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A gambler made three 6-sided dice.
The red die had the numbers 2,4 and 9 each appear twice on its faces.
The blue die had the numbers 3,5 and 7 eachappear twice on its faces.
The yellow die had the numbers 1,6 and 8 each appear twice on its faces.

The total of faces of each dice is the same. In the Gambler's Game  his opponent chooses any die first and then rolls it. The gambler then  chooses a die and rolls it. He is confident he can always choose a die  that would give him a better chance of obtaining a higher score. Explain  why the gambler is correct to be confident and provide calculations to  support your explanation.
Describe the variation on the gambler's game that would give him even a better chance of winning.
Create another set of three dice that would serve the gambler  equally as well, but this time each face on a single die must be a  different number and no number can be used more than once. Again use  calculations to explain why your new dice also work.



6 years ago

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### Answers : (1)

										Dear sunny,
We can ignore the fact that every number appears twice. In the real world, 3-sided dice would be hard to make, which is probably why this is in terms of regular 6-sided dice.  Obviously, the Gambler is sure to lose if the player chooses red and rolls a 9. And he is not likely to win if the players rolls a 7 or 8 on the other dice. So he cannot always be "confident" if he chooses after the player rolls.  So let's say that the order of play is: Player chooses die, Gambler chooses die, they both roll.  Red vs blue: 1/3 chance red rolls a 2 x 100% chance blue wins = 1/3 1/3 chance red rolls a 4 x 2/3 chance blue wins = 2/9 1/3 chance red rolls a 9 x 0 chance blue wins = 0 Total 5/9 chance for blue   Red vs yellow: 1/3 chance red rolls a 2 x 2/3 chance yellow wins = 2/9 1/3 chance red rolls a 4 x 2/3 chance yellow wins = 2/9 1/3 chance red rolls a 9 x 0 chance yellow wins = 0 Total 4/9 chance for yellow vs red, 5/9 for red vs yellow  Yellow vs blue: 1/3 chance yellow rolls 1 x 100 % chance blue wins = 1/3 1/3 chance yellow rolls 6 x 1/3 chance blue wins = 1/9 1/3 chance yellow rolls 8 x 0 chance blue wins = 0 Total 4/9 for blue, 5/9 for yellow.  So, if player chooses red, Gambler chooses blue and has 5/9 chance If player chooses yellow, Gamble chooses red and has 5/9 chance If player chooses blue, Gambler chooses yellow and has 5/9 chance.  No matter what the player chooses, Gambler can choose a die which has a better chance.  2) I don't know what they are looking for  3) 1,1,6,6,8,8 → 1,2,11,12,15,16 2,2,4,4,9,9 → 3,4,7,8,17,18 3,3,7,7,8,8 → 5,6,13,14,15,16 It works out the same.
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6 years ago

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