# Probability with loaded and fair dice

I own five different six-sided dice. Four of the dice are fair dice, meaning they have values 1, 2, 3, 4, 5, 6. However, one of the dice is loaded; thus, it never shows 1, 2 or 3, but is equally likely to show the values 4, 5, or 6. For my experiment, I will pick up one random dice and roll it twice.

The first thing I would like to calculate is the probability of getting two sixes. To calculate this, I first calculated the probability of getting one six and multiplied it by two. Suppose $S$ = event that two sixes are rolled. $$P(S) = 2(\frac45(\frac16) + \frac15(\frac13)) = .4$$ However, I am not sure if this is correct. I need to calculate this because I would also like to calculate $P(L|S)$ where L = event that a loaded die was picked. Additionally, I feel this is incorrect, because if I change the '2' to a '10' to calculate it for 10 rolls instead of 2, I get a value over 1 which makes no sense. To summarize, how can I calculate $P(S)$ properly so I can calculate $P(L|S)$?

• Probability that "two sixes are rolled" is NOT at all equal to 2 * probability of "six is rolled the first time". With your approach you will get that probability of "100 sixes in 100 experiments" is much more that 1, isn't it? – lesnik Feb 10 '17 at 6:46

The $\frac16$ and $\frac13$ should be squared at first, not doubled at the end. The correct calculation for $P(S)$ is $$P(S)=\frac45\cdot\frac1{6^2}+\frac15\cdot\frac1{3^2}=\frac1{45}+\frac1{45}=\frac2{45}$$ The second term above is $P(L\cap S)$, so $$P(L\mid S)=\frac{P(L\cap S)}{P(S)}=\frac{1/45}{2/45}=\frac12$$