Please help solving this Year 5 probability problem

srr on 27/12/2022 - 23:19

My boy got this problem in Year 5:

Ben and Jenny roll a dice once. The first one to roll a five wins. What are the probability of winning?

The teacher gave an answer 1/6 for both players. I mentioned that whoever starts first has an upper hand as the second player doesn't have a chance to roll at all.

I asked Chat GPT, and the AI was confused at first. However, after a few pointers came up with a good reasoning when Ben starts first:

In this game, the players take turns rolling a dice with 6 sides. The first player to roll a 5 wins.

There are 3 possible outcomes in this game: Ben winning, Jenny winning, or a draw.

The probability of Ben winning is 1/6 on his first roll.

The probability of Jenny winning is the probability that Ben does not roll a 5 on his first roll, which is 5/6, multiplied by the probability that Jenny rolls a 5 on her first roll, which is also 1/6. Therefore, the probability of Jenny winning is (5/6) * (1/6) = 5/36.

The probability of a draw is the probability that neither player wins, which is the probability that both players roll a number other than 5. This can be calculated as (5/6) * (5/6) = 25/36.

As a bonus, what are the probabilities of winning if Ben and Jenny keep rolling until someone wins?

Poll Options

  • 70
    1/6 and 5/36
  • 86
    1/6 and 1/6
  • 182
    the problem is too complex for Year 5
  • 7
    your answer
Education & Work

Comments

  • rinco on 28/12/2022 - 19:41

    When I asked ChatGPT about this I got….

    If a person believes that the solution to a 5th grade math problem is a certain number, they may use an AI chatbot to confirm this belief rather than doing the math themselves or seeking out alternative solutions. This may save them time and provide validation, but it may also prevent them from learning and expanding their understanding of math concepts.

    • srr on 28/12/2022 - 22:59

      On the other hand, if the person doesn’t stop there but asks ozbargain community to help out, their understanding expands enormously.

      • rinco on 29/12/2022 - 21:50
        +1

        Surprise twist - the ozbargain community is and always has been ChatGPT!

        • Diji on 29/12/2022 - 22:49

          GPTBargain

  • monky on 28/12/2022 - 19:42

    Someone's probably already said this but you're overthinking it: Who throws first isn't a variable in the original question, so the assumption is they throw at the same time.

  • cloudy on 28/12/2022 - 22:09

    I was a pretty good a maths in school. Now i realized i was just good at getting good school grades.

    The way I approach a problem isn't by looking at the merits of the problem, its looking it through the lens of the teacher. What they have taught prior to this test will yield the answer they want.

  • RolandWaites on 28/12/2022 - 23:06
    +1

    There's an issue with the wording. The children roll /a/ (singular) /dice/ (multiple) once. Does each child have one die each and they're thrown at the same time, or are they going one after the other with a singular die?

    I suspect it's meant to be the former but it can be read as the latter — which is where at least some (if not all) of the confusion stems from.

    • srr on 28/12/2022 - 23:08

      Dice as a singular discussed here https://www.ozbargain.com.au/comment/13170820/redir

      • RolandWaites on 28/12/2022 - 23:24
        +1

        That may be so in colloquial terms, but as it's in an academic paper you might perhaps presume the author was using the term dice for its traditional definition.

        • srr on 29/12/2022 - 07:37

          It was a copied homework sheet for Year 5 children, not an academic paper. Regardless, I don’t have exact wording anyway.

          • RolandWaites on 29/12/2022 - 10:29

            @srr: Academic in the sense it's an educational paper. I was going to clarify that last night but I felt we could all catch the drift of what I was saying.

  • Donaldhump on 28/12/2022 - 23:09

    Scenario 1.

    assuming no matter who starts the other can do the roll after and the result is determined after both have rolled, or both have a dice and roll simultaneously and they repeat till a winner or tie

    A.) ben winning = 1/6 * 5/6 = 5/36
    B.) jen winning = 1/6 * 5/6 = 5/36
    C.) tie = 1/6 * 1/6 = 1/36
    D.) no result = 5/6 * 5/6 = 25/36 goto start and repeat

    thus if D cause a loop until A,B,C
    bens odds of winning is simply D/2 + A = 17.5 / 36 = 48.6%
    jens odds of winning is simply D/2 + A = 17.5 / 36 = 48.6%
    tie = 1 / 36 = 2.7%

    Scenario 2, first to roll wins and one player rolls at a time thus a tie is not possible

    A.) bens chances = 1/6 = 6/36
    B.) jens = 5/6 * 1/6 = 5/36
    C.) no result = 5/6 * 5/6 = 25/36

    bens chances = C/2 + 1/6 = 18.5 / 36 = 51.4%
    jens chances = C/2 + 5/36 = 17.5/36 = 48.6%

    scenarion 3/4 sames 1/2 but only one roll each

    CBF..

    • srr on 28/12/2022 - 23:23

      I don't think your math is right. If a tie is one of the outcomes causing end of re-rolls, then ben winning will be 5/11, Jen 5/11, tie 1/11

      There are a few ways to calculate the second case (see https://www.ozbargain.com.au/comment/13168954/redir for example, end result is 6/11 and 5/11)

      • Donaldhump on 29/12/2022 - 02:35

        You have to explain
        How do you get 5/11 and 1/11
        I read the example it is the same as what I have

        • srr on 29/12/2022 - 07:32

          The only outcomes to end the game are combinations of two numbers where one of them is 5. These combinations are 1-5, 2-5, 3-5, 4-5, 5-5, 6-5, 5-1, 5-2, 5-3, 5-4, 5-6. You can see there’s 11 of them, where if the first number is 5 then Ben wins, the second number is 5 then Jenny wins, and both 5 is a tie. These rolls have a similar probability each, 1/11. Hence 5/11 of winning for both Jenny and Ben, and 1/11 for a tie.
          We only need to consider these 11 combinations because any other counts as a re-roll and effectively removed from the sample space.

  • john71 on 28/12/2022 - 23:31
    +2

    We're cognitively not designed to grapple with statistics and probabilty. That's why we get bamboozled so easily by hucksters peddling bad products and bad politics.

    When I did first-year stats at uni, our tutor introduced the topic thus: "Mary has two children. One of them is a boy. What is the probability that the other is a girl?"

    The answer, of course, is two-thirds. I think 5 students out of the 20-odd in attendance got it right.

    • srr on 29/12/2022 - 01:04

      I have a new sport: get the right answer from ChatGPT.

      Q:

      Mary has two children. One of them is a boy. What is the probability that the other is a girl?
      Consider there are four cases for children. We're only selecting cases with at least one boy. We need to sum those probabilities.

      A:

      If Mary has two children and one of them is a boy, then there are four possible combinations of boys and girls:

      • Both children are boys.
      • Both children are girls.
      • One child is a boy and the other is a girl.
      • One child is a girl and the other is a boy.

      We are only considering cases where at least one child is a boy, so we can eliminate the second combination (both children are girls). This leaves us with three combinations:

      • Both children are boys.
      • One child is a boy and the other is a girl.
      • One child is a girl and the other is a boy.

      The probability of the first combination occurring (both children are boys) is 0, since it does not meet the given condition that at least one child is a boy. The probability of the second combination occurring (one child is a boy and the other is a girl) is 1/3, and the probability of the third combination occurring (one child is a girl and the other is a boy) is also 1/3.

      To find the total probability, we need to sum the probabilities of the three combinations:

      P(both children are boys) + P(one child is a boy and the other is a girl) + P(one child is a girl and the other is a boy)
      = 0 + 1/3 + 1/3
      = 2/3

      Overall, the probability that Mary's other child is a girl, given that we know that one of Mary's children is a boy, is 2/3. This is a conditional probability because it is based on the condition that one of Mary's children is a boy.

      • Kail on 29/12/2022 - 01:36

        The probability of the first combination occurring (both children are boys) is 0, since it does not meet the given condition that at least one child is a boy.

        … what??

        • srr on 29/12/2022 - 01:37

          Yeah I know, but that’s the best I was able to get quickly.

    • p1 ama on 30/12/2022 - 18:54

      When I did first-year stats at uni, our tutor introduced the topic thus: "Mary has two children. One of them is a boy. What is the probability that the other is a girl?"

      The "trick" is that we're specifically not told which of the children is a boy, so having one girl, one boy is twice as likely as having two boys, hence the probabilities are 2/3 and 1/3.

      • john71 on 30/12/2022 - 21:58
        +1

        Of course. But our gut instinct in this case shrieks, “Fiddy percent! Fiddy percent!” And that guy instinct, however wrong, is what 90% of people will go for.

    • DrScavenger on 31/12/2022 - 23:55

      Its like gambler's fallacy come true. I remember something similar with opening boxes/doors and winning money. Always rubbed me the wrong way and I somewhat liked probability.

      • john71 on 03/01/2023 - 17:33

        It’s the Monty Hall problem, which has disgraced many a professional mathematician. Monty Hall is another staple of first-year stats tutors wanting to cut their students down to size.

        The third point of that embarrassing triad concerns disease incidence and testing: a certain test is 95% accurate, yet a positive result has only a 50% chance of being correct. It’s a phenomenon even doctors struggle to wrap their minds around.

        What I find extraordinary about all this is not that we struggle with elementary statistics, but that we seize on the wrong answers with such bravado while refusing to acknowledge our (many) cognitive limitations.

        • DrScavenger on 03/01/2023 - 19:21

          In data science we call it precision as far as I remember. The accuracy of positive results. In that case precision is 50%.

          You can make a test with 99% accuracy to determine the HIV status of random people.

          The test will always give negative as a result and since 99% of people don't have it you will be right 99% of the time. But you will also miss everyone that matters making the test useless but with a high accuracy.

  • Assburg on 29/12/2022 - 02:27

    Assuming both players have a fair chance of winning, whoever rolls first has a chance of winning which approaches 1/2 from above as the number of turns taken for a 5 to occur increases, and whoever rolls second has a chance approaching 1/2 from below.

    You are answering the question as if it is "what is your chance of rolling a 5 on a 6 sided die", and that is a condition necessary for the outcome of winning, but the two outcomes we are concerned with are winning and losing.

    In the event each player only has a single roll of the die each (or none for player 2 if player 1 wins), then your original answer is correct.

  • Daddy Chill on 29/12/2022 - 03:11

    The only thing I got from this was Ben &Jenny ice cream which I love

    • srr on 29/12/2022 - 07:19

      Sorry to report that the ice cream is “Ben & Jerry”

  • Grazz989 on 29/12/2022 - 05:38

    The problem with this problem is that there are too many ambiguities. Let's take a look at the actual question: "Ben and Jenny roll a dice once. The first one to roll a five wins. What are the probability of winning?"

    Okay, so are we to assume that they know those rules and are going to race each other to roll first? Are they equally likely to roll first? It says they roll 'a dice'. Does that mean they each have one, or does that mean they have to take turns rolling a single die? If so: How do they determine who rolls first?

    Does the question ask, 'what is the probability of either player winning?' (if neither roll a 5 on their only turn, both lose it would seem), or does it ask 'what is the probability of winning for each of the two individual players?' … ?

    This has become as absolutely stupid as those Facebook moron-magnet posts which harvest a huge number of replies just by saying 'only a genius can solve this question!' and then presenting a dumb simultaneous equation where instead of 'a+b-c=d', there's four fruit emojis or whatever, and the pair of cherries emoji is sneakily swapped for three cherries on the third line so the question can't actually be answered, but you'll have imbeciles with all sorts of whacky reasoning as a result.

    Tell the teacher to fix poorly worded questions before giving them to the students, and passing feedback up the chain if it came from above.

    • Grazz989 on 29/12/2022 - 06:03
      +1

      For fun, however:

      Assuming both roll one die a single time, and it's random who rolls quicker … Ben has a 1 in 6 chance of rolling a 5. However, there's a 1 in 6 chance that Jenny also rolls a 5, and a further 1 in 2 chance hers lands first. 1/6 * 1/2 = 1/12 chance that she not only rolls a 5, but that her die lands first, beating Ben regardless of his roll. Assuming Ben gets a 5 at all, that means there's an 11/12 chance he doesn't get beaten by a faster 5 from Jenny - any other roll by her, or a slower roll of 5, doesn't matter.

      Therefore: 1/6 (Chance of Ben rolling a 5) * 11/12 (Chance Ben isn't beaten by a quicker roll of 5 from Jenny) = 22/144 = 11/72 (Chance of Ben winning).

  • Heaps for Cheaps on 29/12/2022 - 07:43

    "The teacher gave an answer 1/6 for both players. I mentioned that whoever starts first has an upper hand as the second player doesn't have a chance to roll at all."

    Did you also calculate the probability that the person with the first roll may slip into a coma before being able to complete a roll?

  • brawndo on 29/12/2022 - 07:44
    +1

    Zero. They collectively roll the one dice. The winner can be only one. They're both losers forever. Maybe they should open an ice cream shop instead.

    • Heaps for Cheaps on 29/12/2022 - 07:49
      +1

      I think they both win because both Ben and Jerry discovered in this simple game what it means to be growing up in this new day & age. They learned lessons about gambling, life, luck and love.

      • Diji on 29/12/2022 - 22:53
        +1

        The real reward is the dice we rolled along the way

  • PrettyBrokeTBH on 29/12/2022 - 07:54

    Where does it say they aren't going at the same time?
    Also - why are you doing this on school holidays?

  • peekabooo on 29/12/2022 - 07:57
    +1

    It's 1/6 and 5/36. These roles are independent events. So whoever roles at that point has a higher probability of winning (1/6) and the other person at that point has a lower probability of winning.

  • DarkOz on 29/12/2022 - 09:52
    +1

    Without checking the answers here my instinctive answers are:
    1st to roll: 1/6 of winning
    2nd to roll: 5/6 x 1/6 = 5/36 of winning
    The above will just repeat itself if no one wins after both rolled the dice.

  • hawkshead on 29/12/2022 - 11:21

    Do you also question why the catepillar is hungry hungry???

    • hawkshead on 29/12/2022 - 11:21

      cause there's an answer they arent telling preschoolers. And its absolutely amazing.

      • hawkshead on 29/12/2022 - 11:22

        HOW ARE PRESCHOOLERS MEANT TO UNDERSTAND THE KREB CYCLE IF THEY ONLY LEARN ABOUT TH FOODS THE CATEPILAR EATS???

  • AnophthalmiaCervidae on 29/12/2022 - 13:11

    I mentioned that whoever starts first has an upper hand as the second player doesn't have a chance to roll at all.

    Ok so here is another interpretation…

    Assuming they take turns rolling the dice and Ben goes first. If Ben wins Jenny does not get to roll the dice as the game is over.
    Ben and Jenny roll a dice once.
    For Jenny to get to have a roll Ben must not have rolled a 5 in his turn. We know from the question that they both roll the dice so the probability Ben has won given Jenny also gets to roll is ZERO.
    The probability of Jenny winning is 1/6.

    • srr on 29/12/2022 - 14:15

      Probably a bad wording on my end. Jenny can roll if she wants, but the outcome of her roll doesn’t change anything.

  • berry580 on 29/12/2022 - 14:05

    The teacher couldn't even get his/ her own question right, should at least own up to their own mistake

    • srr on 29/12/2022 - 14:13

      The teacher actually agreed with me during a conversation. I have nothing against the teacher, she is great.

    • Eeples on 29/12/2022 - 23:39
      +1

      How can we know?

      Op admits to not remembering exact wording.

  • Aap Ke Papa on 29/12/2022 - 14:36

    They both roll at the same time. 1/6

  • OzDayZee on 29/12/2022 - 15:13
    +1

    OP's reluctance to post the original question wording makes it look like a troll post.

    • srr on 29/12/2022 - 16:46
      -1

      It is a fair concern. Unfortunately I don’t have exact wording at hand, because the whole situation was about half a year ago. I actually don’t need any help solving the problem, I was curious to learn what people think about its complexity. I was also hoping that maybe there will be someone who can explain what assumptions needs to be made to get to 1/6 each. I was also sharing info about ChatGPT, which is going to revolutionise some industries soon, for example resume writing, generally writing any texts where it needs to be blown in size (women novels come to mind), writing persuasive texts of all sorts, service desks etc.

  • bmerigan on 29/12/2022 - 22:00

    They could each have their own dice and roll at the same time. Then it's the simplest situation with the simplest answer, 1-in-6 each.

  • Eeples on 29/12/2022 - 23:32

    Nowhere in the problem (as stated) it says they take turns.

    If there is one die and they roll it together then they will both win when a 5 comes up; which is 1/6 chance that they win.

    If they roll their own die and are competing against themselves then again each would win with a 1/6 chance.

    If they roll their own die and they are competing against each other; then each would have a 1/6 times 5/6 = 5/36 chance of winning rather than drawing (ie. both getting a 5).

    If they take turns… I’ll leave to others 😀

  • cashless on 30/12/2022 - 11:05

    What are the probability of winning?

    I hope the grammar lessons are working out for the year fives

    • Hangryuman on 30/12/2022 - 12:15

      'I hope the grammar lessons are working out for the year fives'

      yeah bad grandma

      but the OP scenario ('teacher didn't know either') fits with my readings that most school teachers basically failed maths

      • srr on 30/12/2022 - 14:46

        yeah bad grandma

        That turned out funny.

    • srr on 30/12/2022 - 14:45

      You forgot a full stop.

      • cashless on 31/12/2022 - 08:44
        +1

        Sounds like we may need to explain the difference between grandma & punctuation;)

  • syousef on 30/12/2022 - 14:31

    My math is rusty so I hope I don't embarass myself here.
    I was doing double accounting subtracting preivous throws in my last (unpublished) attempt.

    You're right about the first player having an advantage. It's often the case that the first to play has an advantage in games. For example white has an advantage in chess (though that is a game of skill rather than chance). In chess it's called a tempo. So I would say 1/6 for first to roll and then 5/36 are the odds after the first rolls. Then 1/6*31/36=31/216 for player one on the second roll and so on.

    Each time we get to a new round player 2 has a disadvantage throwing second (or alternately you can look at it as player 1 having had an extra chance to win at every point), so the gap grows and the longer the game grows the greater the chance player 1 will win.

    I threw it into Excel.
    2nd and 3rd columns are look at the chances no one has thrown to that point then a 5 is thrown.
    4th column is the advantage for that throw.
    5th and 6th column add up the probabilities to that point that the player has won.
    7th column adds up the advantage for each throw.

    Probability of winning ON that throw. Probability Player has won BY that throw.

    Throw Player 1 Player 2 Advantage of 1st Player Player 1 Player 2 Total advantage
    1 0.166666667 0.138888889 0.027777778 0.166666667 0.138888889 0.027777778
    2 0.115740741 0.096450617 0.019290123 0.282407407 0.235339506 0.047067901
    3 0.080375514 0.066979595 0.013395919 0.362782922 0.302319102 0.06046382
    4 0.055816329 0.046513608 0.009302722 0.418599251 0.348832709 0.069766542
    5 0.03876134 0.032301117 0.006460223 0.457360591 0.381133826 0.076226765
    6 0.026917597 0.022431331 0.004486266 0.484278188 0.403565157 0.080713031
    7 0.018692776 0.015577313 0.003115463 0.502970964 0.41914247 0.083828494
    8 0.012981094 0.010817579 0.002163516 0.515952058 0.429960049 0.08599201
    9 0.009014649 0.007512207 0.001502441 0.524966707 0.437472256 0.087494451
    10 0.006260173 0.005216811 0.001043362 0.53122688 0.442689067 0.088537813
    Totals 0.53122688 0.442689067

    Odds someone won by now 0.973915947
    Advantage of 1st player 0.088537813

    Definitely not year 5 math.

    • srr on 30/12/2022 - 17:17

      Thanks for giving this a shot.

      Then 1/6*31/36=31/216 for player one on the second roll and so on

      This is where you had a small mistake. For Ben probability to win in the second round is 1/6 (second roll) * 5/6 (Ben didn't win in the first round) * 5/6 (Jenny didn't win in the first round) = 5^2/6^3. I.e. the probability of winning in every subsequent roll is 1/6 * P(not winning in the previous roll)

      Better explanation here https://www.ozbargain.com.au/comment/13168954/redir

      • syousef on 30/12/2022 - 18:43

        I was subtracting rhe probability of a win the prior rounds in the table above. I am assuming there are no draws. If either player throws a 5 first, they won. Nothing in the rules said anything about a draw. "The first one to roll a five wins."

        Here's the spreadsheet with show forumulas on.
        Probability of winning ON that throw. Probability Player has won BY that throw.
        Throw/Player Player 1 Player 2 Advantage of 1st Player Player 1 Player 2 Total advantage
        1 =1/6 =(1-B3)(1/6) =B3-C3 =B3 =C3 =G3-H3
        =A3+1 =(1-SUM(B3:C3))        (1/6) =(1-B4-SUM(B3:C3))1/6 =B4-C4 =G3+B4 =H3+C4 =G4-H4
        =A4+1 =(1-SUM(B3:C4))        (1/6) =(1-B5-SUM(B3:C4))1/6 =B5-C5 =G4+B5 =H4+C5 =G5-H5
        =A5+1 =(1-SUM(B3:C5))        (1/6) =(1-B6-SUM(B3:C5))1/6 =B6-C6 =G5+B6 =H5+C6 =G6-H6
        =A6+1 =(1-SUM(B3:C6))        (1/6) =(1-B7-SUM(B3:C6))1/6 =B7-C7 =G6+B7 =H6+C7 =G7-H7
        =A7+1 =(1-SUM(B3:C7))        (1/6) =(1-B8-SUM(B3:C7))1/6 =B8-C8 =G7+B8 =H7+C8 =G8-H8
        =A8+1 =(1-SUM(B3:C8))        (1/6) =(1-B9-SUM(B3:C8))1/6 =B9-C9 =G8+B9 =H8+C9 =G9-H9
        =A9+1 =(1-SUM(B3:C9))        (1/6) =(1-B10-SUM(B3:C9))1/6 =B10-C10 =G9+B10 =H9+C10 =G10-H10
        =A10+1 =(1-SUM(B3:C10))        (1/6) =(1-B11-SUM(B3:C10))1/6 =B11-C11 =G10+B11 =H10+C11 =G11-H11
        =A11+1 =(1-SUM(B3:C11))        (1/6) =(1-B12-SUM(B3:C11))*1/6 =B12-C12 =G11+B12 =H11+C12 =G12-H12
        Totals =SUM(B3:B12) =SUM(C3:C12)

        Odds someone won by now =SUM(B3:C12)
        Advantage of 1st player =B13-C13

        • srr on 30/12/2022 - 21:48

          I see, thanks for sending the spreadsheet, I believe it is correct.

          As for the mistake I was referring to

          Then 1/6*31/36=31/216 for player one on the second roll and so on.

          It should be 1/6 * (1-(1/6+5/36)), or 1/6 * 25/36, or 25/216 = 0.1157407407. In your spreadsheet the number is correct.

  • aDDiKt24 on 30/12/2022 - 16:35

    People arguing over this are embarrassing. It's a simple solution for a year 5 maths problem. Don't read too much into it. Also Shame on the actual adults in this forum arguing like school children themselves. Get real!

    • srr on 30/12/2022 - 17:19

      People arguing over this are embarrassing.

      You can call it arguing or you can call it a discussion. Forum is a place for discussions and it would be strange if they people didn't do it.

      What is your answer actually?

      • aDDiKt24 on 30/12/2022 - 20:00

        My answer is simple, correct and aligns with the teacher. Anything more complicated misses the context of the environment the question is being asked in.

        • syousef on 30/12/2022 - 20:32

          If the reason you do math is to "align with the teacher", you're doing it wrong. Math is beautiful, complex and a wonder. P***ing contests and a race to complying and conforming are not. The original context of a year 5 classroom is different to the context of an online forum. If you don't want to examine the question more deeply, that's fine, but please don't discourage others.

          • Eeples on 30/12/2022 - 21:52

            @syousef: What question? Or which question?

            OP admits to not know the exact wording.

            And the whole post is disingenuous as the Op doesn’t need help as they have admitted a question (not necessarily this question) was posed over 6 months ago and they spoke to the teacher then.

          • aDDiKt24 on 31/12/2022 - 20:05

            @syousef: Context is everything. Context is a year 5 classroom.

  • conza on 30/12/2022 - 19:08

    If the question specifically says "Ben rolls first, then Jenny" etc, you may have a point, otherwise no this is a waste of time, they roll simultaneously and each has a 1/6 chance of getting a 5.

    • srr on 30/12/2022 - 20:24
      +1

      So they both win when both roll 5?

      • conza on 30/12/2022 - 22:38

        They re-roll until only one of them has a 5.

        • srr on 30/12/2022 - 23:27

          Then they both have 5/36 chance on the first roll, which is less than 1/6.

          • conza on 30/12/2022 - 23:42
            +1

            @srr: Not sure I follow. There is a 1/6 chance of any result, and a 1/36 chance they tie, is that what it is? 6/36 - 1/36 = 5/36?

            I'd accept 1/6 from an 11 y/o I think that's fine, 5/36, if I follow, can be persuaded that's more precise.

  • syousef on 30/12/2022 - 20:39
    +1

    If nothing else the big takeaway here is the importance of unambiguously stating a question if you're going to employ math to solve it.

  • kaleidoscope on 30/12/2022 - 23:24
    +1

    how many generations are we ahead for a 5 yr student to tackle such a problem? what’s next, a grade 2 student solving partial differentiation? 😬

  • DrScavenger on 31/12/2022 - 23:42

    Problem isnt too complex. YOU are making it too complex.It is 1/6 for both. No draws are mentioned in question. Nothing to assume game ends when the first to roll 5 wins

    • srr on 01/01/2023 - 07:05

      Thank you for expressing your opinion. What happens with the case of Ben and Jenny rolling a 5 each? They have to both win for 1/6 overall probability of winning.

      • DrScavenger on 01/01/2023 - 07:32

        If they roll 5 at the same time they roll again. This is a Y5 question. I will assume they roll at the same time since no order is given. And if they both roll 5 at the same time they will roll again until it doesn't happen anymore. It's more of an oversight on their part and nitpicking on ours.

        • srr on 01/01/2023 - 08:24

          If they roll 5 at the same time they roll again. This is a Y5 question.

          Just to clarify rules more, they re-roll only in 5-5 case? It’s the same as 5-5 combination removed from the sample space. Then their chances of winning is 5/35 each, with draw 25/35. Still not 1/6.

  • ddilrat on 13/01/2023 - 19:26

    Can you only really work it out if you put a limit on number of rolls

  • srr on 01/02/2023 - 09:56

    !!!Urgent Update!!!

    !!!Screenshot available!!!

    https://www.dropbox.com/s/pcyb74r97uo89my/IMG_5452.jpeg

    Exact wording:

    Harry and Lucy play a game where they roll a die. The winner is the first one to roll a five. The chance of winning is _____

    They play a game where they toss a coin. The winner is the first one to throw 'heads'. The chance of winning is _____

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