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Solutions: consecutive dice

December 30, 2008 pm31 6:57 pm

You hear someone in the next room roll two dice.

“Look at the two numbers. Do you see an odd number?”

“Yes”

Now, what is the probability that he rolled two consecutive numbers? For example, 3-2 or 5-6

Please use the comments section below for posting solutions.

For discussion, clarification, etc, click over to the problem post.

Clarifying Information

You can’t see the dice.

“Do you see an odd number” refers to the face of at least one die, not the sum.

Credit

This is nothing new. But a discussion at If-Then Knots put it in my mind. Then when the Carnival of Mathematics #46 at Walking Randomly linked to Tanya Khovanova’s blog, I visited, poked around, and found this related puzzle with explanation.

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16 Comments leave one →
  1. David P permalink
    December 30, 2008 pm31 8:30 pm 8:30 pm

    (Disregarding Order)

    1-2, 2-3, 3-4, 4-5, 5-6
    Possible “Good” Outcomes: 5

    1-1, 1-3, 1-4, 1-5, 1-6
    3-3, 3-5, 3-6
    5-2, 5-5
    Possible “Bad” Outcomes: 10

    So, probability that consecutives were rolled is 1/3?

  2. December 30, 2008 pm31 9:04 pm 9:04 pm

    With dice probability puzzles, you cannot disregard order. There are two ways to get 2-3, but only one way to get a double number like 3-3. If you ignore ordering, you end up undercounting your options.

    My “obvious” but wrong answer was to let the first die be 1, 3, or 5 and to let the other range from 1-6. That gave 18 possible outcomes, of which 5 were consecutive.

    After reading the related puzzle, I realized my error. (In my defense, I didn’t have to read the solution to the related puzzle. I have learned something over the years!) I made a chart of all the possible rolls of dice, blocked out the ones that couldn’t be, and checked off the winners. Pretty pattern!

    My answer: 10/27.

  3. Prion permalink
    December 30, 2008 pm31 9:11 pm 9:11 pm

    P(consec/odd)=(P(odd|consec)P(consec)*P(consec))/P(odd)

    P(odd|consec)=0.5, because if the numbers are consecutives one is even and the other odd.

    P(consec)=10/36

    P(odd)=0.5

    So, I believe P(consec|odd) is 10/36. Is it?

  4. December 30, 2008 pm31 9:20 pm 9:20 pm

    There are 36 equally likely outcomes from rolling two dice. Knowing there is at least one odd number eliminates 9 possible rolls, since if both dice are even, there are 3 choices for the first die, and 3 for the second die.

    Of those 27 remaining rolls… if the first number rolled is a 1 or a 6, there is only one possible second die roll that gives a consecutive pair (1-2 or 6-5). If the first number rolled is a 2,3,4, or 5, in each of those cases there are two possible second die rolls that yield a consecutive pair. Thus there are 1+2+2+2+2+1 = 10 equally likely ways to get a consecutive pair, out of 27 equally likely ways to roll a pair of dice with at least one odd number showing.

    I get 10/27 too.

  5. December 30, 2008 pm31 9:24 pm 9:24 pm

    Prion: You’re on the right track, although Bayes’ Theorem would say that P(consec|odd) = P(odd|consec)*P(consec) / P(odd)

    However, I disagree with your P(odd|consec) and with your P(odd). (I would assign those the values 1 and 27/36, respectively.)

  6. Rachel permalink
    December 31, 2008 am31 12:10 am 12:10 am

    For clarity, I’m going to assume I have a blue die and a red die and list outcomes as (blue, red).

    There are 36 possible outcomes:

    9 where the blue die is odd and the red die is odd
    9 where the blue die is odd and the red die is even
    –> (1,2), (3,2), (3,4), (5,4) and (5,6) satisfy the consecutive condition
    9 where the blue die is even and the red die is odd
    –> (2,1), (2,3), (4,3), (4,5) and (6,5) satisfy the consecutive condition
    9 where the blue die is even and the red die is even

    So I’m coming up with 10/27 as well.

    Looking at Bayes theorem…

    P(consec|odd) = P(odd|consec)*P(consec) / P(odd)
    = 1 * 10/36 / (27/36)
    = 10/27

  7. December 31, 2008 am31 1:17 am 1:17 am

    Jonathan,
    As always, a wonderful problem! I regarded it as a straight conditional probability problem: P(consecutive|at least one die is odd). I reasoned that there are 27 outcomes in which at least one die is odd (36-9). There are 5 “combinations” involving consecutive dice but we need to double that to account for the different arrangements since order counts here (using colored dice as Rachel did is a nice way to drive this point home). Thus 10/27…
    Great way to end the year and kick off a new one! Happy New Year!
    Dave

  8. Prion permalink
    December 31, 2008 am31 1:49 am 1:49 am

    Ok, but could you explain why?

  9. December 31, 2008 am31 8:58 am 8:58 am

    No one even considers the probability of hearing “Look at the two numbers. Do you see an odd number?”, given that two particular dice were rolled. How sad.

  10. December 31, 2008 am31 9:44 am 9:44 am

    Tanya,

    The report that you receive contains information. As you suggest, it is also possible to have meta-information, information about the report itself, such as the probability of receiving the report. However, the probability that you receive the report is inscrutable unless it is explicitly stated in the set-up of the problem. The link that Jonathan provided to my blog, if-then knots, gives a problem that does give enough meta-information to scrute a probability that one receives a certain report. For that problem, I show that if one updates only on the contents of the report and not on the full evidence, which includes the likelihood of receiving the report (i.e., the meta-information), then one gets an incorrect result. For the class of problems in which the likelihood of receiving the report is inscrutable, however, one should update only on the information contained in the report itself. The comment thread at my post may be clarifying, though I’m a bit embarrassed at how long it took me to get clear.

  11. December 31, 2008 am31 9:48 am 9:48 am

    Tanya,

    I think that Dave did.

    More interesting, no one so far examined the alternate: what happens what is the probability that we see no odd numbers, and what that implies.

  12. Rachel permalink
    December 31, 2008 am31 11:47 am 11:47 am

    The probability that we see no odd numbers is 1/4.

    It implies that the numbers on the dice are not consecutive. Also that their sum is even.

    Were you looking for something subtler than that?

  13. December 31, 2008 pm31 8:32 pm 8:32 pm

    I would like to make three different puzzles from your puzzle depending on the strategy of the first speaker.

    Strategy 1. For any outcome, the first speaker says, “Look at the two numbers. Do you see an odd number?”

    Strategy 2. If there is one odd number he says, “Look at the two numbers. Do you see an odd number?”. If there are two odd numbers he says, “Look at the two numbers. Do you see two odd numbers?”

    Strategy 3. If there is five, he says, “Look at the two numbers. Do you see an odd number?”, otherwise he says something else.

  14. January 1, 2009 am31 12:28 am 12:28 am

    And here’s yet another strategy:

    Strategy 4: If the numbers are consecutive then he says “Look at the two numbers. Do you see an odd number?” Otherwise he says nothing at all.

    If you know that the original speaker is dutifully following strategy 4, then the conditional probability of interest given that he said what he said is 1! The point, however, that must be attended to is that if you don’t know anything about any strategies then you have to update only on the content of the report and should not make any assumptions about likelihoods of having received the report. It is crucially important to distinguish between information in a report and information about a report. In the puzzle Jonathan posted there was no information about the strategies of the speakers in the neighboring room. One must, therefor, update only on the information in the report and not make any assumptions regarding information about the report.

  15. Mohammed Goldberg permalink
    January 10, 2009 am31 1:12 am 1:12 am

    5 out of 30

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