What does this mean for me? You will always be able to play your favorite games on Kongregate. However, certain site features may suddenly stop working and leave you with a severely degraded experience.
What should I do? We strongly urge all our users to upgrade to modern browsers for a better experience and improved security.
We suggest you install the latest version of one of these browsers:
Kongregate is a community-driven browser games portal with an open platform for all web games.
Get your games in front of thousands of users while monetizing through ads and virtual goods.
Learn more »
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?
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?
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.
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.
> *Originally posted by **[HappyAlcoholic](/forums/2/topics/335234?page=1#posts-7019230):***
> > 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.