You can imagine the 12 dice roll as being two sets of the 6 dice roll and the 18 dice roll as being 3 sets.
To win the 6 dice roll, you only have to achieve success once. The 12 and 18 sets require you to achieve that success more than once.
QED the 6 dice dice roll has more chance of success.
Oh, the probabilities are approx 0.66; 0.62 and 0.60.
Thanks Don I always find these probability teasers challenging, so it has distracted me mightily trying to derive your answers when I should have been working .
Finally I got there and the general formula that is useful is the probability of getting r sixes from n die
5^(n-r) /6^n x n!/ ( (n-r)! r! )
For the 6 dice case this gives 66.51%, for 12 dice 61.87%, and for 18 59.73%
brain ache after that one…
These Brain Teasers are meant to be RELAXING not work-distracting or headache inducing Let’s face it, the goat in a field and your falling brick were both a “walk in the Park” (Not !)
but well done for delivering good answers !
However, I haven’t been able to read (*) your published formula in such a way as to generate your published probabilities. I know that this new forum doesn’t help when it comes to incorporating symbols (see my thread in the “New Forum” section). Any chance you could re-write it or spell it out in words ?
Many thanks and Cheers
(*) it’s more likely to be my reading that’s the problem than anything else !
hi, the formula is for getting precisely r sixes form the n dice.
trying to lay it out a bit more clearly
For example the probability of getting exactly 1 six (r=1) with six dice (n=6) is
5^5 /6^6 x 6!/(5! 1!) = 0.401
Getting exactly no sixes is
5^6 /6^6 x 6!/(6! 0!) = 0.3348 ( hence prob of getting one of more sixes is 1-0.3348 = 0.6651 - your result.)
for the 12 die case
The prob of getting no sixes is
5^12 /6^12 x 12!/(12! 0!) = 0.1121
prob of getting exactly 1 six is
5^11 / 6^12 x 12!/( 11! 1!) = 0.2691 so prob of less than 2 sixes is 0.1121 + 0.2691 = 0.3812. so prob of 2 or more sixes is 1-0.3812 = 0.6187
Similarly for the 18 dice case adding the probs for 0 1 and 2 sixes and subtracting the results from 1.
Hope that helps.
Did you have a neater way of doing it?
I just mis-read your formula by using the “x” (multiplication sign) on the wrong numbers !
I notice that Richard has confirmed that the ability to insert symbols isn’t included in this new forum set up. This means doing what I think you have just done, work off-line and copy&paste into the forum. I have usually done that using Powerpoint.
Thank you for persevering.
I think that between us we have cracked all the Brain Teasers posted in the new forum so far.
So time for a few more new ones.
Please feel free to post your favourites, please don’t leave it all to Ian and myself !
But just to get things easily rolling along again…
…A quantity together with its two thirds has one third of its sum taken away to leave 10. What is the quantity ?
Nicely done Eoink !
There are two, and so far as I know only two, whole-number perimeter values for a rectangle which encloses the same numerical area as the numerical length of the perimeter.
So for example, If the area of the rectangle was 55 sqm, the perimeter would be 55m (eg 20m x 7.5m). But of course such a rectangle doesn’t qualify because the area of such a rectangle would be 150sqm rather than 55sqm.
What are the two rectangles that fit the bill ?
4x4 square and 6x3 rectangle - can’t prove there are no more though.
Yes, that’s right Ian.
And neither can I prove the absence of any others. I have looked for a proof to no avail. Just keep getting the same old confirmation.
4 * sqrt(3) if I remember the area of an equilateral triangle correctly.