Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dice, brassknuckles, and guitar

  1. Jan 23, 2009 #1
    Well, dice anyway....

    You make a sequence of rolls with a pair of dice. What is the probability that all six place numbers, 4-5-6-8-9-10, come up in any order before you roll a 7? In his book New Complete Guide to Gambling, John Scarne gave the calculation

    P = (24/30)(20/26)(16/22)(12/18)(8/14)(4/10) (roughly 13.66 to 1 odds)

    without comment as to why it was right. I have my doubts that it is exactly right, but I think it is pretty close. It looks to me like he decided that since each number comes up an average of 4 ways, you can just calculate the probability as if they all came up in exactly 4 ways. I don't think this is right, but it's awfully close to the answer I got using a different approximation: I calculated the probability of getting 4-5-6-8-9-10 in that order before 7

    P(4,5,6,8,9,10 in order before 7) = (3/30)(4/27)(5/23)(5/18)(4/13)(3/9)

    and assumed that all 6! = 720 such sequences have roughly the same probability. This gave me a total probability, expressed as odds, of about 14 to 1.

    Is there an easy way to get the exact answer that doesn't take pages and pages of arithmetic? Thanks.
  2. jcsd
  3. Jan 23, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    My best solution isn't elegant. It requires the calculation of 26 subproblems (well, 25 if you don't count "you rolled everything you need before rolling a 7, probability of success = 1"). That may be short of your "pages and pages of arithmetic", but I have to imagine it can be bettered by the wizards here.

    It also seem that you're both overestimating the probability. I have only ~12:1 odds (7937/92820) of getting 4-5-9-10 before a 7, but you also need 6 and 8.
    Last edited: Jan 23, 2009
  4. Jan 23, 2009 #3
    It's gotta be more elegant than what I was thinking. Out of the 720 sequences that would be successful, it's pretty easy to see that no more than 720/(2^3) = 90 of them have distinct probabilities. Working all those out would be pages and pages of arithmetic. It's quite possible that both Scarne and I are overestimating the real answer. Did my approach make any sense at all or is it completely off? :smile:
  5. Jan 23, 2009 #4


    User Avatar
    Science Advisor
    Homework Helper

    It didn't make sense at first, but then I compared your numbers to mine and saw where you were coming from.

    I did this. Let p(a, b, c) be the probability of succeeding when you haven't rolled a 7 yet and still need a of {6, 8}, b of {5, 9}, and c of {4, 10}. The probability you're trying to find is p(2, 2, 2).

    So p(2, 2, 2) = 10/36 * p(1, 2, 2) + 8/36 * p(2, 1, 2) + 6/36 * p(2, 2, 1) + 6/36 * p(2, 2, 2). If we agree to simplify at each step by rerolling results that don't matter (or we simplify algebraically, they come to the same thing) we get
    p(2, 2, 2) = 10/30 * p(1, 2, 2) + 8/30 * p(2, 1, 2) + 6/30 * p(2, 2, 1).
    Similarly, p(1, 2, 2) = 5/25 * p(0, 2, 2) + 8/25 * p(1, 1, 2) + 6/25 * p(1, 2, 1).

    If you continue far enough you'll get to p(0, 0, 0) which is obviously 1. You can then substitute back into the previous steps as needed to solve. So I got
    p(0, 0, 1) = 3/9 * p(0, 0, 0) = 1/3
    p(0, 0, 2) = 6/12 * p(0, 0, 1) = 1/6, etc.
  6. Jan 23, 2009 #5


    User Avatar
    Science Advisor
    Homework Helper

    I came up with my solution above for p(0, 2, 2) = 7937/92820 by hand, but I didn't want to do the remaining cases that way. I programmed the problem in Pari:
    Code (Text):
    Code (Text):
    time = 0 ms.
    %46 = 832156379/13385572200
    time = 0 ms.
    %47 = 0.06216815886286878345028836347
    So the answer to the original question is about 6.2% (16 to 1) by my calculations.
  7. Jan 23, 2009 #6
    Damn. That is slick. I get it.

    Yes, once a place number comes up, you can ignore it from then on, i.e. eliminate it from the outcome space.

    Awesome. I'll try to work through it to see if I get the same. Thank you! John Scarne was not a mathematician like Ed Thorp, but he was still a pretty smart fellow. His encyclopedic book on gambling is hard to beat, and it's very inexpensive. Incidentally, John Scarne was skeptical of Ed Thorp's system for beating blackjack and tried to arrange a challenge match with a casino for something like $100,000. Unfortunately, that was more than Thorp could afford at the time. Had he had the funds, Scarne might not have liked it.
  8. Jan 23, 2009 #7
    Doing it by hand with the help of a calculator, I got 7.68% or about 12:1. I'm going with your answer since your program is short and thus unlikely to have an error. I also thought of doing it using the Inclusion-Exclusion principle, but got an answer that is way off. I'm not sure yet if it is just an arithmetic error. The idea is to calculate Q = 1 - p(2,2,2), the probability that 7 comes up before all six place numbers come up (i.e. you're still lacking at least one place number when you get a 7). Shouldn't that work, at least in principle?
  9. Jan 25, 2009 #8
    Update: I got hold of a C compiler, brushed up on what little C I used to know, and programmed the Inclusion-Exclusion method for doing this problem. I got the same answer, 6.2%, that your recursive method got. Apparently I really suck at doing arithmetic on paper. Thanks again for the help, CRGreathouse!

    If anyone has other ways of doing the problem, feel free to show and tell. :biggrin:

    BTW, this information might be worth some money. There are probably a few high rollers out there who know for a fact that the correct odds are 13.66 : 1 because they read it in Scarne's book, and thus will take 15:1 on the money.
    Last edited: Jan 25, 2009
  10. Jan 25, 2009 #9


    User Avatar
    Science Advisor
    Homework Helper

    For those of us who know math better than gambling, what game does this apply to?
  11. Jan 26, 2009 #10
    It is just a proposition bet involving dice. It is derived from the dice game called craps. In craps the standard bet is as follows: You roll a pair of dice once, and if you get 7 or 11, you win immediately. If you get 2, 3, or 12, you lose immediately, which is known as "crapping out". If you roll 4, 5, 6, 8, 9, or 10, which are known as place numbers, you keep rolling until you either get the same number again (win) or roll a 7 (lose). Craps players can usually figure (or at least memorize) the odds of rolling a particular place number before a 7, but the idea of this proposition bet is that they will be unable to figure the odds of rolling all six before a 7. This proposition bet is most likely to be made privately between two gamblers. I doubt you could just walk up to a craps table in a casino and make this bet, unless perhaps you are a well known high roller and are willing to bet enormously large amounts. Also, the difference between the correct answer, about 16:1, and Scarne's published answer of 13.66:1, is not huge. You would have to put large amounts of money at risk to be expected to win anything significant, even if you found an 'informed' gambler who was willing to take only 14:1 payout. What a dice hustler hopes is that he will find someone who grossly underestimates the probability of rolling all six place numbers before a 7 and is willing to lay 40:1 or 50:1 odds against it happening. Then the hustler will gladly take those odds and clean up without much risk. John Scarne gives just such a story in his book, although it does not give the names of the people involved and therefore may be apocryphal.
    Last edited: Jan 26, 2009
  12. Jan 26, 2009 #11
    Let me say that Scarne thought 2x odds in dice, was all/more than a casino could afford, since their advantage was so reduced. Yet, we have a casino in Las Vegas that offers 20x odds and has done so for a long time, while 10x odds are not that uncommon.

    In fact, p93 talks about "Free Double Odds Bets," and says "Banks that permit such action woud not stay in business long if all the players made only that type of bet because a casino operation doing fair business requires, on the average, a greater percentage than that on all bets in order to pay its operating expenses before showing a profit."

    Also, I might add here that the type of gentleman gambler that wears suits and ties will almost never play a field bet. BUT most layouts today allow double on the 2 and triple on the 12. (Scarne figures , in many cases, on double on the 2 and 12 giving the house a 5 5/19% advantage.) But with triple on the 12, the house advantage is only 1 part in 36; this results in odds that are better than found in roulette. (I assume high class bettors actually read books on gambling and follow their advice.)
    Last edited: Jan 26, 2009
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook