In a certain game, two unfair six-sided dice are rolled. One dice is red, the other is blue. Both dice have 1,2,3,4,5,6 as values. These dice are rolled at the same time. Both dice have 10 life points each.
Upon rolling, if red shows 2, 2 points are deducted from the blue's life points. Likewise, if blue shows 3, 3 points are deducted from red's life points. Whatever number appears from red, is deducted from blue life points, and vice versa.
We keep rolling both dice, until one dice' life points gets to 0. Last dice standing, wins.
What is the expected number of rolls, for this game? If given the chance to bet which dice will win, what's the chance that red wins? Also, what's the chance that blue wins?
At first I took this as a P(A U B) = P(A) + P(B) - P(A&B) but I am not getting the correct results. I do get P(A) and P(B), but have been stuck getting P(A&B). I may not be using the right formula anyway.
P(A) is the probability space for red, P(B) is for blue.
Sample probability distribution: P(A) -- {1,2,3,4,5,6} = {0.1, 0.15, 0.2, 0.25, 0.05, 0.25} P(B) -- {1,2,3,4,5,6} = {0.15, 0.2, 0.1, 0.25,0.25,0.05}
Any help appreciated, thanks.
You can deduct the health of each dice with its own score and then say that whoever reaches zero first wins.
Basically you have to calculate the expected number of trials it takes for a dice to get to zero or less starting from 10. Let e(10) be the expected number.
e(10) = sum( e(10 - x) + 1 )* pr(x) ; for x in 1 to 6. This can be easily calculated given pr(x) .
Then expected number of trials would be minimum of both values calculated.