1. This site uses cookies. By continuing to use this site, you are agreeing to our use of cookies. Learn More.
  2. Hi Guest, welcome to the TES Community!

    Connect with like-minded education professionals and have your say on the issues that matter to you.

    Don't forget to look at the how to guide.

    Dismiss Notice

Tricky Probability Question

Discussion in 'Mathematics' started by beedge, Nov 9, 2015.

  1. beedge

    beedge New commenter

    So a student of mine is developing a casino game involving rolling dice. The house vs the player. Higher score wins, but the player has to actually beat the house in order to win. Very simple.

    Things get more interesting when there are more than 2 dice involved.

    There are various ways of calculating the probabilities of winning different sorts of games, but I really want to know is what is the most efficient way of calculating a problem such as this:

    What is the probability that if you roll 2 dice, the highest score will be a 3 (for example)

    then with 3 dice, 4 dice, and better still, what happens with N dice.

    J
    ust to clarify, this doesn't mean that all the dice are 3 or less. At least one of them has to be a 3.

    I've found a way of doing it, but it seems very inefficient particularly if N is bigger than 2!

    For my original question, there are 3 situations if you call the 2 dice A and B:

    a) dice A is a 3, dice B is less than a 3
    b) dice B is a 3, dice A is less than a 3
    c) both 3's

    Now if you had 6 dice, then parts a and b would be similar to calculate.

    But part c is what's bothering me. You've have to consider ALL of the cases in which at least 2 of the dice would be 3's. You'd be there for ever.

    That's why I'm convinced there must a far more efficient way of calculating this.

    So if anyone has any good ideas, please let me know!

    Thanks very much.
     
  2. Piranha

    Piranha Star commenter

    Off the top of my head, would the probability that the highest dice is a 3 = P(all dice are 3 or less) -P(all dice are 2 or less)? So, with n dice, (3/6)^n-(2/6)^n . Or am I missing something?
     
  3. beedge

    beedge New commenter

    Thank you!!! Yes, you're right. Thank you very much.
     
  4. Piranha

    Piranha Star commenter

    Glad to be of service. It's nice to know that retirement has not removed all of my skills!
     
    Casy likes this.

Share This Page