### Stop or Dare

All you need for this game is a pack of cards. While you play the game, think about strategies that will increase your chances of winning.

### Game of PIG - Sixes

Can you beat Piggy in this simple dice game? Can you figure out Piggy's strategy, and is there a better one?

### Tricky Track

In this game you throw two dice and find their total, then move the appropriate counter to the right. Which counter reaches the purple box first? Is this what you would expect?

# Roll These Dice

##### Stage: 2 Challenge Level:

* I have found out that all the different possible answers are between $-4$ and $11$ including $11$ and $-4$. It is not possible to get any answers over $11$ and below $-4$.
I recorded all of this data in a table.
* I have found out that if all $3$ dice are the same the total will become the value of one of the die ( Eg. $1+1=2-1=1$ ).
* this would be the same conclusion as above if any $2$ of the die were the same. (E.g. $2+1=3-1=2$)
* I know there is a $50/50$ chance of the answer being odd or even because;
odd + odd - odd = odd
even + even - even = even
odd + even - odd = even
odd + odd - even = even
even + even - odd = odd
odd + even - even = odd
These are all the posible ways of adding the dice.
Thank you for reading my solution I hope all is correct.
Molly        ;-)
Indeed Molly it is very good and I am impressed that you did this and came to those conclusions. You could of course extend the exploration by wondering about using $4$ dice and deciding whether to subtract just $1$ of those or maybe $2$.

Ben, Harry, Will and Lucas from Tarporley Church of England School also worked on this activity and this was their report:

There are four of us so two of us wrote ALL of the combinations down [one from $6+6-1$ and one from $1+1-6$]. There were $216$ possible calculations. At the same time the other two of us worked out which is the most likely answer[which is $4$]. Once we did that we were done.

Sion from the same school added this extra piece of information;

There are $225$ ways and your answer is the numbers $3$ and $4$. By finding all the $225$ calculations you then make a tally chart to show the most popular number.  Finally you count up the number and then your answer should be $3$ and $4$.

We also had a number of good ideas from North Molton, namely, Michael, Jack, Beth, James and Sam.
Bram  from the British School of Bucharest  in  Romania , sent in what I think is the first from Romania, - well done and thanks - saying;

There is a higher probability to get $6$ than $2$ eg. there are fewer ways to get $2$ because there are $13$:

$1+2-1=2$ , $1+3-2=2$ , $1+4-3=2$ , $1+5-4=2$ , $1+6-5=2$ , $2+3-3=2$ , $2+4-4=2$, $2+5-5=2$ , $2+6-6=2$ , $3+3-4=2$ , $3+4-5=2$ , $3+5-6=2$ , $4+4-6=2$

and for $6$ there are:

$1+6-1=6$ , $2+5-1=6$ , $2+6-2=6$ , $3+4-1=6$ , $3+5-2=6$

Thanks you all, a great effort.