[Math] Probability of visiting a city before another

probability

A businesswoman in Philadelphia is preparing an itinerary for a visit to six major cities. The distance traveled, and hence the cost of the trip, will depend on the order in which she plans her route.

a) How many different itineraries (and trip costs) are possible?

b) If the businesswoman randomly selects one of the possible itineraries and Denver and San Francisco are two of the cities that she plans to visit, what is the probability that she will visit Denver before San Francisco?

I think I understand that the number of itineraries is simply 6! because there are six cities and each time we select one we have to take that into account.

But for the second part, I am not sure what to do, does it start with findingthe probability of visiting Denver and San Francisco?

Any help or hint is well appreciated, thank you in advance!

Best Answer

Hint: presumably she visits all six cities and the only question is what order. For every order that has Denver before San Francisco, the opposite order has San Francisco before Denver.

Related Solutions

[Math] Probability of visiting $4$ cities

Let's start with part $(i)$ (the other parts will use the same idea):

If Veena is to visit $A$ before she visits $B$, then we are looking to count the number of strings using $A,B,C,D$, where $A$ precedes $B$. Such strings could have the form

$$AB__$$ $$A_B_$$ $$A__B$$ $$_AB_$$ $$_A_B$$ $$__AB$$

We didn't actually need to list these; we could have simply recognized that there are 3 possible positions for $A$ (any but the last), and that if $A$ is in position $i$ ($i = 1,2,3$), then there are $4-i$ possible positions for $B$. Therefore the number of such strings of this general form is $\sum_{i=1}^3(4-i) = 3+2+1 = 6$, which agrees with the list above.

Now all we have to do is place $C$ and $D$, and there are no restrictions on how we may do this. There are two ways to place $C$ and $D$ in the empty spaces in each string listed above, therefore we get a total of $2\cdot6 = 12$ unique strings where $A$ precedes $B$.

Thus the probability that Veena visits $A$ before she visits $B$ is $$\frac{\text{# of strings where $A$ precedes $B$}}{\text{total number of strings}} = \frac{12}{4!} = \frac{1}{2}$$

For parts $(ii)$ - $(v)$, the same general method is employed, we are just putting different restrictions on the particular types of strings we are looking for.

Help calculating probability of dice rolls in a board game

Although there is surely a closed-form solution as @BGM alluded to, you can also simulate the situation. Following is code to simulate a sampling distribution.

set.seed=126790
sims=100
probs=numeric(sims)
for(k in 1:sims)
{
  reps=1000
  rolls=matrix(NA,nrow=reps,ncol=8)
  for(i in 1:reps)
  {
    dice=c(NULL)
    for(j in 1:8)
    {
      dice=c(dice,sample(c('a','a','b','c','d','e'),1))
    }
    dice[sample(1:8,1)]=sample(c('a','a','b','c','d','e'),1)
    rolls[i,]=dice
  }
  rolls

  counter=0
  for(i in 1:reps)
  {
    if(length(grep('a',rolls[i,]))==6||
       length(grep('b',rolls[i,]))==6||
       length(grep('c',rolls[i,]))==6||
       length(grep('d',rolls[i,]))==6||
       length(grep('e',rolls[i,]))==6){
      counter=counter+1
    }
  }
  probs[k]=counter/reps
  cat(k,'\n')
}
probs
mean(probs)
sd(probs)
hist(probs)

In my run of the simulation, the mean probability of getting a 6 character match was $1.959$% with standard deviation $0.397$%. Because the standard error is the standard deviation of the sampling distribution, in 95% of repetitions of 100 sets of 8 rolls with one being replaced, between 1.18% and 2.74% will have a 6-character match. Of course this assumes a random replacement roll, but most people would try to behave strategically. However, I doubt it would change the results by very much. Additionally, this assumes the rolls are normally distributed. Visual analysis says the rolls are normally distributed.

Best Answer

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[Math] Probability of visiting $4$ cities

Help calculating probability of dice rolls in a board game

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