Understanding Dice Throwing Probabilities: A Layman's Guide

[Peterconfused]

Hello
I am involved with a Rotary Club which uses a dice game to raise funds at various events. We want to look carefully at the probabilities of throwing dice. Can you tell me please in very basic layman's terms the odds on the same number being thrown on 4 out of 7 dice in one throw (which number doesn't matter). Then for 5, 6 and 7 the same, and finally 7 sixes.

We would like to use the formula then to look also at changing the number of dice and the number of the same digit required e.g. 5 out of 8 dice. Can you help please in real layman's terms? There are no mathematicians amongst us!

Many thanks in advance

Peter

 

[tkhunny]

...the probabilities of throwing dice.

There is no probability of throwing dice. You either throw them or you don't.

You can try counting...

If you threw the dice one at a time, this might be close...

5/7
Any of the 6 numbers on the first die.
The next 4 have to be the same as the first.
The last 2 can be anything EXCEPT what you've been throwing.

6 1 1 1 1 5 5

That's not quite the idea. Since you are NOT throwing them sequentially, you must also include all shuffles of that arrangement.

Why do you think there are "laymen's terms" for this sort of thing?

 

[Peterconfused]

Hello

Thank you very much for replying. Possibly we are at cross purposes here. I am wanting to know the odds or probabilities of, say, 7 dice being thrown and the same number showing on 4 of them. I was given to understand that there is a formula which can be used to calculate the odds, involving binomial coefficients. This is not something I learned at school and, as a layman, I simply don't understand it. What I am seeking is a simple formula which could then be applied to variations on the number of dice thrown and the number showing the same digit.

That would be extremely helpful

Many thanks again for your interest

Peter

 

[tkhunny]

Well, if the right answer is binomial coefficients, you can quit looking for layman's terms.

What formula have you discovered? Perhaps it's use can be described without much trouble.

 

[Peterconfused]

Hello

Thanks for the reply. It was someone from MHB who quoted binomial coefficients. I don't understand them. Anyway, I think I will just have to soldier on.

Best regards

Peter

 

[HallsofIvy]

The rules of your game are not clear to me. You ask only about four dice being the same. What about the other three? Does, say, 3333666 count the same as 3333456? What about 3333225? That is, is the crucial thing having four the same without regard for possible duplicates in the other three?

What if, when looking for four the same, five happen to be the same? Does 3333356 count the same as 3333456?

 

[Peterconfused]

Thanks for your reply. The game being played is simply that you throw the 7 dice and if 4 land showing the same number you win a prize. 5, 6 or 7 showing the same will give better prizes still. We are wanting to look at how the odds would change using a different number of dice and a different number showing the same being required to win a prize. Hope this makes sense.

Regards

Peter

 

[HallsofIvy]

With 7 dice, the first die may be anything. The probability the second die is the same as the first is 1/6. The probabiity the third die and fourth die are also the same is, of course, 1/6 also. The probability the fifth sixth and seventh dice are one of the other 5 number, not the same as the first four numbers, is 5/6 each. So the probability that the first four numbers are the same but the last three are different, in that order, is $$\left(\frac{1}{6}\right)^4\left(\frac{5}{6}\right)^3= \frac{5^3}{6^7}$$.

But we want any order, not just "1111324" but "1311412", etc. There are $$\frac{7!}{4!1!1!1!}= \frac{7!}{4!}= 7(6)(5)= 210$$ (the "multinomial coefficient") ways to order those so the probabiity of 4 out of 7 the same is $$\frac{210(5^3)}{6^7}= [FONT=Verdana,Arial,Tahoma,Calibri,Geneva,sans-serif]0.09377[/FONT]$$.

 

[Peterconfused]

Thank you very much. The problem I and my colleagues have is that none of us are mathematicians. Hence we don't understand what is being said. You have very kindly set out a formula. Is there any way you can set that out in very simple terms that would help us? None of it is anything that we learned at school. It is very frustrating.

Hoping you can help

Best regards

Peter