1. ## Probabilities

Roll 2 dice and sum the spots. Then flip a coin that number of times. Find the exact probability of rolling a total of two if the experiment produced exactly as many heads as tails.

I recognize that it's a conditional probability but I don't know how to find the probability of getting exactly as many heads as tails.

2. Originally Posted by fulltwist8
Roll 2 dice and sum the spots. Then flip a coin that number of times. Find the exact probability of rolling a total of two if the experiment produced exactly as many heads as tails.

I recognize that it's a conditional probability but I don't know how to find the probability of getting exactly as many heads as tails.
Probability of heads is the same as tails (50%).

3. but the probability of getting exactly the same number of heads and tails is not .5

That's what I'm having trouble finding

4. Originally Posted by fulltwist8
but the probability of getting exactly the same number of heads and tails is not .5
Exactly!
In order to have an equal number of head & tails, E, you must have rolled an even sum: 2 4 6 8 10 or 12.
Here is an example. The probability of E and a sum of 4 is
$P(E4) = P(E|4)P(4) = \left( {\frac{\binom {4}{2}}{{2^4 }}} \right)\left( {\frac{3}{{36}}} \right)$.

So you must work out the following:
$P(E) = P(E2) + P(E4) + P(E6) + P(E8) + P(E10) + P(E12)$.
That is your denominator in the answer to this question which is:
$P(2|E) = \frac{{P(E2)}}{{P(E)}}$ .

5. Hello, fulltwist8!

This will take a while . . .

Roll two dice and sum the spots.
Then flip a coin that number of times.
Find the probability of rolling a total of 2
if there were exactly as many heads as tails.

We want:. $P[\text{(sum of 2) }|\text{ (H = T)}] \;=\; \frac{P[\text{(sum of 2)} \wedge \text{(H = T)}]} {P[\text{H = T}] }$

The numerator is: . $P[\text{(sum of 2) } \wedge \text{ (H = T)}]$

. . $P[\text{sum of 2}] \:=\:\frac{1}{36}$
. . With two tosses, $P[\text{H = T}] \:=\:\frac{2}{4} \:=\:\frac{1}{2}$
. . . . Hence: . $P[\text{(sum of 2) }\wedge\text{ (H = T)}] \;=\;\frac{1}{36}\cdot\frac{1}{2} \:=\:\frac{1}{72}$

The denominator is: . $P[\text{H = T}]$
We must consider all twelve oucomes of the dice.

$\begin{array}{ccccccccc}
P[\text{sum of 2}] &=& \frac{1}{36} & & P[\text{H = T}] & = & \frac{1}{2} \\
P[\text{sum of 3}] &=& \frac{2}{36} & & P[\text{H = T}] &=& 0 \\
P[\text{sum of 4}] &=&\frac{3}{36} & & P[\text{H = T}] &=&{4\choose2}\left(\frac{1}{2}\right)^4 & = & \frac{3}{8} \\
P[\text{sum of 5}] &=&\frac{4}{36} & & P[\text{H = T}] &=& 0 \end{array}$

$\begin{array}{ccccccccc}P[\text{sum of 6}] &=&\frac{5}{36} & & P[\text{H = T}] &=&{6\choose3}\left(\frac{1}{2}\right)^6 &=& \frac{5}{16} \\
P[\text{sum of 7}] &=&\frac{6}{36} & & P[\text{H = T}] &=& 0 \\
P[\text{sum of 8}] &=&\frac{5}{36} & & P[\text{H = T}] &=& {8\choose4}\left(\frac{1}{2}\right)^8 &=& \frac{35}{128} \\
P[\text{sum of 9}] &=&\frac{4}{36} & & P[\text{H = T}] &=& 0 \end{array}$

$\begin{array}{ccccccccc}P[\text{sum of 10}] &=& \frac{3}{36} & & P[\text{H = T}] &=&{10\choose5}\left(\frac{1}{2}\right)^{10} &=& \frac{63}{256} \\
P[\text{sum of 11}] &=&\frac{2}{36} & & P[\text{H = T}] &=&0 \\
P[\text{sum of 12}] &=&\frac{1}{36} & & P[\text{H = T}] &=&{12\choose6}\left(\frac{1}{2}\right)^{12} &=& {\color{blue}\frac{231}{512}}\end{array}$

Hence: . $P[\text{H = T}] \;=\;\frac{1}{36}\!\cdot\!\frac{1}{2} + \frac{3}{36}\!\cdot\!\frac{3}{8} + \frac{5}{36}\!\cdot\!\frac{5}{16} + \frac{5}{36}\!\cdot\!\frac{35}{128} + \frac{3}{36}\!\cdot\!\frac{63}{256} + \frac{1}{36}\!\cdot\!\frac{231}{512} \;=\;\frac{2941}{18432}$

Therefore:. $P[\text{(sum of 2) }|\text{ (H = T)}] \;=\; \frac{P[\text{(sum of 2)} \wedge \text{(H = T)}]} {P[\text{H = T}] } \;=\;\frac{\frac{1}{72}}{\frac{2941}{18432}} \;=\;\boxed{\frac{256}{2941}}$

I hope my arithmetic is correct . . . this time
$\begin{array}{ccccccccc}P[\text{sum of 10}] &=& \frac{3}{36} & & P[\text{H = T}] &=&{10\choose5}\left(\frac{1}{2}\right)^{10} &=& \frac{63}{256} \\