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I've got a puzzle for you, OT.

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You are given the choice of playing one of three different games.


The first game is played with 2 six-sided dice numbered one through six.


The second game is played with 3 four-sided dice numbered one though four.


The third game is played with 6 coins, where heads is counted as one and tails is counted as two.


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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0.

 
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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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1337

 
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I suck at math :(

 
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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 six-sided dice with seven as my chosen number is my best bet. I didn’t do the math on the three four-sided 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 6-12.

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, 7-8 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:

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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91238576

 
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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 2-12

The second 4-12

The third 6-12

So I would pick the last since I am most likely to get higher numbers.