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You are given the choice of playing one of three different games.
* * *
The first game is played with 2 sixsided dice numbered one through six.
![](http://www.thebigquestions.com/wpcontent/uploads/2011/08/dice.jpg)
* * *
The second game is played with 3 foursided dice numbered one though four.
![](http://www.shamusyoung.com/twentysidedtale/images/four_sided1.jpg)
* * *
The third game is played with 6 coins, where heads is counted as one and tails is counted as two.
![](http://www.murderousmaths.co.uk/books/6coins/6coins1.gif)
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In each game, you must choose a number from one through twelve. The dice/coins are then cast and if the values of the dice/coins total up to the number you chose, you win.
So, OT, which game and which number will give you the greatest probability of winning?


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Not really using any math, but I’m guessing numbers 6~8 will yield the highest possibility with the second game.


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First game’s potential results: 2 – 12
Second game’s potential results: 3 – 12
Third game’s potential results: 6 – 12
I’m thinking the last game has the best chance of rolling what you guess, since there’s only six possible outcomes, whereas the first and second have ten and nine respectively.
* * *
Is there a correct answer here?


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I’m assuming that the two sixsided dice with seven as my chosen number is my best bet. I didn’t do the math on the three foursided dice so I could be wrong.
Forget the math though, I place all of my money of the six quarters with five as my chosen number.


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> Forget the math though, I place all of my money of the six quarters with five as my chosen number.
It’s impossible to get five on the last game.
The lowest you can get is a six.


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So, the numbers we can choose are 612.
In the 1st game, we have 1/36 for 2 and 12, 2/36 for 3 and 11, 3/36 for 4 and 10, 4/36 for 5 and 9, 5/36 for 6 and 8 and 6/36 for 7.
In the 3st game, we have 1/64 for 6, 6/64 for 7, 15/64 for 8, 20/64 for 9, 15/64 for 10, 6/64 for 11, 1/64 for 12
The 2nd game is the hardest to calculate. But the possibilities are like this: 7=8\>6=9\>5=10\>4=11\>3=12
The best number for 1st game is 7, 78 for 2nd game and 9 for 3rd game.
The best number to choose is 8 if we want the highest possibility of getting the number we chose in all games.


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> *Originally posted by **[HappyAlcoholic](/forums/2/topics/335234?page=1#posts7019230):***
> > Forget the math though, I place all of my money of the six quarters with five as my chosen number.
>
> It’s impossible to get five on the last game.
> The lowest you can get is a six.
That was the joke. It is obvious that I have no chance of winning that bet.


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I just came back from school very tired, and i’m too lazy to think right now.


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I’m too lazy to think today.
I shall return tomorrow and submit my answer.


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wtf i have no time for this but i say 4.


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I choose the coins and the number 7
The dice could be loaded.


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The first game at least 212
The second 412
The third 612
So I would pick the last since I am most likely to get higher numbers.
