1  SNAIL 1 WINS !!!  
2  SNAIL 2 WINS !!!  
3  SNAIL 3 WINS !!!  
4  SNAIL 4 WINS !!!  
5  SNAIL 5 WINS !!!  
6  SNAIL 6 WINS !!!  
7  SNAIL 7 WINS !!!  
8  SNAIL 8 WINS !!!  
9  SNAIL 9 WINS !!!  
10  SNAIL 10 WINS !!!  
11  SNAIL 11 WINS !!!  
12  SNAIL 12 WINS !!! 

This is a visual aid designed to be projected onto a whiteboard for whole class exposition. A printable version is here.
Two dice are rolled and the sum calculated. The snail with that number takes one step forwards. The first snail to the end of the race track wins. Before playing the students could be asked to predict which snail will win. Twelve students could be asked to volunteer to be the 'trainers' for each of the snails, their names can be typed into the spaces provided above.
This activity could be followed by the students producing the possibility space for the sums of two dice (printable empty version).
Students could then design their own race boards and make predictions:
A wonderful way to follow this activity is playing Two Dice Bingo
How did you use this resource? Can you suggest how teachers could present or develop this resource?
Do you have any comments? It is always useful to receive feedback and helps make this free resource even more useful for those learning Mathematics anywhere in the world. Click here to enter your comments.
Teacher, do your students have
access to computers? 

Here is the URL which will take them to a student version of this activity.
Ben, Wellington, New Zealand
Friday, March 11, 2016
"I am a young Maths teacher in Wellington, NZ, love your site and it regularly to engage my students. Thank you so much for your amazing efforts in helping students worldwide achieve in Mathematics. I currently teach a class of 1213 year olds and we have been analyzing the Snail Racing game for our topic on probability. We are touching on the basics of probability in our class and this game suits us perfectly.... However.... This is where it gets fun…
My Mathematical brain wants to know more about the game. We have noticed that snail 7 wins most of the time. Yes, it is the favorite, but just becait has a 1/6 chance of moving forward at each diceroll (higher then all others) it obviously does not mean it has a 1/6 chance of winning the entire race. I have put my evenings and freetime into trying to do a full analyse of snail 7's chances of winning.
The awesome part about this is it all started becaa student asked "what chance does 7 actually have of winning?" so our class is invested and I’m at my wits end with an answer.
I have made some calculations but as yet not been able to come up with a reasonable answer.
I have spent plenty of time on this, and would love to engage in a maths conversation with someone who is keen to shed some light on this problem."
Presh Talwalker, mindyourdecisions.com
Wednesday, March 16, 2016
"This is an interesting problem. Off the top of my head, this is an example of a Negative Binomial Distribution (https://en.wikipedia.org/wiki/Negative_binomial_distribution).
I think it can be calculated explicitly, but I do not know of any shortcut. For example, let's say the race needs 2 steps to win. What is the probability that snail 7 will win?
One way is to get 2 rolls for 7. (Prob = (1/6)(1/6) = 1/36)
Another way is to get 2 rolls for 7, 1 roll for any other number (The prob of two 7s is 1/36. Now we have to count the probabilities for all the combinations of getting the other rolls. This is equivalent to the number of subsets from the {2, 3, 4, 5, 6, 8, 9, 10, 11, 12}. As each of the 9 elements can be in a particular subor not (2 ways), there are 2^9 = 512 subsets to count! For example, the sub{2, 5, 9} means getting 1 roll each for 2, 5, and 9).
I suspect this could be programmed and solved. And the method could be generalized for finding other parameters of the game.
Hope this helps! ."
Richard Morpeth, Wellington
Wednesday, June 15, 2016
"Hey Ben, here's my take on this.
The chance of rolling any combination for the two dice is 1/36 which is 1/6 (probability of rolling any number from the first dice) multiplied by 1/6 (probability of rolling any number from the second dice). So to roll 1 with the first dice and 1 with the second is 1/36 and you have the same probability of rolling a 4 with a 3 etc.
So if you start looking at what the probability of getting a certain total, you need to add up the probabilities for the number of combinations for the total, ie 4 could be (1,3) (prob of 1/36), (3,1) (prob of 1/36), or (2,2) (prob of 1/36). So the probability of a 4 is 1/36 + 1/36 + 1/36 = 3/36 or simplified to 1/12.
And as for getting a total of 7, there are 6 possible combinations giving a probability of 6/36 or 1/6.
Below are a couple of links to sites that may help explain to the kids and give a visual representation.
I'm going through teacher training in Wellington and am enjoying some of the awesome resources available.
Cheers,
http://www.edcollins.com/backgammon/diceprob.htm
http://www.awesomedice.com/blog/206/ddandtheprobabilitycurve/."