Rolling Dice

Dross · 2007-02-26 02:09:51

Here's a problem someone asked me if I could do. I'm no good at probability (was never really interested, to be honest) but I gave it a shot and got an answer I'm fairly confident of - just wanted to see other approaches (and if I was right or not ).

Here's the problem:

You roll m dice. What is the probability that any one of them is a six?

Toast · 2007-02-26 02:16:54

Well, I have to go... but I got

99% Incorrect.

Last edited by Toast (2007-02-26 02:17:40)

Dross · 2007-02-26 02:22:33

But then you're saying that if you roll six dice, you're garunteed that one of them will be a six, and if you roll 9 dice the probability is 1.5 that one of them will be six

Maelwys · 2007-02-26 02:30:36

pi man · 2007-02-26 03:01:14

You roll m dice. What is the probability that any one of them is a six?

By this, do you mean: one and only one of the dice is a six OR at least one of the dice is a six?

If you mean "one and only one", the number of ways to roll one 6 is

. That's out of different possible rolls.

If you mean at least one, the answer Maelwys gives is the correct probability.

luca-deltodesco · 2007-02-26 05:19:05

let X be the number of 6's rolled

P(X>=1) = 1-P(X=0)

for each die roll, the probability that it doesn't roll a 6 is 5/6, so the probabilty that every single die doesn't roll a six is (5/6)^m, therefore

Last edited by luca-deltodesco (2007-02-26 05:19:55)

Dross · 2007-02-26 21:04:52

So it really is much simpler than I had it made out to be - thanks very much, all!

mathsyperson · 2007-02-27 02:49:14

It is quite a hard problem in general though, it's just that for the particular case of you only looking for one 6 you can look for the complement probability instead, which is much easier to find.

If you wanted to know how likely it was to get 2, 3, or even k sixes with m dice, then the problem becomes substantially trickier.

JaneFairfax · 2007-02-27 04:52:55

mathsyperson wrote:

It is quite a hard problem in general though, it's just that for the particular case of you only looking for one 6 you can look for the complement probability instead, which is much easier to find.
If you wanted to know how likely it was to get 2, 3, or even k sixes with m dice, then the problem becomes substantially trickier.

Actually, its a binomial distribution. The probability of getting exactly k sixes (k = 0, 1, m) is

To find the probability of getting at least k sixes, just sum the probabilities for k, k+1, m OR find the probability of getting fewer than k sixes and subtract from 1 (depending on the value of k one would be easier than the other).

John E. Franklin · 2007-02-27 05:39:35

Now what is the big X in the following binomial equation, is that what you called it?

JaneFairfax · 2007-02-27 06:16:38

Technically, the big X is called a random variable. In this case, it represents the number of sixes from the throwing of m dice.

Dross · 2007-02-27 08:28:44

And you make it look so easy - I can see exactly why that solution works, but it would never have occured to me to put it down like that. Thanks, Jane Fairfax!

mathsyperson · 2007-02-27 10:32:05

Ah, of course. I completely forgot about that. I actually meant throwing at least 2, 3 or k sixes, but the Binomial distribution can still sort that out. It will involve summing them, so it might take a bit longer to calculate, but it's still got the same complexity.

Math Is Fun Forum

#1 2007-02-26 02:09:51

Rolling Dice

#2 2007-02-26 02:16:54

Re: Rolling Dice

#3 2007-02-26 02:22:33

Re: Rolling Dice

#4 2007-02-26 02:30:36

Re: Rolling Dice

#5 2007-02-26 03:01:14

Re: Rolling Dice

#6 2007-02-26 05:19:05

Re: Rolling Dice

#7 2007-02-26 21:04:52

Re: Rolling Dice

#8 2007-02-27 02:49:14

Re: Rolling Dice

#9 2007-02-27 04:52:55

Re: Rolling Dice

#10 2007-02-27 05:39:35

Re: Rolling Dice

#11 2007-02-27 06:16:38

Re: Rolling Dice

#12 2007-02-27 08:28:44

Re: Rolling Dice

#13 2007-02-27 10:32:05

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