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. \(16 =4 \times 4\)):

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


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

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

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