Two dice are biased, so that the probability of getting a 6 on each dice equals 0.3. The two dice are rolled, and by using a tree diagram or otherwise, I would like to find:
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P(two 6's)
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P(exactly one 6)
How can this be achieved?
[Math] Probability on biased dice
diceprobability
Related Solutions
For a single die, let's find the expected value for the number of hits.
You have a $\dfrac{1}{2}$ chance of not getting any hits on one die.
You have a $\dfrac{2}{6} + \dfrac{1}{6}\cdot \dfrac{1}{2} = \dfrac{5}{12}$ chance of getting exactly 1 hit on one die.
You have a $\dfrac{1}{6}\cdot \dfrac{1}{3}+\dfrac{1}{6^2}\cdot \dfrac{1}{2} = \dfrac{5}{72}$ chance of getting exactly 2 hits.
In general, for $h>0$, the probability of exactly $h$ hits is $$\dfrac{5}{2\cdot 6^h}$$
I am not going to go into the proof of this, but I will demonstrate that the total probability is preserved.
$$\sum_{h\ge 1} \dfrac{5}{2\cdot 6^h} = \dfrac{5}{12}\sum_{h\ge 0} \left(\dfrac{1}{6}\right)^h = \dfrac{5}{12}\cdot \dfrac{6}{5} = \dfrac{1}{2}$$
Plus the probability of $h=0$, which is $\dfrac{1}{2}$.
So, this gives the expected value:
$$\dfrac{5}{12}\sum_{k\ge 0}(k+1)\left(\dfrac{1}{6}\right)^k = \dfrac{3}{5}$$
So, that is the expected value for the number of hits.
Continuing, let's create some notation.
Let $P(n,h)$ be the probability of rolling $n$ dice and getting exactly $h$ hits. Let $P(n,h^+)$ be the probability of rolling $n$ dice and getting at least $h$ hits.
$$P(n,h) = \sum_{\sum_{i=1}^n a_i = h}\prod_{i=1}^n\begin{cases}\tfrac{1}{2}, & a_i = 0 \\ \tfrac{5}{2\cdot 6^{a_i}}, & a_i>0\end{cases}$$
Basically, you are summing over all nonnegative integer solutions to the Diophantine equation: $$a_1+\cdots + a_n = h$$
This tells you which dice rolled hits, and the total number of hits is $h$. Then, multiply the probabilities that the chosen die scores that many hits.
This is an impractical approach. Instead, we can try to break down the probabilities.
Example:
$$P(4,2^+) = 1-P(4,0)-P(4,1) = 1-\left(\dfrac{1}{2}\right)^4-\dbinom{4}{1}\left(\dfrac{5}{12}\right)\left(\dfrac{1}{2}\right)^3 \approx 72\%$$
What I am doing here is for $P(4,0)$, this is the probability that not a single die rolled a hit. For $P(4,1)$, this is the probability that exactly one die rolled exactly one hit.
$$P(4,5^+) = 1-P(4,0)-P(4,1)-P(4,2)-P(4,3)-P(4,4)$$
This is a bit trickier to calculate.
$$\begin{align*}P(4,4) & = \dbinom{4}{1}P(1,4)P(3,0)+\dbinom{4}{2}\dbinom{2}{1}P(1,3)P(1,1)P(2,0)+\dbinom{4}{2}P(1,2)^2P(2,0)+\dbinom{4}{3}\dbinom{3}{1}P(1,2)P(1,1)^2P(1,0)+P(1,1)^4 \\ & = 4\left(\dfrac{5}{2\cdot 6^4}\right)\left(\dfrac{1}{2}\right)^3+12\left(\dfrac{5}{2\cdot 6^3}\right)\left(\dfrac{5}{12}\right)\left(\dfrac{1}{2}\right)^2+6\left(\dfrac{5}{72}\right)^2\left(\dfrac{1}{2}\right)^2+12\left(\dfrac{5}{72}\right)\left(\dfrac{5}{12}\right)^2\left(\dfrac{1}{2}\right)+\left(\dfrac{5}{12}\right)^4 \\ & = \dfrac{865}{6912}\end{align*}$$
Essentially, what I am calculating is the probability of the following:
Only one die hits, but four times
2 dice have hits, one rolls 3 hits one rolls 1 hit
2 dice hit, both roll 2 hits
3 dice hit, one rolls 2 hits, two others roll 1 hit each
4 dice hit, one hit each
$$P(4,3) = \dbinom{4}{1}\left(\dfrac{5}{2\cdot 6^3}\right)\left(\dfrac{1}{2}\right)^3 + \dbinom{4}{2}\dbinom{2}{1}\left(\dfrac{5}{72}\right)\left(\dfrac{5}{12}\right)\left(\dfrac{1}{2}\right)^2+\dbinom{4}{3}\left(\dfrac{5}{12}\right)^3\left(\dfrac{1}{2}\right) = \dfrac{205}{864}$$
$$P(4,2) = \dbinom{4}{1}\left(\dfrac{5}{72}\right)\left(\dfrac{1}{2}\right)^3+\dbinom{4}{2}\left(\dfrac{5}{12}\right)^2\left(\dfrac{1}{2}\right)^2 = \dfrac{85}{288}$$
$$P(4,5^+) = \dfrac{55}{768}$$
Edit: How to build an Excel spreadsheet to perform these calculations:
On Sheet1 - Probability of getting exact number of hits: Cell C1: Number of Dice Cell C2: 0 Cell D2: 1 Cell E2: 2 . . . Cell M2: 10 (go as high as you like) . . . Cell A3: Number of Hits Cell B3: 0 Cell B4: 1 . . . Cell B13: 10 (go as high as you like) . . . Cell C3: =POWER(0.5,C2) Copy this cell and paste formula to all cells C3:M3 Set Cells C4:C13 to 0 Cell D4: =5/(2*POWER(6,B4)) Copy this cell and paste formula to all cells D4:D13 Cell E4: =E2*$D4*POWER(0.5,D2)) Copy this cell and paste formula to all cells E4:M4 Cell E5: =SUMPRODUCT($D$3:$D5,N(OFFSET(D5,ROW(D$3)-ROW(D$3:D5),0))) Copy this cell and paste formula to all cells E5:M13 At this point, you should have a table filled out with all probabilities for getting the exact number of hits (0 to 10) on any number of dice (0 to 10). Rename the spreadsheet: NumberHitsCalculator (or whatever else you want to name it, but I am using this name below). Add a new worksheet. I called it ThresholdCalculator. Copy everything from NumberHitsCalculator to get the same general layout. Change cell A3 to: Minimum Number of Hits Change cell C3:M3 to be 1 across the board (you will always get at least zero hits). Cell ThresholdCalculator!D4: =D3-NumberHitsCalculator!D3 Copy this cell and paste formula to all cells: ThresholdCalculator!D4:M13 This will give you all of the probabilities of hitting specific thresholds given the number of dice. Examples we have already gone over, like P(4,2+) gives 0.729167 as expected. P(4,5+) gives 0.071615 as expected, as well.
The correct answer should be to guess $\frac{25}{3}$ plus your total number of "sixes." Noting that this isn't an integer, round to the nearest, so $8$ plus your total.
More generally, you should guess $\frac{1}{3}$ times the number of dice that you didn't throw plus the amount of sixes (or ones) that you did throw and rounding to the nearest integer.
This is again a trivial application of linearity of expected value and the binomial distribution along with basic properties of expected value.
"Fake" sixes versus natural sixes doesn't matter.
Best Answer
The probability of getting two 6's equals $0.3 \cdot 0.3 = 0.09$
The probability of getting one 6 equals $0.3 \cdot 0.7 + 0.7 \cdot 0.3 = 0.42$