1
$\begingroup$

Question

There are $40$ games in a season.

$30$ are against class A teams which you have a $.4$ probability of winning and $.6$ of losing.

$10$ are against class B teams which you have a $.7$ probability of winning and $.3$ of losing.

The season schedule is completely random in regards to what type of team you play.

What is the probability that you:

  • win $20$ games against class A teams (and lose $10$ games)

  • AND win $3$ games against class B teams (and lose $7$ games)


My Attempt

First I created my own notation...

For example: "A a b b B..." means your first game was against a "class a team" and you won, then you played another "class a" and lost, and then you played 3 "class b teams" and lost, lost, won. I hope that makes sense.

If the probability of winning is always $.5$ regardless of the class, then I believe you'd get the answer by dividing the following:

Numerator: $\frac{40!}{20!10!3!7!}$

Denominator: $\sum_{B=0}^{10} \sum_{A=0}^{30} \frac{40!}{A!(30-A)!B!(10-B)!}$

My thinking here is it's like getting the number of unique permutations of the letters PEPPER.

But since the probability is not $.5$ you take the quotient above and multiply it by:

$.4^{20} .6^{10} .7^{3} .3^{7}$


Is this the right track? I don't think it is. That "Denominator" above is such a big number I can't do any calculations to do a sanity check. Thanks.

$\endgroup$
  • 2
    $\begingroup$ ${30 \choose 20} 0.4^{20}0.6^{10}{10\choose3}0.7^30.3^7$ $\endgroup$ – Doug M Dec 22 '17 at 2:35
  • $\begingroup$ Thanks for your help. Does that answer factor in the different schedules you could have? Can you give me the intuition why that doesn't imply your season must be all 30 type a games first and then 10 type b games at the end? That's why I thought you couldn't do that. $\endgroup$ – HJ_beginner Dec 22 '17 at 2:37
  • 2
    $\begingroup$ The order in which you play the games is completely irrelevant, as long as there are $30$ type A games and $10$ type B games, because the win probability for a given game depends only on the type and not on what other games you have played or are going to play. $\endgroup$ – Robert Israel Dec 22 '17 at 3:01
  • $\begingroup$ In the answer Doug M gave the order matters within each type of game doesn't it? For example 30 choose 20? There's only one way to have 30 wins against type A teams (30 choose 30). So is it accurate to say that order matters within type A teams and type B teams but you can mix them freely without worry of order? $\endgroup$ – HJ_beginner Dec 22 '17 at 6:07
1
$\begingroup$

Between Comments and Answers we have two different "answers" to the Question. Unfortunately, the formal Answer seems to compute to something greater than 1, which can't be the probability of anything. (It has somewhat the form of a multinomial probability, except that the probabilities don't add to 1.)

I believe the Comments are correct, and the numerical answer indicated by @DougM computes to $1.79808 \times 10^{-05}.$

In case more explanation is required, let $A \sim \mathsf{Binom}(n=30,\, p= .4)$ denote the number of Class A games won and $B \sim \mathsf{Binom}(n=10,\, p= .7)$ the number of Class B games won.

Then we seek

$$P(A = 20) \times P(B = 3) = {30 \choose 20}.4^{20}.6^{10} \times {10 \choose 3}.7^{3}.3^7 = 0.001997491 \times 0.009001692.$$

In R statistical software:

dbinom(20, 30, .4) * dbinom(3, 10, .7)
## 1.79808e-05

dbinom(20,30,.4);dbinom(3,10,.7)
## 0.001997491
## 0.009001692
$\endgroup$
  • $\begingroup$ As you indicated A is a Binomial RV... and order would matter wouldn't it? Because of the 30 choose 20. So order in some sense matters. When you mix A and B, order suddenly doesn't matter? Sorry I'm a novice and I'm trying to build intuition. Thanks. $\endgroup$ – HJ_beginner Dec 22 '17 at 6:08
  • $\begingroup$ In a binomial probability order doesn't matter, only the number of Successes matters. (To derive the binomial PDF, one may temporarily keep track of order, but then probabilities of outcomes with various orders are merged. For example, with 3 fair coins, P(2 heads) = P(HHT, HTH, THH) = 3(1/8) = 3/8.) $\endgroup$ – BruceET Dec 22 '17 at 6:37
  • $\begingroup$ Hmmm... what you wrote makes sense to me... a part of me feels that you can't just multiply the two binomial RVs A & B together. I get this tugging feeling that that almost assumes all type A games are played first then all type B games are first. I believe you are correct... I just have very bad intuition when it comes to probability... does my concern make any sense to you? Thanks brother $\endgroup$ – HJ_beginner Dec 22 '17 at 6:42
  • $\begingroup$ Would it be different if the order were 15 type A then 10 type B, then 15 type A--as long as there must be 20 type A wins altogether? $\endgroup$ – BruceET Dec 22 '17 at 6:47
  • 1
    $\begingroup$ Okay, I think I understand this more. Let's say you play two games total. First against a type A and then against a type B. Then the probability of winning both is intuitively $0.4*0.7$. What if I said the order of games is random. The probability of winning does not double because you can do AB and BA... it is still $0.4*0.7$. Now in the case of winning 20 out of 30 type A games, the order does matter and then is merged... because the 30 trials are part of a Binomial RV. Whereas the two game scenario is not a binomial RV, but... a joint Bernoulli... $\endgroup$ – HJ_beginner Dec 22 '17 at 8:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.