Conditional Probability Help?

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$B$ is the event that the first and second ace in the pack are adjacent.   Since we are only concerned with placement of the aces, consider a deck of 4 ace of hearts, and 48 jokers.

$C_j$ is the event that the first ace is at position $j$.   When given that event, there are $3$ remaining aces among the $52-j$ remaining cards, so the probability that one from these aces is the next card in the deck is: $\mathsf P(B\mid C_j) = \Box/\Box$


The Law of Total Probability says $\mathsf P(B)=\sum_{j=1}^{49} \mathsf P(C_j)\mathsf P(B\mid C_j)$.

We have $\mathsf P(B\mid C_j)$, so then, what is $\mathsf P(C_j)$?   The probability that the first ace will be at position $j$?

There are $\Box$ unbiased ways to select places in a deck of fifty two for four aces and of these, $\Box$ are ways to select places for the remaining aces among the places after position $j$.

Then put the sum together.


To find a simpler expression for $\mathsf P(B)$, we argue similarly.

There are $\Box$ unbiased ways to select places in a deck of fifty two for four aces and of these, $\Box$ are arrangements where the first and second ace are adjacent.

Therefore $\mathsf P(B)= \Box/\Box $?

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Yilan B.
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Yilan B.

Updated on December 28, 2022

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  • Yilan B.
    Yilan B. 10 months

    A standard deck of cards will be shuffled and then the cards will be turned over one at a time until the first ace is revealed. Let $B$ be the event that the next card in the deck will also be an ace.

    (a) Intuitively, how do you think $P(B)$ compares in size with $1/13$ (the overall proportion of aces in a deck of cards)? Explain your intuition. (Give an intuitive discussion rather than a mathematical calculation; the goal here is to describe your intuition explicitly.)

    (b) Let $C_j$ be the event that the first ace is at position $j$ in the deck. Find $P(B|C_j )$ in terms of $j$, fully simplified.

    (c) Using the law of total probability, find an expression for $P(B)$ as a sum. (The sum can be left unsimplified, but it should be something that could easily be computed in software such as R that can calculate sums.)

    (d) Find a fully simplified expression for $P(B)$ using a symmetry argument. Hint: If you were deciding whether to bet on the next card after the first ace being an ace or to bet on the last card in the deck being an ace, would you have a preference?

    Intuitively, it makes more sense that the probability of getting another ace is larger, since, after drawing one ace, $P(B) = 1/12$. However, I'm having issues determining the value of $C_j$. For instance, the probability that $C_j$ is at position $j$ is just $13/52$, right? There's only one place $j$, and there's only $13$ possible aces there.

    • Yilan B.
      Yilan B. about 6 years
      Either way, though, P(B) is always smaller?
  • Yilan B.
    Yilan B. about 6 years
    Am I correct, though, assuming then that Cj|B = 3/51?
  • neonpokharkar
    neonpokharkar about 6 years
    We can apply the same argument,
  • neonpokharkar
    neonpokharkar about 6 years
    I will explain more
  • neonpokharkar
    neonpokharkar about 6 years
    Suppose the card at $j^{th}$ and the next position are $X$,$Y$
  • neonpokharkar
    neonpokharkar about 6 years
    So $X$ can be anyone of $51$ other cards (minus the ace at $Y$)
  • neonpokharkar
    neonpokharkar about 6 years
    So total no of outcomes for $X=51$
  • neonpokharkar
    neonpokharkar about 6 years
    No of outcomes we want $X=3$
  • neonpokharkar
    neonpokharkar about 6 years
    So probability is $=\frac{3}{51}$
  • neonpokharkar
    neonpokharkar about 6 years
    @Yilan B. Bingo! Your answer is right.
  • Yilan B.
    Yilan B. about 6 years
    So, then, would the probability of P(B|Cj^c) be 4/(53-j)?
  • Graham Kemp
    Graham Kemp about 6 years
    No, @neonpokharkar . If the first ace is at position $j$ then all of the remaining three aces are somewhere among the $52-j$ cards after that first ace.