First, I am conducting a very short survey into sleep as part of the Edinburgh International Science Festival and would love you to take part.  The link is here.  Thank you for that – much appreciated.

OK, to the puzzle.  I want to play a game with you.  We have three unbiased dice, and each of the two of us have a target number.  My target number is 9 and your number is 10.  On each round of the game we throw all three dice and add up the numbers.  If I get my target number I get 1 point, and if you get your number you get a point.  The winner is the person with the most number of points after 20 rounds.  There are 6 ways of getting 9 (1,2,6:  1,3, 5:  1,4,4:  2,2,5:  2,3, 4:  3,3,3) and 6 ways of getting 10 (1,3,6: 1,4,5: 2,2,6: 2,3,5: 2,4,4,: 3,3,4) and so it seems to me that it is a fair game.  Want to play?

As ever, please do NOT post your answers, but do say if you think you have solved the puzzle and how long it took. Solution on Monday.

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available for theKindle (UK here and USA here) and on the iBookstore (UK here in the USAhere). You can try 101 of the puzzles for free here.

85 comments

  1. Easy, and almost instantaneous. There is a fundamentally obvious fact which immediately splurged itself from the wording of the puzzle directly into the corner of my brain which deals with probability. Clue: think about a large number of monkeys trying to type Shakespeare.

  2. If the question is do I want to play, the obvious answer is yes. But in the World of Wiseman, nothing is as it seems.

    So I looked at the number combinations and changed my answer. Yes if I get 9 and you get 10.

  3. I’m glad to hear you think it’s a fair game. I’d be happy to play it with you.

    But I wouldn’t want to play it if we switched target numbers.

    1. I totally agree with you.
      But I knew this one already so it was quite fast for me…

      I did this one to my pupils last year and it was really fun ^^
      In France it’s called “Le paradoxe du Duc de Toscane” which means “Duke of Toscane’s Paradox” and I like this name 🙂

  4. Now, I wonder, how will Dr. Wiseman post the solution to this problem on Friday? Is there even a solution in the first place, given the way the question is framed? One has to make assumptions and such… meh.

  5. Hah! I knew that years of RPGs would pay off some day. My answer was “No” before I had finished reading the question. Finding an explanation that would convince somebody who doesn’t roll dice a few hours each week took about 30 seconds.

  6. I’m really curious about the split in answers. Has Richard’s habit of posing trick questions finally confused people into over thinking this?

    1. I think some people understand the nature of the problem, but are declaring their answer without reading the question fully.

    1. Depends on how you think of the dice ?

      So if you play this game and think about the dice in a certain way, you can prevent 10s from appearing 8% more often than 9s ?

      Are you talking about prayer ?

    1. Are you a member of the Petropolitan Molice? (Kettlers of Satan). All right, I’ll keep it to myself.

  7. I would play for small stakes.

    But what if we introduce a twist. Before each round, the players take turns deciding how many dice to use that round. Anyone want to play against Richard?

  8. I’d play, mostly so that I could have the opportunity to chat with you while we play.

    Do I think it is a FAIR game? No, but that wouldn’t be my reason for playing/not playing.

  9. Sure I’d play game, but – as with others – only for small stakes. Give it 100 turns and I’d bet large, 1,000 turns I’d bet the farm.
    10 seconds, then 3 mins back-of-envelope checking.

    We know the dice must be six-sided with the sides having 1 through 6 dots from the list of combinations he provides. But his use of “ways”…

  10. No. I will not play your game. You listed only the permutations not the combinations; 9 is much more likely. About 2 seconds, but proving it takes a good deal longer.

    1. Umm,

      You listed only the permutations not the combinations

      Nope. The opposite.

      9 is much more likely

      Nope. The opposite.

      but proving it takes a good deal longer

      That I can agree with 🙂

    2. SPOILER ALERT (but I think the puzzle is already spoiled before here.)

      Anonymous is right. Richard listed combinations, not permutataions, and 9 is slightly less likely. The proof is easy if you list all 216 (equally likely) permutations and count the number of 9s and 10s. But there’s a less tedious way of proving it.

  11. I wanted to play, but unfortunately Richard is too far away from me. Fortunately, I can create a virtual Richard to play against a virtual me. And, since virtual people can play faster than real people, I challenged virtual Richard to a best-of-1000-sets-of-six-throws-each match.

    Have fun with this.

    ***********************

    #!/bin/bash

    #simulation for Richard Wiseman’s Friday Puzzle, 03.30.2012

    cuentadetres=0
    turno=0
    total=1

    nuevetotal=0
    dieztotal=0

    while [ $dieztotal -o $nuevetotal -lt 1001 ]; do
    cuentadenueve=0
    cuentadediez=0
    cuentadetiradas=0
    clear
    echo Playing set $total
    echo Sets won so far: Richard $nuevetotal — $dieztotal Luis
    while [ $cuentadenueve -o $cuentadediez -lt 6 ]; do
    dados=()
    while [ $turno -lt 3 ]; do
    tirada=$(echo $RANDOM | egrep -o ‘.$’)
    dados=(“${dados[@]}” “$tirada”)
    if [ $tirada -lt 7 ]; then
    if [ $tirada -gt 0 ]; then
    cuentadetres=$((cuentadetres + $tirada))
    turno=$((turno + 1))
    fi
    fi
    done

    cuentadetiradas=$((cuentadetiradas + 1))
    turno=0

    if [ $cuentadetres -eq 10 ]; then
    cuentadediez=$((cuentadediez + 1))
    echo -e “Set $total, throw $cuentadetiradas: ${dados[0]} + ${dados[1]} + ${dados[2]} = $cuentadetres, LUIS WINS \t Richard $cuentadenueve — $cuentadediez Luis”

    elif [ $cuentadetres -eq 9 ]; then
    cuentadenueve=$((cuentadenueve + 1))
    echo -e “Set $total, throw $cuentadetiradas: ${dados[0]} + ${dados[1]} + ${dados[2]} = $cuentadetres, RICHARD WINS \t Richard $cuentadenueve — $cuentadediez Luis”
    else
    echo -e “Set $total, throw $cuentadetiradas: ${dados[0]} + ${dados[1]} + ${dados[2]} = $cuentadetres, null throw \t Richard $cuentadenueve — $cuentadediez Luis”
    fi

    if [ $cuentadenueve -eq 6 ]; then
    break
    fi

    if [ $cuentadediez -eq 6 ]; then
    break
    fi

    cuentadetres=0

    done

    if [ $cuentadenueve -gt $cuentadediez ]; then
    nuevetotal=$((nuevetotal + 1))
    total=$((total + 1))
    echo ” ”
    echo Richard wins this set $cuentadenueve — $cuentadediez
    echo Sets won so far: Richard $nuevetotal — $dieztotal Luis
    sleep 1
    elif [ $cuentadediez -gt $cuentadenueve ]; then
    dieztotal=$((dieztotal + 1))
    total=$((total + 1))
    echo ” ”
    echo Luis wins this set $cuentadenueve — $cuentadediez
    echo Sets won so far: Richard $nuevetotal — $dieztotal Luis
    sleep 1
    fi

    done

    echo ” ”

    if [ $dieztotal -gt $nuevetotal ]; then
    echo Final result:Luis wins $nuevetotal to $dieztotal
    elif [ $nuevetotal -gt $dieztotal ]; then
    echo Final result: Richard wins $nuevetotal to $dieztotal
    else
    echo Final result: Tie! Both player A and player B have $dieztotal round wins
    fi

    echo ” “

  12. Well, if the stakes weren’t too high, I might be willing to play just to have the opportunity to play a game with Richard Wiseman regardless of who targets 9 or 10.

    though it is not specified, I assume all three dice we would be playing with would be conventional D6 dice. From the wording of the puzzle, it seems possible that while there must be at least one D6, but there could be one or two D4s.

    With one D4 and two D6s, I’ll play as long as the house doesn’t take a cut.

    With Two D4s and one D6, I’ll play and bet big.

  13. The survey was annoying. First, I don’t work, so the questions about when I go to bed and get up on work days don’t apply. Second, I virtually never dream, so only having answers to the dream questions that assume I do was also irritating.

    1. I too was irritated, I keep my own hours to suit myself. I wasn’t sure if I should lie or attempt an average so I just abandoned it.

  14. With regards to the sleep survey (which I have just filled out) it might be worth factoring in those of us who have had a baby in the last six months!

  15. As a mathematician the idea was immediate. But it was a fun puzzle (which is likely to be favorable to those who have had high school lever probability theory).

  16. Got it in a few minutes. My natural suspicion of being cheated nearly made me come down on the wrong side though of yes/no until I checked the wording of the offer again.

  17. obviously the higher the number the more chances the three dices have of getting to that number. that’s like giving a runner more time to complete a race.

    therefore 10 will be reached more often than nine, though there is no easy way of checking the precise odds.

    1. “Obviously, the higher the number, the more chances…of getting to that number”? I don’t think so… Consider 18, the highest possible number. Is it reached very often? You need to roll all 6’s!

  18. I put this aside when I could not immediately see the answer. The next day, I looked at it again and saw the answer immediately. Amazing what a good night’s sleep will do.

  19. This was actually funny, psychologically. Richard offered to be on the losing side of the proposition (choosing 9 over 10), and several people replied, offering to switch sides! Not a good plan.

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