Motor vehicles arriving at an intersection can turn right or left or continue

straight ahead. In a study of traffic patterns at this intersection over a long

period of time, engineers have noted that 40 percents of the motor vehicles

turn left, 25 percents turn right, and the remainder continue straight ahead.

For the next five cars entering the intersection, what is the probability that

at least one turn right?

# [Math] Motor vehicle at an intersection probability

probabilityprobability distributions

#### Related Solutions

My simulation program works by first shuffling the entire $52$ card deck and then searching the entire deck for candidate wins for each player, then simply selecting the candidate win with the lowest drawn card position. Normally there is only $1$ candidate win per winning hand as can be seen by only $769$ candidate ties. A candidate tie is when the program detects a possible win for both players, which are normally at different final drawn card positions, but very rarely will be the same. In $22$ cases they were the same, thus resulting in a tie. It was easier for me to let the program search the entire $52$ card deck then to just draw $1$ card at a time looking for winners after each card (after the $6$th card is drawn which is the minimum # of cards needed for a player D win).

I have some results from a simulation program. Also it is interesting that since the game is so close to $50/50$, as the simulation runs, the partial results show C sometimes winning. This is why $1$ million hands is not enough.

$1,000,000,000$ : simulated hands.

$24$ hours : approx runtime using interpreted language.

$11,500$ : approx # of hands played per second.

$3.06$ Ghz dual core Intel : CPU speed but only $50$% used by simulation.

$152,981$ : wins for player C.

$157,822$ : wins for player D.

$49.22$% : $50.78$% ... approx ratio of C wins to D wins.

$99.9689$% : approx percentage of nonwinning hands (deck exhausted without a winner).

$41.1453$ : approx # of average cards drawn for C win (only counting when C actually wins).

$28.9136$ : approx # of average cards drawn for D win (only counting when D actually wins).

$769$ : candidate ties (both players have possible win in entire shuffled deck).

$22$ : actual ties.

$35:1$ approx candidate tie to actual tie ratio.

$6537$ : approx average hands per C win.

$6336$ : approx average hands per D win.

$3217$ : approx average hands for any win (or tie).

$45,454,545$ : approx average hands for a tie.

I am hoping someone else can do some analysis too to help confirm these numbers. Perhaps some mathematical calculations to help show that player D has a slight advantage. Also if someone would like some other stats tracked in the simulation program just let me know in the comment section and I can try to put it in and then post those results.

Here are the patterns we need to check for a possible C win. Note each length $3$ straight can appear in any order so for example, $234678TJQ$ and $678234TJQ$ are equally good winners for player C. Those are just $2$ of the $6$ possible permutations of each of these $10$ main patterns. So there are $ 60$ actual patterns including permutations. The comma indicates any number of cards (including $0$ cards) may appear as long as the pattern continues. For example, in pattern $1$, if the drawn cards are $K234J7ATJQ2495678$, that is a winner for player C cuz it follows (a permutation of) a winning pattern.

$~~1)~234,678,TJQ$

$~~2)~234,678,JQK$

$~~3)~234,678,QKA$

$~~4)~234,789,JQK$

$~~5)~234,789,QKA$

$~~6)~234,89T,QKA$

$~~7)~345,789,JQK$

$~~8)~345,789,QKA$

$~~9)~345,89T,QKA$

$10)~456,89T,QKA$

The winning patterns for D are (no permutations are allowed):

$~~1)~234567$

$~~2)~345678$

$~~3)~456789$

$~~4)~56789T$

$~~5)~6789TJ$

$~~6)~789TJQ$

$~~7)~89TJQK$

$~~8)~9TJQKA$

$~~9)~223344$

$10)~334455$

$11)~445566$

$12)~556677$

$13)~667788$

$14)~778899$

$15)~8899TT$

$16)~99TTJJ$

$17)~TTJJQQ$

$18)~JJQQKK$

$19)~QQKKAA$

Note that for player D, NO permutations of these $19$ patterns are allowed. They must appear EXACTLY as shown here. For example, $QQJJKK$ is NOT a win for player D. This is a clarification not a rule change.

It is somewhat amazing to me that wins are rare for both C and D (based on the average number of hands between wins), but they are almost equally as rare, making it almost $50/50~$ like a fair coin toss.

In case anyone is interested, I had the program display the cards for all $22$ ties. It displays the entire deck in the shuffled order for that hand so you can just trace where the tie occurred. For example, in the first tie, the relevant cards are $,345,89T,9TJQKA$. Here is a cropped partial screenshot for the simulation output for the ties: (T= ten).

I will now attempt to solve this mathematically but in a very simple way.

Looking at player's D's ways to win, there are only $2$ patterns. Pattern $1$ which is a $6$ card straight has probability $32/52 * 4/51 * 4/50 * 4/49 * 4/48 * 4/47$ but this is the probability of getting the $6$ card straight on the first $6$ cards drawn so we multiply that by $47$, assuming the straight can appear equally likely anywhere in the deck. So far that is about $1 / 9518$.

Next we have player D's 2nd pattern such as ($223344,334455, ... ,QQKKAA$) so that is $44/52 * 3/51 * 4/50 * 3/49 * 4/48 * 3/47$ but again we have to multiply that by $47$ since that $6$ card pattern can appear anywhere in the deck so we get $1/16407$. Adding the 2 probabilities we get $1/6024$ which is fairly close to the probability reported by the simulation program which was $1/6336$. The slight difference could be a combination of interaction with C's wins, ties, and the patterns not being uniformly distributed in all $47$ positions, but considering the simplicity of the calculations, it is quite a good approximation, only roughly $5$% off. For example, imagine we get $234567$ very late in the deck. It is quite possible that $234$, $345$, or $456$ can give C the win instead, or $567$ can give C and D a tie. Thus we would expect D's win chances to be lower than $1/6024$ which means the denominator has to be larger, more like $1/6336$. Also we must consider cases where multiple $6$ card straights appear. For example, it maybe possible that $234567$ appears early in the deck but something like $6789TJ$ appears much later (towards the end of the deck). Using my simple math, I would overcount these. I think worst case there could be $8$ straights of length $6$ in the same deck but that would be an astronomically rare event. Perhaps something like $234567,~234567,~234567,~234567,~89TJQK,~89TJQK,~89TJQK,~9TJQKA$. In my simulation program, I could count up how many times a straight of length $6$ appears multiple times in the same shuffled deck and add that to the table of results.

So I have done some of the work for you. I just need someone to somehow compute the interactions and account for those in the final probability. I will attempt to calculate C's win probability, also using simple math if possible.

Now let's look at C's win pattern. Let's take $234,678,TJQ$ as an example. Any one of those $3$ card patterns can appear first. Let's suppose the $234$ appeared first. The first part of this pattern MUST appear by card draw $46$ to allow ample time for the other part of the pattern to appear. To get $234$ on the first $3$ card draw we have probability $4/52 * 4/51 * 4/50$ = $1/2072$. From here it gets more complicated. Placing the three $3$ card straights might be one of those stars and bars type problems. Just find the probability of getting the $234,678,TJQ$ in order (without completing a $6$ card straight) and then multiply by $6$ to account for the possible permutations of them. Then multiply by $10$ to account for the $9$ other similar win patterns for C. This should be a rough approximation. Treat the $234$ as one entity and just place it anywhere in the $52$ card deck and do the same for the $678$ and $TJQ$. That is stars and bars. I think one way to solve this is to use stars and bars with a gap space of at least $2$ (such as $2,3,4,i,j,6,7,8$ so we are guaranteed not to have a $6$ card straight, then handle the cases where the gap space is $1$ separately so that we can make sure we only place cards there that will not complete the $6$ card straight. Also if anyone needs a certain count of some pattern that will help them to use math to solve this just let me know and I can try to create a bucket for it in my sim prog. I would love to get an exact answer to this problem but based on the lack and answers here (even with a $150$ bounty), I am assuming it is not easy.

As you said, the mean rate of arrival is $\lambda = \frac{1}{4} \frac{\text{vehicles}}{s}$. By the problem statement, the probability of $n$ cars passing in a time interval $t$ is $$ P_n(t) = \frac{(\lambda t)^n}{n!}e^{-\lambda t}.$$ If $F_n(t)$ is the probability that $n$ cars have passed at time $t$, the probability that $n$ cars have also passed at time $t + \delta t$ is $$ F_n(t+\delta t) = Pr(\text{n cars have passed in t})Pr(\text{no cars have passed between } t \text{ and } t+ \delta t)$$ $$ F_n(t+\delta t) = F_n(t) (1-\lambda \delta t).$$ Taking $\delta t \rightarrow 0$ and using the definition of the derivative gives the differential equation $$ \frac{d}{dt}F_n(t) = -\lambda F_n(t)$$ for the probability that only $n$ cars have passed at time $t$. This has the solution (it's separable) $$ F_n(t) = \lambda e^{-\lambda t}.$$ Therefore, the distribution of times between cars passing is an exponential distribution with mean $\lambda$. The probability that the gap is greater than $8$ seconds is $$Pr(t>8) = \int_{8}^{\infty}\frac{1}{4} e^{-\frac{1}{4}t},$$ which you can easily evaluate.

## Best Answer

Let $L$ be the event that the vehicle turns left, $S$ be the event that the vehicle goes straight, and $R$ the event the vehicle turns right. Thus, the probability that the vehicle turns left OR goes straight is $P(L) + P(S)$ since the events are mutually exclusive and independent. Call this probability $P(LS)$. Then $P(LS)$ = $P(L)$ + $P(S)$ = 0.40 + 0.35 = 0.75.

Given the above, the probability that the next 5 cars go straight or left is equal to $P(LS)^5$ due to independence. So, the probability that at least 1 car goes right is simply the complement of this; that is, $P(R) = 1 - P(LS)^5$ = 1 - $0.75^5$ = 0.7626953.