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The Lady or the Lions

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A Dicey Paradox

Four fair dice are marked differently on their six faces. Choose first ANY one of them. I can always choose another that will give me a better chance of winning. Investigate.

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Nines and Tens

Explain why it is that when you throw two dice you are more likely to get a score of 9 than of 10. What about the case of 3 dice? Is a score of 9 more likely then a score of 10 with 3 dice?

Non-transitive Dice

Stage: 3 Challenge Level: Challenge Level:2 Challenge Level:2

Victor, from St Paul's school in Brazil advised Alison to choose dice in the following way:

If Charlie has selected the red die, Alison is better off with the green die because it is better than the red in two numbers, if Charlie has selected the Green Alison is better choosing the Blue because it is better than the green in two ways. If Charlie chooses the blue Alison chooses the Red because it is better than the blue in two ways.

Emmanuel from Loreto College  and Chris, Natalie and Gino from St Andrews School in Thailand came to the same conclusion.

George from Canberra Grammar School explained his answer using some probabilities:

 

It was a good idea to allow Charlie to go first. This allows Alison to make a choice for her die that increases her chances of winning as follows:

1. Charlie will probably rush to pick the green die to have a chance at rolling 9. He only has a $\frac{1}{3}$ chance of rolling 9. Alison should pick the blue die as in the case Charlie does not roll a 9, Alison has at least a $\frac{2}{3}$ chance of rolling a higher number.

2. If Charlie picks the red die, Alison should pick the green die. Alison has a $\frac{1}{3}$ chance of winning the game if she rolls a 9. If Charlie rolls a 1, Alison is certain to win.

3. Lastly, if Charlie picks the blue die, Alison should pick the red die. If Charlie rolls a 7, Alison has a $\frac{1}{3}$ chance of winning. If Charlie rolls any other number, Alison has at least a $\frac{2}{3}$ chance of winning.

 

Josh from Wilson's School listed all the possibilities:

Red vs Green: 1-2, 1-2, 1-4, 1-4, 1-9, 1-9 (x2), 6-2, 6-2, 6-4, 6-4, 6-9, 6-9 (x2), 8-2, 8-2, 8-4, 8-4, 8-9, 8-9 (x2)

Green wins 20 times, Red wins 16 times.

Red vs Blue: 1-3, 1-3, 1-5, 1-5, 1-7, 1-7 (x2), 6-3, 6-3, 6-5, 6-5, 6-7, 6-7 (x2), 8-3, 8-3, 8-5, 8-5, 8-7, 8-7 (x2)

Red wins 20 times, Blue wins 16 times.

Blue vs Green: 3-2, 3-2, 3-4, 3-4, 3-9, 3-9 (x2) 5-2, 5-2, 5-4, 5-4, 5-9, 5-9 (x2), 7-2, 7-2, 7-4, 7-4, 7-9, 7-9 (x2)

Blue wins 20 times, Green wins 16 times.

 

Komal from Alexandra School in India explained:

It is not a good idea to go first because whichever die Charlie selects there is another die which has a higher probability of winning against him.

Alison can select her die using the following logic -

1. If Charlie selects the red die ${1,1,6,6,8,8}$ then Alison should select the green die ${2,2,4,4,9,9}$:

The probability of Charlie rolling a 1 is 2 from 6 and the probability of Alison rolling a greater number is 6 from 6.

The probability of Charlie rolling a 6 is 2 from 6 and the probability of Alison rolling a greater number is 2 from 6

The probability of Charlie rolling a 8 is 2 from 6 and the probability of Alison rolling a greater number is 2 from 6

Therefore the total number of ways Alison can win is (6 + 2 + 2) = 10 out of 18.

 

The blue die is ${3,3,5,5,7,7}$. 

The probability of Charlie rolling a 1 is 2 from 6 and the probability of Alison rolling a greater number is 6 from 6.

The probability of Charlie rolling a 6 is 2 from 6 and the probability of Alison rolling a greater number is 2 from 6.

The probability of Charlie rolling a 8 is 2 from 6 and the probability of Alison rolling a greater number is 0 from 6

Therefore the total probability of Alison winning is (6 + 2 + 0) = 8 out of 18.

There are similar arguments to show that Alison should choose the blue die if Charlie chooses the green die.