A deck of cards contains 6 cards numbered from 1 to 6. If three cards are randomly selected one-by-one from the deck, what is the probability that the numbers on the cards drawn are consecutive and in ascending order?
Soln: The possible combinations to draw 3 cards in consecutive and in ascending order are:
{1,2,3},{2,3,4},{3,4,5},{4,5,6}. There are a total of 6*5*4=120 ways of picking 3 cards. So our answer is 4/120=1/30.
I am trying to use another method to solve this problem. I am trying to understand when to use the "place-holder" method and when not too.
Let say I picked a "1" for my first card. So 1 __ __ with two slots available. The Prob("2" is selected for the 2nd slot)=1/5. Prob("3" is selected for the 3rd slot)=1/4. Similarly, if I had already picked a "2" for my first card, so 2 __ __. Prob("3" is selected for the 2nd slot)=1/5. Prob("4" is selected for the 3rd slot)=1/4. Etc.
So combined the prability is =(1/5)(1/4)+(1/5)(1/4)+(1/5)(1/4)+(1/5)(1/4)=4*(1/5)(1/4)=1/5 What am I doing wrong? Why can't I just make a place holder for the first card and then consider the probabilities for the remaining two?
You forgot to take into account the probability of getting that first $1$ (or $2$, or $3$, or $4$) in the first place: after all, your first card might be a $5$ or $6$, in which case you’re out of luck right away. If your first card is $1$, then the probability of drawing the $2$ and $3$ in that order is $\frac15\cdot\frac14$, but the probability of drawing the $1$ in the first place is $\frac16$, so the overall probability of the $123$ sequence is $\frac16\cdot\frac15\cdot\frac14$. Thus, you really end up with $4\cdot\frac16\cdot\frac15\cdot\frac14=\frac16\cdot\frac15=\frac1{30}$, just as before.
You could also argue like this: the probability of getting a usable card (meaning $1,2,3$, or $4$) on the first draw is $\frac46=\frac23$. Once you’ve drawn it, the probability of drawing the next card in ascending sequence on the second draw is $\frac15$, and the probability of drawing the right card on the third draw is $\frac14$, so the overall probability of success is $\frac23\cdot\frac15\cdot\frac14=\frac1{30}$.