Question

Probability Question ("text" questions):
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You are betting on 12 matches. Each match has three possible outcomes; win, draw and loss.

a) How many possible outcome are there? How do you get to the answer?

b) What is the probability of guessing correctly on all 12 matches?

c) What is the probability of guessing correctly on all 11 matches?

d) What is the probability of guessing correctly on all 10 matches?

e) Draw out a formula for guessing x matches.

Answer 1

I think the answer to a) is \[3^{12}\], since we have 3 options per match.

I think the answer to b) is \[\frac{ 1 }{ 3^{12} }\] since the chance to guess correctly per match is 1/3. Multiply that 12 times and you get the above.

What I am trying to learn is not just the answer; but to understand the questions as well as getting confirmation of the answer I believe to be correct :) Any help appreciated.

Answer 2

Your answer for (a) is correct.

Each match has three outcomes. The next match's outcome is independent of the previous result (presumably).

Consider this diagram to show why. I use 2 matches with 2 possibilities:

For \(n\) matches and \(k\) possible outcomes per match, you would have \(n^k\) total possible outcomes overall.

Answer 3

@Ivanskodje does that make sense?

Answer 4

Almost... Need to process the information... I mean digest it. :)

Answer 5

Doesn't that mean the chance to get 11 correct would be \[\frac{ 3^{2} }{ 3^{12} } = \frac{ 1 }{ 3^{10} }\] ?

with formula \[\frac{ 1 }{ 3^{k-1} }\] with k being the number of games you want correct?

Answer 6

Just a second, for part (b), your answer IS correct. I had misinterpreted the question...

As before, there are \(3^{12}\) total outcomes to the 12 matches, but guessing the exact outcome would mean that you guess every outcome correctly regardless of whether it was a win, loss or draw. This means the probability is indeed \(\dfrac{1}{3^{12}}\). Sorry about that!

Answer 7

For part (c), you must guess all but one of the matches correctly. For any of the 11 matches, you have a \(\dfrac{1}{3}\) chance of guessing correctly, so the probability of guessing 11 correctly is \(\dfrac{1}{3^{11}}\). The 12th match doesn't matter, so we leave that alone.

However, there is more than one WAY of guessing 11 matches correctly. Consider this example: Suppose you bet on 11 wins, so the sequence would look like this:
\[\underbrace{WW...WW}_{11\text{ matches}}\underbrace{\_}_{12\text{th match}}\]
The last match can take one 3 possibilities, so that your guess sequences could be
\[WW...WW\underline{W}\\
WW...WW\underline{L}\\
WW...WW\underline{D}\]
This means there are 3 ways to guess correctly, and so you have a \(\dfrac{3}{3^{11}}=\dfrac{1}{3^{10}}\) chance of guessing correctly.

Answer 8

The difference between the guess for part (b) and part (c) is that part (b) has *only* one possible way to guess all 12 matches correctly, whereas (c) has three possibilities.

Answer 9

Whoops, another typo. For (c), I meant to say \(\color{red}{\dfrac{1}{3^{12}}}\) where I said \(\dfrac{1}{3^{11}}\), so the final probability would indeed be \(\dfrac{3}{3^{12}}=\dfrac{1}{3^{11}}\).

Part (d) is similar, but this time the 11th and 12th spots are now free to take on any outcome:
\[\underbrace{WW...W}_{10\text{ matches}}\underbrace{\_~\_}_{\text{11th & 12th match}}\]
The last match can take one of \(3^2=9\) possibilities, so that your guess sequences could be
\[WW...W\underline{W}~\underline{W}\\
WW...W\underline{W}~\underline{L}\\
WW...W\underline{W}~\underline{D}\]
and so on. The other 6 come from switching the first underlined W with D and L.

The probability of getting 10 right is then \(\dfrac{3^2}{3^{12}}=\dfrac{1}{3^{10}}\).

Answer 10

As for the formula, you're not too far off, but notice that each probability so far has had the form
\[\frac{3^{12-n}\text{ ways to guess }n\text{ of 12 games}}{3^{12}\text{ total possible outcomes}}=\dfrac{1}{3^n}\]

Answer 11

@kirbykirby would you mind checking this? I get the feeling my mind might be in the wrong place...

Answer 12

Without thinking about it too much, I feel like this problem is a multinomial distribution, where you have 3 outcomes (win/loss/draw) Denote "win" by \(x_1\), "loss" by \(x_2\) and "draw" = \(x_3\) and each of their associated probabilities by \(p_1, [_2, p_3\) (although they are all \(1/3)\)
\[\large \frac{n!}{x_1!x_2!x_3}p_1^{x_1}p_2^{x_2}p_3^{x_3}\] where \(x_1+x_2+x_3=n\)
(I didn't completely read all of your thought processes yet.. let me check :) )

Answer 13

Hm actually according to e), I think my answer defeated the purpose of the problem.. oye!

Answer 14

sorry I will think about this for a bit. I don't think this is a true multnomial .. I'll be back

Answer 15

Ok I think I somewhat figured it. I tried thinking of a simple case then generalized my thought process. I think it's related in a way to a multinomial, but modified.

Let's just assume that your guess will be only "Wins" (but it can be anything, without loss of generality)... I think it will just simplify notation. W=win, D=draw,, L=Loss

For 12 correct guesses, yes I think we all agree on \(\dfrac{1}{3^{12}}\) , or represented as:
\(WWWWWWWWWWWW\)

For 11 correct guesses, you may consider these options:
\(WWWWWWWWWWWL\) -- (1)
and
\(WWWWWWWWWWWD\) -- (2)

However, a draw or a loss shouldn't be unique to the last position (at least, it is not specified), so you can essentially have all unique permutations using L, and all unique permutations using D)
Since L can occupy any of the 12 spaces, giving 12 permutations of (1) and 12 permutations of (2), so 24 in all. so:
\[\frac{24}{3^{12}} \]

I think another way of seeing this:
for (1): you have \(\large {12 \choose 11}\) ways of choosing a space for "W", and then \({1 \choose 1}\) ways to choose a space of L.
for(2): it is exactly the same process
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Consider 10 correct guesses. You can have:
WWWWWWWWWWDD and all its permutations (1)
WWWWWWWWWWDL and all its permutations (2)
WWWWWWWWWWLL and all its permutations (3)

(1) and (3) would have the same number of permutations, and (2) is different since you don't have repeating elements on the end.

For (1) and (3): You may chose \({12 \choose 10}\) spaces for W, and \({2\choose 2}\) for D., giving \({12 \choose 10}{2\choose 2}=\dfrac{12!}{10!2!}\)

For (2): You may chose \({12 \choose 10}\) spaces for W, and \({2\choose 1}\) for D and \({1 \choose 1}\) for L, giving \({12 \choose 10}{2\choose 1}{1\choose 1}=\dfrac{12!}{10!1!1!}\)

I think from now on it may be easier to view each case like a "MISSISSIPPI" problem, where you are asked to find how many ways are there to arrange each of those letters, which has repeating letters. It's I think a faster way then to keep thinking of the combinations, where it is really just a simplification of all the "choose" combinations multiplied together (In my initial post, this number of arrangements is that factor \(\dfrac{n!}{x_1!x_2!x_3!} \))

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Consider 9 correct guesses: you can have the permutations of:
\(WWWWWWWWWDDD\)
\(WWWWWWWWWDDL\)
\(WWWWWWWWWDLL\)
\(WWWWWWWWWLLL\) etc...

I'm having trouble though establishing a formula right away. Let me know what you think. @SithsAndGiggles @Ivanskodje

Answer 16

@kirbykirby I was assuming that the 12 matches were independent of each other, like the first 12 games in a tourney bracket, as opposed to a sequence of games, so that order doesn't matter. Fortunately we should be able to account for either combination or permutation by factoring in or out some factorial term.

@Ivanskodje could you let us know what level of probability is this? Like have you learned about multinomial coefficients, permutations, etc?

Answer 17

It seems that I get something like this for the number of outcomes:
for 12 correct guesses:
\[ \frac{12!}{12!0!0!}\]
For 11:
\[ \frac{12!}{11!1!0!}+\frac{12!}{11!0!1!}\]
for 10:
\[ \frac{12!}{10!2!0!}+\frac{12!}{10!1!1!}+\frac{12!}{10!0!2!}\]
for 9:
\[ \frac{12!}{9!3!0!}+\frac{12!}{9!2!1!}+\frac{12!}{9!1!2!}+\frac{12!}{9!0!3!}\]

Etc.

If I can somehow find a formula for this.. ?

Answer 18

@SithsAndGiggles I don't think the order doesn't matter, but I tried accounting for all of those arrangements, like for 11: we could have as the number of possible outcomes
WWWWWWWWWWWL , WWWWWWWWWWLW, WWWWWWWWWLWW, etc?

Answer 19

As for a formula (according to what you just posted), for \(k\) correct guesses,
\[\sum_{i=0}^{12-k}\frac{12!}{k!(12-k-i)!i!}\]

Answer 20

oh of course lol

Answer 21

Does "guessing correctly" also mean we can bet that a game also ends in a loss (L) or draw (D)?

Answer 22

I believe so, But I don't think it matters what you choose as a correct guess. In a simple case, there is 1 game. I guess with the team will WIN

The probabilty that I guess correctly is 1/3

What if I guess the team will LOSE?
The probability is 1/3

So I think that's why I just considered "wins" to visually see what's going on? But.. as I think of it now from what you said, I am doubting myself.. oye maybe I did misinterpret this problem :S Bcause if you guess 11 games correctly, are you assuming that you only guessed ONE "sequence" of guesses correctly, or can you consider any sequence of wins?

Answer 23

this problem is really hurting my brain xD

Answer 24

Haha we must definitely be overthinking it ...

Answer 25

I guess so. I would love for you @Ivanskodje to tag me again when you get the solution for this :o

Answer 26

@ganeshie8 Another opinion maybe?

Answer 27

In match 1, there are only 1 out of 3 outcome that can be correct. The same goes for all other matches (singled out).

What I originally thought (without much time or effort, thus making me believe it is wrong!)
was; guessing correctly 12 out of 12 matches: 1/3^12
If guessing correctly 11 out of 12 matches I thought it would be 1/3^11, however, when thinking you still have that last match that could increase the odds in your favour... made me think that it wouldn't be 1/3^12, since the odds should be "a tiny bit" better than that, since you are ALWAYS guessing 12 matches regardless of outcome. (so getting at least 1 correct, should be very likely!) The order of them does not matter, as long as X are correct. Each match is individual (a 1/3 chance of getting correct guess). However, (which I don't understand) according to my teacher the order matters? I don't get that.

He did confirm that the chance of getting 12 out of 12 correct is 1/3^12 though.

win is not us guessing correctly. win as the individual match win. If we bet on the match to lose (well, specific team) then we win as well. If both teams win and we bet draw, then we guessed correctly as well. (In that case, left team, draw and right team might be better description of our choices?)