### CONDITIONAL PROBABILITY:

From this post:

If I toss a biased coin with $$2/3$$ chance of landing on heads, given that there was at least one head in $$3$$ flips, what is the probability that there is only $$1$$ head?

There are $$8$$ possible outcomes for the three tosses. As the probabilities for $$T,H$$ are not the same, these outcomes are not equally probable.

There are seven outcomes that have at least one $$H$$. They are $HHH,HHT,HTH,THH,HTT,THT,TTH$

The probability that we have at least one $$H$$ is $P_{≥1}=1-P(TTT)=1-\left(\frac 13\right)^2=0.962962963$

How do we compute the conditional probability? Well, in this case we simply divide the unconditional probability by $$P_{≥1}$$. Why? Well, in general, conditional probability is defined by $P(A\,|\,B)=\frac {P(A\cap B)}{P(B)}$ Here $$A$$ is the event “you get exactly one $$H$$” and $$B$$ is the event “you get at least one $$H$$”. Of course, in this case $A\cap B=A\implies P(A\cap B)=P(A)=P(HTT)+P(THT)+P(TTH)$ Thus our answer is the ratio $\frac {P(HTT)+P(THT)+P(TTH)}{P_{≥1}}$

Of course all three of those events, $$HTT,THT,TTH$$ have the same probability, namely $$\left( \frac 13 \right)^2\times \frac 23$$. Thus our numerator is $$3\times \left( \frac 13 \right)^2\times \frac 23=\frac 29=.222\cdots$$

Finally the answer we seek is $\frac {.2222\cdots}{0.962962963}= {0.230769231 }$

What is the probability of getting a poker hand where all the cards are $$7$$ or higher if we see that AT LEAST one of the cards is higher than $$10$$ (i.e. $$\text{J, Q, K, A}$$)?

The event $$\text{A = at least one card }>10$$ can occur in the complement of the event that all the cards are less or equal to ten ($$36=9\times 4$$ lower cards), i.e. $$2,3,4,5,6,7,8,9,10:$$

$\binom{52}{5} - \binom{36}{5}$

over the total sample space:

$\Pr(A) = \frac{\binom{52}{5} - \binom{36}{5}}{\binom{52}{5}}$

Now, naming the event $$\text{B = all cards }\geq 7$$, $$\Pr(A\cap B)$$ will be calculated by considering all the hands with cards equal or greater than $$7$$: i.e. $$7,8,9,10,J,Q,K,A$$ ($$32 = 8 \times 4$$), but subtracting those that don’t contain any cards higher than $$10$$ - i.e. hands composed of only $$7,8,9,10$$ (i.e. $$16 =4 \times 4$$):

$\Pr(A\cap B) = \frac{\binom{32}{5} - \binom{16}{5}}{\binom{52}{5}}$

Therefore,

$\Pr(B\Vert A) = \frac{\Pr(A\cap B)}{\Pr(A)}= \frac{\binom{32}{5} - \binom{16}{5}}{\binom{52}{5} - \binom{36}{5}}$

Here is the simulation in R:

cards = rep(2:14, 4)
m = replicate(1e5, sample(cards, 5, replace = F))
one_ten = m[, colSums(m > 10) > 0, drop=FALSE]
ncol(m[, colSums(one_ten > 6)  == 5, drop=FALSE]) / ncol(m)
## [1] 0.08833