1 against.


    Of course, you can't multiply odds in this way: you should be multiplying probabilities. But this probably isn't really far off, and it's what I thought of first.


    Roulette odds are boring. On average, you lose 2 chips out of every 38, no matter what you do, unless you make the 5 number bet in which case you lose 3. :tongue:

    (Americal roulette, of course)
     
    Last edited: May 26, 2006
  10. May 26, 2006 #9

    shmoe

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    I worked out the probability of getting a 4,5,6,8,9, and 10 (in no particular order) before a 7 to be 832156379/13385572200~.06216815... or about 16.085.. to 1. It was pretty straightforward, find the probability of the other guy winning on the nth roll then sum from n=1 to infinity (this probability is zero for n<6 of course). To find the probability of a win on the nth roll, break into 6 cases the last roll being a 4, 5, 6, 8, 9 or 10 (some symmetry here if you finish this off by hand of course), then use inclusion/exclusion to find the probability that the previous n-1 rolls contain the other 5 numbers and no 7.

    To test, I also ran a monte carlo simulation of 10 million runs, getting 6,208,111 wins for the other guy, so pretty close to the above. (always the possibility that I messed both up though, but this makes me much more confidant).

    I'm not sure how to interpret rolling these numbers in 'a specific order', there seems to be a few options here. It seems like the probability will be very low if you wanted an uninterupted sequence of 4, 5, 6, 8, 9, 10... meaning rolling 4, 5, 6, 8, 9, 2, 10 wouldn't count for example. How was the game played? Would this second sequence count as a win for him?
     
  11. May 26, 2006 #10

    mathwonk

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    i learned most of what i know of probablility from john scarne, in scarne on dice. he tells an interesting story in one of his books, of a guy who owned a gambling emporium in vegas and made 3 million dollars over his career from the suckers who gambled at his place, then retired and had nothing to do so started gambling himself.

    it only took him about 3 years to lose all he had gained in his whole life of running a casino.

    don't be a sucker. take a hint.
     
  12. May 26, 2006 #11

    mathwonk

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    if the guy who cheated you was doing well, wehy was he hanging out in bars at night? instead of home on his yacht. he was as big a sucker as you.l
     
  13. May 26, 2006 #12

    mathwonk

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    forgive me for being rude, but you are an intelligent guy longing to become an imbecile.
     
  14. May 27, 2006 #13

    The way we were playing it, the numbers did not have to come in any particular order. It's okay to roll numbers other than the six place numbers or seven, and also okay to repeat numbers that have already come up. So 4, 5, 6, 8, 9, 2, 10 would count. So would 2, 9, 5, 6, 8, 8, 8, 6, 4, 12, 10. He just has to make each of those six numbers appear at least once before a seven is rolled. I like your idea of doing a computer simulation. That's probably the most effective way to get the odds right.

    As Hurkyl pointed out, the guy probably had a pretty good idea of what the real odds were....at least close enough to know when he has a big edge. How he figured it out or where he learned it from, I wish I knew.
     
  15. May 27, 2006 #14

    shmoe

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    What you've described is what I had assumed for my calculation.

    The most effective way to get the correct odds is to work them out exactly! The simulation was just as a check to make sure I wasn't horribly off. Typcially a simulation is simpler to code but can take a long time to run to get a healthly number of trials. It's handy in a pinch if you can't figure out how to work out the probablities exactly, but no substitute for being able to do that kind of calculation.
     
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