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 Subject: Probability of 3-digit numbers Category: Science > Math Asked by: davidj-ga List Price: \$6.00 Posted: 08 Feb 2004 15:43 PST Expires: 09 Mar 2004 15:43 PST Question ID: 304812
 ```I need to know the probability of any 3-digit number randomly matching with an established 3-digit number outcome. I'll explain: If any number in the random sequence matches any number in the outcome sequence, but NOT in the same placeholder spot, it gets an X score. If a number in the random sequence matches a number in the outcome sequence IN THE SAME placeholder spot, it gets a P score. Examples: Random Outcome Score 130 242 - - - 829 950 X - - 340 741 P - - 973 396 X X - 542 522 P P - 014 401 X X X 952 259 P X X 179 190 P X - 568 568 P P P I need the probability (listed like .3566 or .1000) for all possible scoring options (10 i think!), which would be: - - - X - - X X - X X X P - - P X - P X X P P - P P X P P P Please provide the formula or method used to calculate these. Thank you!``` Request for Question Clarification by mathtalk-ga on 08 Feb 2004 16:58 PST ```Hi, davidj-ga: These probabilities actually depend on the number of repeated digits in the "established 3-digit number outcome". In other words the probability of the various X & P combinations is different when the number of repeated digits allowed in the outcome changes. I notice that some of your outcomes have these and some do not. One can "integerate over" these variations if a distribution for the outcomes is assumed. For example, one might assume that all 3-digit outcomes are equally likely, just as presumably you mean for all "random" 3-digit numbers to be considered equally likely. regards, mathtalk-ga``` Clarification of Question by davidj-ga on 08 Feb 2004 17:53 PST ```Hi, All these calculations are just for ONE guess (one random number) for ONE outcome only. Like right now, if you guessed "427" and the actual answer was "734", then what was the probability that you got that score of X X - ? In other words, what were the chances of you selecting 2 out of those 3 numbers right, but both are in the wrong placeholder. Hope that helps. Thanks.``` Clarification of Question by davidj-ga on 08 Feb 2004 17:54 PST ```To clarify also: The random guess CANNOT have repeating digits. The actual outcome CAN have repeating digits.``` Clarification of Question by davidj-ga on 08 Feb 2004 17:58 PST ```Hi mathtalk, For the question purposes, you are right, we *can* assume that all 3-digit outcomes are equally likely, and can integrate over these variations. That would be find to assume that distribution for calculation purposes. Thanks.```
 Subject: Re: Probability of 3-digit numbers Answered By: mathtalk-ga on 08 Feb 2004 21:00 PST Rated:
 ```Hi, davidj-ga: The simplest approach (to my devious mind) is to turn the problem upside down. Since you've agreed to treat all the "outcomes" as having a uniform distribution, and to find the expected probability of each "random guess" giving certain results (subject to the restriction that the random guess has no repeats), it makes sense to treat the "random guess" as if it were any fixed 3-digit number without repetition and do the probabilities based on the "random selection" of the outcome. Specifically let the "guess" be 123 (or any other three-digit number with distinct digits). Then the "outcome" has three digits that are randomly and independently selected (allowing in particular repeats to occur with natural frequencies). For each digit of the outcome, there is a 0.1 chance of getting a P, a 0.2 chance of getting an X, and a 0.7 chance of getting a _. Combining the results is now a simple "multinomial" probability, to wit: Pr( - - - ) = (0.7)^3 = 0.343 Pr( X - - ) = 3 * (0.2) * (0.7)^2 = 0.294 Pr( X X - ) = 3 * (0.2)^2 * (0.7) = 0.084 Pr( X X X ) = (0.2)^3 = 0.008 Pr( P - - ) = 3 * (0.1) * (0.7)^2 = 0.147 Pr( P X - ) = 6 * (0.1)(0.2)(0.7) = 0.084 Pr( P X X ) = 3 * (0.1) * (0.2)^2 = 0.012 Pr( P P - ) = 3 * (0.1)^2 * (0.7) = 0.021 Pr( P P X ) = 3 * (0.1)^2 * (0.2) = 0.006 Pr( P P P ) = (0.1)^3 = 0.001 ------ 1.000 The fact that these all add up to probability 1 suggests that at the least we have likely not made any arithmetic errors! But have we understood the problem correctly? Here's a way to remember how a multinomial calculation works. The three digits of the outcome each provide a P, an X, or _, and their selection is by independent, identical uniform distributions of digits (1 in 10). Therefore for each single digit of the outcome: p = Pr( P ) = 0.1 x = Pr( X ) = 0.2 t = Pr( _ ) = 0.7 and these single digit "outcomes" are mutually exclusive and encompassing, so: p + x + t = 1.0 The three digits of outcome are then a repeated series of trials for these one digit probabilities, a fancy name for which is "Bernoulli process". In any case it clearly also holds that: (p + x + t)^3 = 1.0 and this is one reason we refer to the calculation as "multinomial". Expanding the left hand side above gives us terms with factor p^i * x^j * t^k, where the exponents i + j + k add up to 3. Much like a binomial expansion, we can express the formula's constant coefficients in terms of factorials: (p + x + t)^3 = SUM (3!/(i! j! k!)) p^i x^j t^k where the sum is over all possible ways for nonnegative exponents i,j,k to add up to 3. Basically the "multinomial" coefficient 3!/(i! j! k!) simply counts the number of arrangements in which P can occur i times, X can occur j times, and _ can occur k times. Here the permutations can done mentally, as if all the symbols are the same there is only 1 arrangement, if all are different there are 6 arrangements (permutations of three distinct things = 3!), and otherwise there are 3 possible arrangements (because we have two symbols alike and one distinct). Please let me know whether this "devious" approach to computing the various probabilities is convincing or not. If it does not convince you, then I will break it down from the other side, i.e. what if all the outcome digits are distinct, what if two are alike and one is different, what if all three are the same. That way will give the same answers but it necessarily involves a number of "sub-cases". regards, mathtalk-ga``` Request for Answer Clarification by davidj-ga on 09 Feb 2004 07:47 PST ```Thank you for the detailed "devious" answer. That makes sense to me - however, as you offer, I would appreciate seeing the results from the "sub-classes" as well. Thanks!``` Clarification of Answer by mathtalk-ga on 09 Feb 2004 17:11 PST ```Hi, davidj-ga: Glad to oblige, but let me ask you to clarify a point about scoring involving repetitions in the "outcome". You listed as one of the examples: Guess Outcome Score 542 522 P P - Clearly the first digit 5 counts for a P, but I could see either counting the third digit 2 as one correct guess (giving it P) or as two correct guesses (both as a P and as an X) by virtue of the repetition. What rule do you intend in these cases? To take an extreme example, suppose the outcome had three of the same digit and the guess includes that digit. Would that count as P _ _ or as P X X ? thanks, mathtalk-ga``` Clarification of Answer by mathtalk-ga on 09 Feb 2004 22:30 PST ```In the interest of expediency and completeness, having seen your new Question, let me go ahead and provide further breakdowns here. First the framework. There are three "outcome" types we distinguish: A := all three digits of outcome number are different B := two digits of the outcome number are the same, one different C := all three digits of outcome number are the same A priori the chances of these (mutually exclusive) outcomes are: Pr( A ) = (10*9*8)/1000 = 0.720 Pr( B ) = (10*9*3)/1000 = 0.270 Pr( C ) = 10/1000 = 0.010 For example, in A one has 10 choices for the first digit, 9 for the second digit, and 8 for the third digit (as they must all be distinct). In B we can choose the "doubled" digit 10 ways, the "single" digit 9 ways, and the position of the single digit (out of three) in 3 ways. Finally for C we can choose a tripled digit in 10 ways. Of course these choices added up to the "universe" of 1000 three-digit numbers. As before there are "on paper" ten different possible scores, but depending on the type of outcome and the rule for scoring with matched repetitions, some of these scores are impossible. E.g. your observation in the other thread that if each guess digit contributes at most once to the score, then PPX would not be possible. So let's breakout "conditional probabilities" first for that scoring rule, ie. each digit of the guess counts at most once: - as P if it matches the outcome digit in the same position - as X if it matches an outcome digit but not the one in its position - as _ if it does not match any outcome digit We then have the following table of conditional probabilities for combined scores, where outcome event O is one of A,B,C as displayed in these columns: O = A O = B O = C Pr( _ _ _ | O ) 210/720 336/720 504/720 Pr( X _ _ | O ) 252/720 168/720 0 Pr( X X _ | O ) 63/720 16/720 0 Pr( X X X | O ) 2/720 0 0 Pr( P _ _ | O ) 126/720 168/720 216/720 Pr( P X _ | O ) 42/720 16/720 0 Pr( P X X | O ) 3/720 0 0 Pr( P P _ | O ) 21/720 16/720 0 Pr( P P X | O ) 0 0 0 Pr( P P P | O ) 1/720 0 0 A quick check on our arithmetic is to verify that each column adds to 1, and I've done that. It is worthwhile going through the computation of each entry in detail, but first let's apply the results above, using the definition of conditional probability, to find the absolute probability of each score. This formula is: Pr( Score ) = Pr( Score | A ) * Pr( A ) + Pr( Score | B ) * Pr( B ) + Pr( Score | C ) * Pr( C ) because the events A,B,C "partition" the possible outcomes. Plugging in the values for Pr( A ), Pr( B ), Pr( C ) obtained earlier: Pr( _ _ _ ) = 0.210 + 0.126 + 0.007 = 0.343 Pr( X _ _ ) = 0.252 + 0.063 + 0.000 = 0.315 Pr( X X _ ) = 0.063 + 0.006 + 0.000 = 0.069 Pr( X X X ) = 0.002 + 0.000 + 0.000 = 0.002 Pr( P _ _ ) = 0.126 + 0.063 + 0.003 = 0.192 Pr( P X _ ) = 0.042 + 0.006 + 0.000 = 0.048 Pr( P X X ) = 0.003 + 0.000 + 0.000 = 0.003 Pr( P P _ ) = 0.021 + 0.006 + 0.000 = 0.027 Pr( P P X ) = 0.000 + 0.000 + 0.000 = 0.000 Pr( P P P ) = 0.001 + 0.000 + 0.000 = 0.001 ------ 1.000 Now let's go through the columns of conditional probabilities in detail. In each case the "universe" of possibilities is the choice of three distinct digits for the "guess", so we wind up with a common denominator of: 10 * 9 * 8 = 720 that I did not try to reduce, knowing that the decimal answers would work out cleanly in the end. Let's go through the first column, conditional on A, having all three digits of the outcome distinct, esp. since this part of calculation is valid for either scoring rule: Pr( _ _ _ | A ) = (7 * 6 * 5)/720 since we have 7 choices for the first digit of the guess, to avoid all three digits of the outcome, then 6 choices for the second digit (since it must also differ from the first digit of the guess), times 5 choices for the final digit of the guess. Pr( X _ _ | A ) = (3*2*7*6)/720 because we can pick which of the outcome digits to match in 3 ways, and the position in which to put that digit in 2 ways, times 7 ways to pick the next digit of the guess and 6 ways to pick the final digit of the guess (to avoid repetition and further matches). Pr( X X _ | A ) = (3 * 3 * 7)/720 since we have 3 choices for which digit of the outcome will not be matched, times 3 positions for the digit of the guess that does not match anything, times 7 choices for the value of that unmatching guess digit (the positions of the matching guess digits are then forced to avoid a "position" match). Pr( X X X | A ) = 2/720 because all three outcome digits are matched but none in its own position; there are only 2 such "derangements" (rotate the three digits once to the left or once to the right). Pr( P _ _ | A ) = (3 * 7 * 6)/720 since there are 3 choices for the position match and then 7 * 6 ways to choose the remaining unmatched guess digits. Pr( P X _ | A ) = (3 * 2 * 7)/720 since there are 3 choices for the position match, times 2 choices for other "non-position" match, times 7 choices for the remaining unmatching guess digit. Pr( P P _ | A ) = (3 * 7)/720 because we can choose the unmatching position in 3 ways and unmatching guess digit in that position in 7 ways. Pr( P P X | A ) = 0 because as you pointed out, it's not possible to have an "out of position" match if the other two positions do match (assuming the three guess digits are all distinct). Pr( P P P | A ) = 1/720 because there is only one choice for each position. Next we consider the conditional probabilities given B, having two digits in the outcome which are equal and a third digit which is distinct. Pr( _ _ _ | B ) = (8 * 7 * 6)/720 since it is only necessary to avoid the two digits that appear in the outcome. Pr( X _ _ | B ) = (3 * 8 * 7)/720 since we have 3 choices for the location of the (non-positional) match, times 8 * 7 ways to fill the other two guess digits without matching the outcome (if the location of the non-positional match is chosen, the value for that guess digit is forced). Pr( X X _ | B ) = (2 * 8)/720 because we have a choice of 2 locations (the positions of the equal outcome digits) for our unmatched guess digit, times 8 possible values for that guess digit (the other two guess digits are forced). Pr( X X X | B ) = 0 because we can't have three matches with only two distinct digits in the outcome. Pr( P _ _ | B ) = (3 * 8 * 7)/720 since again we can choose the position of the (positional) match in 3 ways, times 8 * 7 ways to fill the other two guess digits without matching (all the values except the two that appear in the outcome are available to choose from). Pr( P X _ | B ) = (2 * 8)/720 because we must choose one of the 2 locations in which the outcome digits are equal to put a positional match (we cannot have a non-positional match there if the outcome's "odd" digit has a positional match) and then we have 8 choices for the unmatched guess digit in the other one of those locations (the value for the guess digit in the outcome's "odd" digit position is forced). Pr( P X X | B ) = 0 because again it's not possible to have three matches. Pr( P P _ | B ) = (2 * 8)/720 because one of the 2 locations where the outcome has equal digits must be unmatched (we cannot match both as the guess digits are distinct) and that position can be filled 8 ways. Pr( P P X | B ) = 0 because three matches are not possible. Pr( P P P | B ) = 0 because three matches are not possible. To complete these conditional probabilities it remains only to consider the outcomes C in which all three digits are equal. One quickly sees that under the scoring rules considered here, either there are no matches or there is exactly one match and that one is positional: Pr( _ _ _ | C ) = (9 * 8 * 7)/720 since there are 9 choices for the first guess digit (that avoid the triple digit in the outcome), 8 choices for the second guess digit, and 7 choices for the third one. Pr( P _ _ | C ) = (3 * 9 * 8)/720 since there are 3 positions for the matching digit (which necessarily matches also in position) and 9 * 8 choices for the remaining unmatched digits. All the other conditional probabilities involving outcomes C are zero, as it is not possible to have two matches or to have a match that does not agree in position. This completes the calculation of the scoring probabilities with the interpretation that you evidently intended (at most one "score" per each digit of the guess). To see that the numbers I originally provided are consistent with this interpretation of scoring: - a P for each guess digit matching an outcome digit in its position, and - an X for each guess digit matching an outcome digit not in its position we need only revise the conditional probabilities in the columns for B and C, because the scoring changes only affect outcomes with repetitions. For the column under B the basic differences are that any positional match associated with a repeated digit gives and X as well as a P, and also opens up a number of ways to obtain three X's (which was not possible given B under the scoring system you intended). For the column under C there are still only two possible scores, but these are now "_ _ _" and "P X X" because a positional match in the latter case also draws in two nonpositional matches. If there is interest I can supply the gory details, but the final numbers would revert to those I originally posted. Sorry for the confusion. regards, mathtalk-ga```
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