maths help

Sorry, my maths isn't very good, so I'd be grateful if someone could work this out for me! (or, better still, give the formula (in layman's terms) for an n-sided dice, t times)

Me and my friend both throw a six-sided dice ten times. What are the chances that the same number comes up *none* of the ten times?

btw, is there a good un-moderated maths newsgroup?

Thanks,

If you roll one die once the chance of rolling one is

1/6

If you roll one die twice the chance if rollling one twice is

1/6 * 1/6 = 1/36

(This is because the two rolls are independent events). Likewise if you roll one die ten times the chance if rolling ten ones is

1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 1/6 *1/6 * 1/6 * 1/6 * 1/6

You can multiply these numbers to get 1/60466176

Now you can Make a table. Suppose you roll one die 10 times then The chance if rolling ten ones is 1/60466176 The chance if rolling ten two is 1/60466176 The chance if rolling ten threes is 1/60466176 The chance if rolling ten fours is 1/60466176 The chance if rolling ten fives is 1/60466176 The chance if rolling ten sixers is 1/60466176

The chance of rolling either ten ones or ten twos or ten threes or ten fours or ten fives or ten sixers is

1/60466176 +1/60466176 +1/60466176 +1/60466176 +1/60466176 + 1/60466176

You can add these numbers to get

1/10077696

This number is the answer to your problem

sci.math is an unmoderated math group, but I think that

will answer further questions over there.

Niels

Because I'm not sure that my problem has been solved by the answers Niels and Orion have kindly provided (Thanks, guys!), I'll reword it to make it clearer.

I roll a die twice. The outcome is two different numbers, ie. *not* 1, 1 or 2, 2 or 3, 3 or 4, 4 or 5, 5 or 6, 6. What is the probability of this outcome?

I repeat this operation ten times. What is the probability of this outcome occurring all ten times?

Adrian Bailey:

I merely observe that the two answers that have reached here so far (by Niels Ellegaard and 'Orion Ryder' both belong to questions quite different from the one asked. As to the actual question, there doesn't seem to be a simple formula, and I don't have time to work out an answer mathematically now.

But I do have time to do a brute-force search. It turns out that of 6^10 = 60,466,176 possible sequences of die rolls, 16,435,440 contain all 6 different numbers at least once each. So the answer for Adrian's specific problem is 44,030,736/60,466,176 = 101,923/139,968, or about 72.819% numerically. (Tantalizingly close to 72+9/11, isn't it?) But Adrian also asked for a generalized answer.

) Because I'm not sure that my problem has been solved by the answers Niels ) and Orion have kindly provided (Thanks, guys!), I'll reword it to make it ) clearer. ) ) I roll a die twice. The outcome is two different numbers, ie. *not* 1, 1 or ) 2, 2 or 3, 3 or 4, 4 or 5, 5 or 6, 6. ) What is the probability of this outcome?

Let's reword it for you, to make it easier:

I throw a one on my die. What's the probability that you will throw the same number as me ?

And if I'd thrown a two ? Or a three ? Or some other number ?

) I repeat this operation ten times. ) What is the probability of this outcome occurring all ten times?

How much do you know about multiplying probabilities ? Do you know what the probability of throwing two ones on a die is ? Three ones ?

SaSW, Willem

Adrian Bailey:

No, this is a completely different question.

5/6.

This question is asking about *consecutive* repetitions. This is

If you roll two n-sided dice, there are n^2 outcomes, of those n outcomes have both dice with the same number. This gives you a 1/n chance of rolling two dice with the same number showing. You therefore have a 1-1/n = (n-1)/n chance of not rolling dice with the same numbers.

For your example, you have 36 possible outcomes with 6 of them having the same number on both dice. This means you have a 1/6 chance of rolling two identical dice and a 5/6 chance of not doing this.

On each trial you have a (n-1)/n chance of 'success', therefore if you repeat the trial t times, you'll have a ((n-1)/n)^t chance of 'succeeding'. The answer to your question is that there is a (5/6)^10 chance of you and your friend rolling different numbers all 10 times.

LOL Well, in my head it's the same question! afaiac, I just got rid of the second die-roller. Maybe I need to take an online course in expressing myself mathematically. Any suggestions? (I'm usually quite good at expressing myself in other spheres, but I've always regretted, and suffered because of, not continuing with my maths studies beyond the school-leaving certificate.)

Oh, and thanks for your reply, Mark. The question was prompted by your competition, wondering how safe my money would be if I bet someone that they would not give the same answer as me to any of 't' MSB-type questions, each having 'n' possible answers. Maybe I should've just put it that way, since you would've understood me immediately...

My brain must work at the same bizarre frequency as yours, because your second wording was the meaning I understood from the initial problem statement. Perhaps the following wording would be more clear:

A friend and I each roll a 6-sided die. What is the probability that we will not roll matching values in any of 10 such trials?

EVen that may be misinterpreted, but it seems clear enough to me. But then again, so did your initial statement...

I'm not sure what Mark was getting at when he emphasized 'consecutive'. You're concerned about a number of events where each event is independent of the others. Whether they're consecutive or not shouldn't matter. I would have given the same answer as Kevin. 10 trials means you need to raise the probability for a single trial to the 10th power. So in the case of 2 dice, there would be a 16.1% chance of never rolling matching values. In other words, unless you're prepared to make the value of n sufficiently large that it would seem intuitive to your friends not to make the bet with you, I'd try another method of getting their money. If we stick to 10 trials, we can solve for n such that your chances would be at least 50%. I get that n must be 15 or greater, and even then you only barely come out ahead. Unless my calculations are off...

Hmm, the same number comes up *none* of the tem times?

Are you asking the probability of getting a different number on each roll?

such as 1,2,3,4,5,6,7,8,9,10

or 4,5,7,8,9,2,1,3,10,6 ? This I will call a straight.

or the odds of not getting '10 of a kind'?

The odds of *not* getting '10 of a kind' is equal to (1 - the answer that Niels gave you)

The odds of a straight are