## Put me in coach

Billy and Tommy are in the soccer finals (football if you prefer, but it gets confusing here in the States). They are up against their biggest rival, and need this win.

There are 12 players on their team, and only 9 from each team are allowed on the field at one time.

Their team decides to play in a 3-3-2-1 formation. That is to say there are 3 forwards (left – center – right), 3 midfielders (left, center, right), 2 defenders (left & right), and a goalie. The other 3 players will sit on the bench until they can sub in.

The team only has 2 players that are any good at the goalie position (Tommy and his buddy, John), but anyone can play any other position, including the one goalie that is not in goal.

What are the chances that Billy and Tommy will play on the same line at the same time? (eg. Billy is left forward, Tommy is center forward, and Joey is Right forward)

March 3rd, 2011 at 1:02 pm

chance of being together on the first row – 9/144

2nd row – 9/144

3rd row – 4/144

ans = 11/72

I’m not sure this is the right way to do it but it’s a guess

March 3rd, 2011 at 4:00 pm

We want Billy and Tommy to be in the same line. So we must assume that John is in goal, that has a probability of 1/2 of occurring. I’ll factor that in at the end.

There are 12 availiable players, one of which is John and he is in goal. There are 8 places left to fill with the remaining 11 players. The probability that both are selected for a particular line of 3 is 3/11*2/10. The probability that they are both selected into the line of 2 is 2/11*1/10. Putting all that together gives 1/2*(2*3/11*2/10 + 2/11*1/10) = 7/110.

March 3rd, 2011 at 4:39 pm

I’ll try it with the card dealing formulae. But as life is short, I’ll assume that John is in goal and can be ignored, after allowing for the 1/2 probability.

There are 11 remaining players. The total number of ways of filling the field is C(11,3,3,2) = 11!/((11-8)! 3! 3! 2!). The probability of Billy and Tommy being in the first row of 3 is C(11-1-1,1,3,2)/C(11,3,3,2) = C(9,1,3,2)/C(11,3,3,2) = C(9,1)/C(11,3) = 9/165 = 3/55, the same for the second row of 3. For the row of 2 we have C(9,3,3,0)/C(11,3,3,2) = C(9,0)/C(11,2) = 1/55.

So altogether, 1/2 (2*3/55 + 1/55) = 7/110

March 3rd, 2011 at 6:13 pm

Chris,

Please explain why you divide by two. I am not much of a soccer fan, but doesn’t there always have to be a goalie, leaving 11 players to fill all the other positions? at all times. I wouldn’t think it would matter which goalie is playing. Otherwise I would come to the same number as you.

March 3rd, 2011 at 6:56 pm

Hi JB. I don’t like soccer either. I think you didn’t read the question properly. Assuming that John and Tommy are equally likely to be picked to be the goalie, then Tommy can’t be in one of the lines for half of the games.

Aside: I’ve ignored the possibility that a sub will be called in during a game.

March 3rd, 2011 at 7:22 pm

Chris,

You are right. I didn’t notice that Tommy was both a goalie and one of the pair with Billie. So now I do agree. However, If neither Billie nor Tommy were goalies it should not matter at all how many of the others were goalies; the odds would always be the same.

March 5th, 2011 at 8:30 pm

I’m confused, when is joey introduced?

March 7th, 2011 at 9:11 am

Brandon, Joey and John are just some random guys on the team that represent players other than Billy or Tommy.

March 7th, 2011 at 9:40 am

Let’s first consider the total number of possibilies.

n: number of players

g: number of total goalies

b: number of players on the bench

total possibilities = (n-1)! / b! * g

the 1 is for the current goalie. since there is one currently in goal, there will be 11! ways of arranging the rest of the players. But, because the order of players on the bench doesn’t matter, we divide by b!. then simply multiply by how many goaies to increase the arrangements by that factor.

solving the equation: total possibilities = (12-1)! / 3! * 2 = 11!/3!*2 = 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * (2) = 13305600

Now there are 3 places where Billy and Tommy could be on the same line; forward , mid, and defense. for the first two, there are 3! arrangements, and for defense there are 2!, or simply 2. so they have 3! + 3! + 2! = 6 + 6 + 2 = 14 possibilities.

so they have 14 out of 13305600 chances of being placed together. 1/950400 is my answer. Can someone correct me if I am wrong?

March 7th, 2011 at 1:34 pm

Hi mightbewrong. I’ve already corrected you – see post 2. Your analysis is a collection of errors.

March 7th, 2011 at 2:50 pm

Hi Chris

I follow the logic for the card deck as well, but get a different answer … 7/90 rather than your 7/110.

Like you, I consider initially that the goalie position is taken by John.

My view is different at this point, though.

There are now 8 positions to be had by 10 players which gives us C(10,3,3,2) = 25200 combinations.

The scenarios that the buddies are both …

Defenders: C(8,3,3,0) = 560

Midfield: C(8,3,1,2) = 1680

Forwards: C(8,1,3,2) = 1680

Total chances are:

(560 + 1680 + 1680) / (2 * 25200) = 3920 / 50400 = 7/90

March 7th, 2011 at 2:52 pm

Arggg … see an error already … 12 total players, not 11.

Reworking…

March 7th, 2011 at 2:57 pm

I’m with you now!

Please excuse my grey matter flatulence.

Exactly as you stated…

C(11,3,3,2) = 92400 combinations.

The scenarios that the buddies are both …

Defenders: C(9,3,3,0) = 1680

Midfield: C(9,3,1,2) = 5040

Forwards: C(9,1,3,2) = 5040

Total chances are:

(1680 + 5040 + 5040) / (2 * 92400)

= 11760 / 184800

= 7/110

March 7th, 2011 at 3:43 pm

Hi cazayoux. Your version was much neater than mine

March 9th, 2011 at 1:46 pm

@ chris: i am trying to follow your solution. i understand it for the most part. could you explain where i went wrong?

March 10th, 2011 at 6:49 am

Hi mightbewrong. I don’t know which bit you’re having difficulties wiith, so I’ll try to cover everything.

First There are only two players that can be in goal; John and Tommy. But we want Tommy to be in a line with Billy. Assuming that John and Tommy are equally likely to be in goal, that means that for half the games, Tommy is in goal and then the probability of him being in a line with Billy is 0. If Tommy isn’t in goal, then he’s just an available ordinary available player of the 11 remaing men.

Now for the bit that I suspect is really stumping you. I’ll do this in a way that should be clearer. Let a line have 7 players. I’ve chosen 7 so that it’s easier to see what’s going on. There are 11 possibie choices for the first man that is picked. For each of those there are 10 possible choices for the second man, etc. Altogether that’s 11*10*9*8*7*6*5 possible ways to fill the line. That 7 term multiple can be written more neatly as 11!/4!. It will prove very convenient if we say that we don’t care which order the 7 players were chosen. For a given 7 players, there are 7! possible orders that we could have picked them in. So combining the two results gives 11!/(4! 7!) = C(11,7) unique sets of players.

If Billy and Tommy have been picked, that leaves 5 places to fill from the remaining 9 men. There are 9*8*7*6*5/5! unique ways of picking them. That expression can be simplified to 9!/(4! 5!) = C(9,5).

So the number of line ups with Billy and Tommy in, divided by the number of possible line-ups is C(9,5)/C(11,7), and that is the probability of picking Billy and Tommy into the line. That gives 7/11*6/10. Formally, that expression suggests that you can make a short-cut to get the answer. It’s as if you had 7 chances at picking the first player from the 11 =>7/11, and assuming that happened, you have 6 chance of picking the second player from the remaining 10. I’ll admit that putting it that way sounds wrong; but it does produce the correct result. See “Bags I go nth” http://trickofmind.com/?p=819 for another way to shed some light on that.

In the problem, it was 3 and 2 rather than 7. It is pretty obvious that we’d have got 3/11*2/10 and 2/11*1/10 if we’d used 3 and 2 rather than 7.

We have 3 lines, and we simply add the probabilities for each, to get the total. So the final probability is 1/2 *(3/11*2/10 + 3/11*2/10 + 2/11*1/10) = 7/110. The 1/2 is because Tommy is in goal for half of the games.

March 14th, 2011 at 8:17 am

Here’s some more calculations; they are only examples though.

The probability of picking Billy and Tommy into the first line of 3, assuming that’s the first line to fill is found as follows: The possible ways we can pick B and T is BT?, TB?, B?T, T?B, ?BT or ?TB. The first two have a probability of 1/11*1/10*9/9, the second two, 1/11*9/10*1/9 and the third two 9/11*1/10*1/9. So, for each case, the probability is 1/110. As there are six equally likely cases that gives 6/110 = 3/11*2/10, as before.

If the first row of 3 didn’t get Billy or Tommy, the examine the probability of picking them for e.g. the second row of 3. The probability of not picking Billy or Tommy into the first row of three is 9/11*8/10*7/9 = 56/110. The probability of selecting Billy and Tommy into the second row of 3, from the remaining 6+2 men is found as follows. We could pick BT?, TB?, B?T, T?B, ?BT or ?TB. Similarly to the previous case, we find each of those 6 possibilities has a probability of 1/8*1/7*6/6 = 1/56. So, the probability of picking Billy and Tommy into the second row of 3 is 56/110*6*1/56 = 3/11*2/10, as before.

No matter what sequence you chose to do it in, you will always find that the probability of picking Billy and Tommy into a row of 3 is 3/11*2/10, and into the row of 2 is 2/11*1/10.

I do find that amazing.

March 14th, 2011 at 11:38 am

DONT KNOW

March 16th, 2011 at 11:56 am

@ chris: Thank you for explaining to me. i am still not coming up with the same number for the row of 2 though. I keep getting 4/110 no matter how I calculate it.

Logic would say that if in a row of 3 (where there are 6 combinations that include B & T) there is a probability of 6/110, that in a row of 2 (where there are only 2 combinations) it would be 2/110. But I can’t seem to prove it.

Say they are picked first. there is a 2/11 chance of being picked for the first position, and a 1/10 for the second. However, there are 2 equally likely cases (BT and TB). this makes for 2*2/11*1/10 = 4/110.

If that row were picked second, we go back to your 56/110 chance they are not picked in the first row. 2/8*1/7 = 2/56 chances for the 2 spots, but again there are 2 ways of getting that. so 56/110*2*2/56 = 4/110.

And if they are picked last, 2/5*1/4 = 2/20. And the final 3 are on the bench. Now the chances they are not picked in the first 2 lines is (9/11*8/10*7/9)*(6/8*5/7*4/6) = 20/110

so 20/110*2*2/20 = 4/110.

What am I missing this time?

Even though I mightbewrong, if I am correct this would make the final answer 1/2 * [2*(6/110) + 4/110] = 8/110

March 16th, 2011 at 5:10 pm

Hi mightbewrong. Kick yourself. The 2/11*1/10 already implies that you pick either B or T first, then the other next. Otherwise you’d have done BT => 1/11*1/10, TB => 1/11*1/10, then add to get 2/110.

March 16th, 2011 at 5:20 pm

… if the row had been n long, then you’d have got P(n,2)/110, where P(n,2) = n!/(n-2)! and is the number of permutations (ordered picks) of 2 objects from n.

March 17th, 2011 at 6:54 am

OK now I agree with your answer.

Something that still comes to mind is that I know there are only 14 ways of B & T being in the same line (6 + 6 + 2). That means that there are 14/(7/110) = 220 possible ways of arranging the 12 players, with only 2 able to be in goal, and 3 on the bench (where order doesn’t matter).

I’m just going to make up a scenario and proove why I think it should be a different answer. Let’s say 3 players get sick one day and can’t show up. Joey, George, and John (our other goalie). This puts only Tommy in goal, and no one left on the bench. We now have 8 players on the field.

How many different ways are there to arrange them? Well the first position can be filled with any of 8 players, then the second with any of the remaining 7 players, then 6…8*7*6… = 8! = 40320.

Now add back in the 3 benched players and the number of possible arrangements goes way up. Now factor in the goalie and we double our number again.

Where is my logic flawed?

March 17th, 2011 at 10:33 am

Hi mightbewrong. That’s quite a few different scenarios to examine.

In the following I’m using combinations rather than permutations as I don’t care where in a row the players are. If you did care, then repace all my C(…) with P(…) Where e.g. P(n,a,b,c) = n!/(n-a-b-c)!. But all that will happen is that you’ll get incredibly large numbers, and it would also be slightly more difficult to discuss Billy and Tommy in the same row. See “Bags I go nth”, http://trickofmind.com/?p=819 for an explanation of the C(…) stuff.

If anyone can be a goalkeeper, then we have 3,3,2,1 rows. The number of arrangements (with 12 available players) is C(12,3,3,2,1) =

12!/((12-3-3-2-1)! 3! 3! 2! 1!) = 12!/(3! 3! 3! 2! 1!) = 1108800 (or 79833600 permutations) – not 12/(7/110) = 220. Then the probability of Billy and Tommy being in the same row is

(C(10,1,3,2,1)+C(10,3,1,2,1)+C(10,3,3,0,1))/C(12,3,3,2,1) =

(2*50400+16800)/1108800 = 117600/1108800 = 14/(12*11), and that is what I would have written (almost) down straight away (as in my post 2)

If John is in goal, then we have C(11,3,3,2) = 11!/((11-3-3-2)! 3! 3! 2!) = 11!/(3! 3! 3! 2!) = 1108800/12 = 92400 ways of arranging the players (or 6652800 permutations). The same for Tommy in goal. We must (in terms of the original problem) realise that introduces a factor of 1/2 to the probability of Billy and Tommy being in the same row (over many games).

If you make John be sick, then Tommy has to be in goal; then Billy and Tommy cannot be in the same row. However, if we disregard the original question and consider 3 sick players, and Tommy in goal, then we have C(8,3,3,2) = 8!/(8-3-3-2) 3! 3! 2!) = 560 ways of arranging the players (or 40320 permutations).

You ended with a poorly defined phrase “Now factor in the goalie and we double our number again”. It can’t lead to different answers – there isn’t a shortcut way to factor back the now not sick players. You also factored in the goalie and only allowed for a doubling – that’s wrong, you have increased the C(11,…) to C(12,…) as well; that alone is a factor of 12 – but at this point I’m not even sure what scenario you are thinking of.

March 17th, 2011 at 10:56 am

I guess the point that I am trying to make is that there are many more arrangements for the players than 220, which disagrees with your answer. Do you disagree that there are 14 ways of getting Billy and Tommy in the same row? Probably not. There are 6 ways in the rows of 3 players, and 2 ways in the row of 2.

Let’s look at the expression:

# of ways B&T are paired / Probability of B&T paired

where Probability of B&T paired = # of ways B&T are paired / Total # of arrangements

You get: Total # of arrangements

Substitute in numbers from your solution and the 14 I just talked about:

Total # of arrangements = 14 / (7/110) = 220

But you just said yourself that there are far more than 220 possible combinations of 12 players with only 2 that can be in goal, and 3 on the bench.

March 17th, 2011 at 11:35 am

Hi mightbewrong. You said 220. For the team of12 with all players equal, I said 1108800 combinations or 79833600 permutations.

I said (or implied) there are 6 permutations for Tommy and Billy in a row of 3, and 2 permutations for the row of 2; that’s 14 permutations altogether – but that’s only a sub-permutation and it ignores all the other players. In the case of a row of 3 that’s 6 permutations with a given third man. There are 10 possible third men, and 9*8*7*6*5*4 = 9!/3! permutations for the other 3 rows (of 3,2 and 1). So the number of ppermutations of the team with Billy and Tommy in a given row of 3 is 6*10*9!/3! = 6*10!/3! = 3628800. The probability of that is 3628800/79833600 = 6/(12*11).

You have too many variations oon the number of players. 12 men, 11+1 10+2, 8. That just makes for lots of calcualtions. I’ve given more than enough worked examples. You are shooting from the hip instead of thinking.

March 17th, 2011 at 12:28 pm

… I’ve go to go out for a couple of hours. I’ll try to see how to answer in your terms. But can we please define just one team, as I’m not sure which one, of the many that you’ve introduced, that you’re talking about at any particular moment. All the extra teams simply cause confusion, give different numbers, and don’t shed any useful extra light on anything.

March 17th, 2011 at 2:17 pm

OOOOHHHHHH.

You’ve showed me the way. I wasn’t factoring in all of the remaining players while B&T were in a specific possition in each line. So for each time say Billy is left forward and Tommy is center forward like in the example posted in the problem, there are 10 other players that could be right forward. And even then there are several ways of possitioning the rest of the team.

I’m sorry that I am a little bit slower on here. At least I’m learning things!

Ok. Again, I will agree to 7/110 as the answer. And I’m glad to see that you used factorials to shoot my answers down. At least that makes me think I wasn’t completely nuts when trying to solve it that way. Thanks for the lesson.

March 17th, 2011 at 2:42 pm

Hi mightbewrong. LOL, I like your opening line.

I have a sort of answer to the other way of counting. The number of ways of picking 2 players from 12 (where order matters) is P(12,2) = 12!/10! = 12*11 = 132. So for a given pair of slots, there are 132 ways of selecting pairs of players, and 2 correspond to Billy and Tommy (i.e. BT and TB). For a row of 3, we have 3 pairs of slots (let the slots be a,b,c => ab, bc, and ba). So that’s 3 chances, each with with a probability 2/132 => 6/132 altogether. That’s easy to carry over to the 14/132 = 14/(12*11).

The above assumes 12 peer players. For the original problem with John in goal I’d have said P(11,2) = 11*10 = 110 and so 14/(11*10). Then divide by 2 as Tommy is in goal for half of the matches.

So I thank you for causing me to see another way of doing the calculation.

March 17th, 2011 at 2:51 pm

.. If I’d used combinatiosn instead of perutations, I’d have said the number of ways of choosing 2 players from 12 is C(12,2) = 12!/(10! * 2!) = 12*11/2. But then I’d say only 1 corresponds to Billy and Tommy, so I’d still calucalte the probability of selecting them into that slot pair is 1/(132/2) = 2/132. Again 3 pairs of slots so 6/132.

I had meant to remove the “I have a sort of answer to the other way of counting” in my last post. I hadn’t fully worked out my story when I started writing it.

March 18th, 2011 at 7:48 am

Ok. I think this one has gone on long enough. Chris has provided more than enough evidence that his solution in post #2 is correct. I guess this one could have stopped pretty early on.

I do like the discovery in post #17 that no matter when Billy and Tommy are picked the answer is still the same.

Sorry riggers, 3^2/12^2 is not the probability of 2 guys out of 12 being picked in a line of 3. Hopefully Chris and cazayoux’s soultions have cleared that up for you.

Good work guys. Maybe next time mightbewrong. Or maybe you’ll be wrong again