I need the statistics/figures for the probability of throwing (x)
number of sixes with six dice. So, because that's as clear as mud, how
about, how many throws would I need, using SIX dice, to get 4 sixes, 5
sixes and 6 sixes? Still doesn't sound right. OK, if it's 50p a throw
of six dice, what should the prize values be for getting 4,5 or 6
sixes with one throw? Oh, and I need to know by Saturday lunchtime!
Cheers, oraccomp (Dave)

Hi oraccompdave-ga,

Assuming you toss six fair dice, there are six to the sixth possible
outcomes, i.e., 6 x 6 x 6 x 6 x 6 x 6, or 46,656.

Of these, only one outcome is all sixes, so the chance of that
occuring is one in 46,656.

There are 30 possible ways of throwing five sixes (because there are
six different dice which might be "non-6", and the "non-6" die could
show 1, 2, 3, 4 or 5), so the chance of throwing exactly five sixes is
30 in 46,656.

There are 375 possible ways of throwing four sixes (because there are
15 possible pairs of dice which might be "non-6", and the "non-6" dice
pairs could show any of the 25 combinations of 1, 2, 3, 4 or 5), so
the chance of throwing exactly four sixes is 375 in 46,656.

If it's 50 pence per throw, the notional "prize pot" if every possible
combination is thrown (all 46,656 of them) would be £23,328. You could
divide that into three separate "prize pots" of £7776. During 46,656
throws, you would expect one throw of six sixes - you can give them
£7776. You would expect 30 throws of five sixes - you can pay them
each one-thirtieth of £7776, i.e. £259.20. You would expect 375 throws
of four sixes - you can pay them each one-375th of £7776, i.e. £20.73
and six-tenths of a penny.

If this is a fund-raising event, you will probably be making much
smaller payouts to provide a profit margin and to help cover the
uncertainty of random events. And if there's big money involved, do
double-check these calculations before running the competition. The
relevant math is here:

"Bernoulli Trials"
http://www.ds.unifi.it/VL/VL_EN/bernoulli/bernoulli1.html

Finally, remember that there's a one in 46,656 chance that the very
first toss will result in six sixes, so you will need a sizeable
float. However, you won't be paying out too often - fewer than one in
a hundred tosses is expected to win something.

Regards,
eiffel-ga


Google Search Strategy:

combinatorial dice
://www.google.com/search?q=combinatorial+dice

combinatorial bernoulli trials
://www.google.com/search?q=combinatorial+bernoulli+trials

Well guys, I'll rate it FIVE stars for the answer as you did your
best! However, it was useless! We had the event and in 450 throws, we
only had 14 x 2's, and 3 x 3's. Then as we packed up (chaos theory
steps in) a kid chucked them on the floor and got 5 x 6's up! No stats
or maths could have demonstrated that! So thank's very much to ALL,
(except the guy who's piggybacking his chance of dying!), until my
next question.

As your question is not very clear I post this as a comment.
The probability  of getting a certain number is:
1/6 for a single dice
1/6*1/6 for two, thus
1/6^n for n dices.
It´s impossible to say how many throws you will need to get 4,5 or 6
sixes with one throw, all you can say is the mathematical probability
of such a result.

till-ga

There are six to the sixth power possible ways to throw six six-sided
dice.  That's 46,656 possible throws, of which 6-6-6-6-6-6 is one
possibility, so the odds are 46,655-1 against.

There are people of great athletic skill who can throw dice in such a way as to 
greatly improve their odds of a certain number.

There are also dice that have been shaved and weighted and have extra
sixes added. And people who can switch the altered dice into a game.

You might want to buy an experience pit boss a drink and get some
advice on how to do the game without inviting the cheats.

Hi guys.

I have a reaaaaally simple question. What is the probability that ill
die tomorrow? It certainly isnt 0% and certainly not 100%. So what
value is it?

Hi oraccompdave-ga,

Thanks for your interesting comments and kind tip.

I had been wondering how the event had turned out. It was interesting
to hear about the outcome, fortunate as it was for the competitor and
unfortunate as it was for the fund-raisers.

If the outcomes were certain, we wouldn't need to call this field
"statistics". But in the long run, in events designed to exploit
statistical outcomes, the "house" should always win...

Regards,
eiffel-ga