I think you could be right about some sort of complicated graph based on the randomness of the distribution of Heads and Tails during the 300 flips. This might provide some measure of the confidence with which the coin can be said to be ‘fair’.
But as you say, let’s not go there !
Now you’ve tempted me!
Mathematically we can use combinatorics or Pascal’s Triangle. These are both equivalent to using a binomial distribution.
Using the combinatoric notation C(n, m) to mean choose any m from n (I used COMBIN(n, m) in Excel), then:
Prob(heads between 123 and 127 inclusive from 250) = (C(250, 123) + C(250, 124) + C(250, 125) + C(250, 126) + C(250, 127)) / 2^250 = 25%
Hence it is quite likely (75%) to get a result outside of that range. The 0.49 proportion of 300 is quite similar.
The result flipping (pun intended) from mostly heads to mostly tails is a bit more unusual, namely 19 heads out of 50 as Mike noted. I used the Pascal’s Triangle method to find that there is a 12% chance of such an outcome (or worse). Not statistically enough to infer unfairness.
Nicely posed Mike !!
And … thanks to Ravvie for taking us all there !
Every day is a school day, as they say 
I did the maths bit quite quickly, but getting Excel to show the graph took a bit longer!
It does show that the percentage of heads gets closer to 50% as the number of tosses increases.
Bullseye …
The maximum score using the usual three darts on a standard dart board is 180, ie treble twenty.
There are a few scores that you can’t get with three darts (sets in which darts miss the dartboard, or fall out, thus leaving only one or two darts, are null and void)
What is the lowest score you can’t get with three darts ?
PS (other than the most obvious scores of either 1 or 2)
I make it 159?
All multiples of 3 are possible. 159/3 = 53 which is 20+20+13 for example.
My Dad was a good pub darts player, he would have known the answer instinctively!
Just to refresh memories …
Inner Bull = 50
Outer Bull = 25
Outer Track = 2xValue
Inner Track = 3xValue
Highest score possible = 3 darts into the Inner Track on 20 = 180
I should eliminate the (trivial ?) answers ie you can’t score 1 or 2. (apologies for not making this statement in the initial post (I will rectify this now)
Close Mike, but as Ravvie has noted … treble 20; treble 20; treble 13
PS I have added a note to the initial teaser to exclude a score of either 1 or 2. Mrs D claimed a victory with both of these
The lowest score, and Mike was close with 159, is 163.
Cheers
Don
Twisting Algebra
To help my grandson revise for his up-coming GCSE maths exams, I wrote out a range of typical questions on individual cards. He or I could then select a card at random from the pack and we worked through about a dozen questions each evening.
I have replicated one of the cards/questions below. As you can see, it’s a straightforward bit of algebra to solve for ‘X’
I gave him a couple of minutes whilst I went to the fridge to get us each a beer. When I returned with the beers, he had solved the equation, but instead of my answer, he had obtained two new answers, neither of which were the same as mine !!
He insisted that his answers were correct, and he was right, they were !
My answer was also correct !
What were the three answers that we had obtained ?
Clever teaser!
I suspect your answer is X = 21, based on the equation as displayed.
Your grandson read the equation upside down. This reduces to X squared = 9, giving X = 3 or X = -3.
Brilliant !
Well done Steve.
Door Keypads
You might need to ‘click’ the ‘Enlarge Image’ icon, bottom right of the picture, in order to read the text
I’m going have a stab at this and say the same number of codes are available on both pads. The & is a symbol and not a digit.
Well done Pete !
Not quite the answer I was expecting, but I have to agree that the ‘&’ symbol is not a digit and therefore your answer certainly justifies a “Well Done”
I will modify the diagrams and re-post the teaser showing only well-worn keys with digits
Well done Pete.
Door Keypads
You might need to ‘click’ the ‘Enlarge Image’ icon, bottom right of the picture, in order to read the text
The diagram in this post is a (slightly) modified version of the original. It shows Key ‘7’ on Door One as being worn, rather than the ‘&’ symbol as shown on the original post.
(Thank you Pete)
I would have thought more with 4.
For example, on the first keypad. 2479 would be one distinct code, and 2497 another one. I think there are 4! or 24 different combinations. If you were unlucky, you would need to test all 24 before you got the correct one.
On the second keypad, there are also 24 combinations, but some of them are repeated, so you would need fewer than 24 attempts (half as many I think).
They are, but any one of the three digits can be repeated.
Mrs R used the following logic:
Door 1: 4x3x2x1 = 24
Door 2:
3 choices of doubles
6 positions for doubles, this is C(4,2) or choose any two from four for mathematicians that like combinatorics
2 ways for the remaining non-doubles.
Hence 3x6x2=36, so better to have chosen door 1.
