Sports / Maths Puzzle(s)...

[Bob De Caux]

FWIW Matt, there's a much simpler way of doing the first part (working out prob of a London team playing in the quarter finals).
Prob of Arsenal playing a London club = 2/7. Prob of Arsenal not meeting a London club but Chelsea and Fulham meeting = 5/7*1/5=1/7. That's all the possibile London combos, so prob of London derby in QF = 2/7 + 1/7 = 3/7. Should be able to generalise the answer but will take a bit of thought.

I actually had to solve a similar problem last year. Sporting Index were offering odds on an all-English Champions League final before the QFs were drawn, with four English teams left, so I wanted to check for arbitrage opportunities. Obviously couldn't use the 50:50 match outcome assumption, but otherwise similar (although a bit easier).

[Ian Volante]

Right. Now for some easy maths. When two London teams are drawn together in the quarter-finals (3/7 of the time), the probability of at least one London derby in the remainder of the competition is clearly 1.

I have to question this assumption. Firstly, the teams on being drawn in the quarters aren't restricted to a top or bottom half of the draw since the semis are drawn randomly as well, although I'm not sure if this is a moot point mathematically.

Secondly, surely this occurrence is (prob of the other London team getting to semis)x(prob of them being paired)+(prob of them not being paired)x(prob of both winning through to final), or

(1/2)x(2/3!)+(4/3!)x(1/4)=1/3

[Matt Bayfield]

Ian: I think my explanation was badly worded, rather than wrong. When I wrote "remainder of the competition", I meant quarter-finals onwards, not semi-finals onwards. Obviously, if there's a London derby in the quarter-finals, then the probability of a London derby in the quarter-finals, semi-finals or final, is 1.

The fact that the semi-finals are not drawn until after the quarter-finals are played, doesn't affect the calculation, as the act of drawing the teams randomly, whenever it takes place, fixes their positions within the bracket.

[Ian Volante]

Ian: I think my explanation was badly worded, rather than wrong. When I wrote "remainder of the competition", I meant quarter-finals onwards, not semi-finals onwards. Obviously, if there's a London derby in the quarter-finals, then the probability of a London derby in the quarter-finals, semi-finals or final, is 1.

The fact that the semi-finals are not drawn until after the quarter-finals are played, doesn't affect the calculation, as the act of drawing the teams randomly, whenever it takes place, fixes their positions within the bracket.

Ah right, that makes much more sense!

[Michael Wallace]

And we’re nearly there. The overall probability of a London derby must then be p(London derby given that London teams are paired in qtr-finals) + p(London derby given that no London teams are paired in qtr-finals). And this is:

(3/7 x 1) + (4/7 x 5/16) = 17/28.

From A-Level Maths, I vaguely remember there are fancy symbols in probability with bars in, or something like that, to denote conditional probability, which I think is what we’ve calculated here.

Yeah - you can write P(A|B) for "Probability of A given B".

One thing I'd observe (not that it's particularly important): you've written p(London derby given that London teams are paired in qtr-finals) = (3/7 x 1), which isn't quite right. This probability is just 1 - as you note above if you're assuming that two London teams meet in the quarter finals, then the probability there is a London derby is 1, so this conditional probability is also just 1. A slightly more accurate way to phrase it would've been p(London derby in quarter finals) + p(London derby | no London derby in quarter finals). But like I say, not especially important in this instance.

[Matt Bayfield]

Thanks for that Michael - of course, that's correct. It's been a while since I've done A-Level Maths (I won't be too specific about the number of years), so occasionally I write something unintentionally misleading by using a technical term wrongly.

[Bob De Caux]

Had a crack at the general case. Think you have to do it as a recursive formula. If you don't like mathematical notation, look away now! This assumes that n = 2^z, where z is an integer (i.e. it's a power of 2)


Should work for all cases. If anyone wants an explanation, I'll give it a go!

[Matt Bayfield]

Very impressive Bob! I'll see if I can make sense of that when I'm next at home (I'm currently at work, and images are blocked).

[Howard Somerset]

As an extension to the London derbies football puzzle, I had a look last night, before dropping off to sleep, at the probabilities of getting any specific number of London derbies givin the situation of having three clubs in the quarter finals. And consquently the expectation of the number of London derbies there will be. I was surprised to find that my solution for the expectation turned out to be a relatively uncomplicated fraction.

Anyone else want to try this extension, and prove me wrong?

[Bob De Caux]

Anyone else want to try this extension, and prove me wrong?

I agree Howard, I get the expectation to be 3/4

[Howard Somerset]

Glad you got that, Bob. Agrees with my answer.

[Dinos Sfyris]

Here's a little maths puzzle based on The Krypton Factor. Not really a sport I know but then neither is darts (I'm kidding, Kirk!) plus I didn't really fancy starting a Game Show/Maths puzzles thread. So here goes:

Puzzle D2: The Krypton Factor

In the popular ITV gameshow The Krypton Factor there are 4 contestants (Red, Green, Blue and Yellow) who compete to be the most super-awesome. In the first 2 rounds, mental agility and observation, each player's score is converted into a Krypton Factor. 10 points for winning the round, 6 for coming 2nd, 4 for 3rd position and 2 for 4th. There is always an outright winner in each round as tie breaks are resolved by who gave the fastest answers or another observation question. After which ties may occur ie scores given out for each of the 1st 2 rounds could be:

10 6 4 2
10 6 6 6 (for a 3-way 2nd place tie)
10 6 6 2 (for a 2-way 2nd place tie)
10 6 4 4 (for a 2-way 3rd place tie)

Question: After 2 rounds, how many possible total score combinations are there, given that one player won both the mental agility and the observation?

NB: Red 20, Green 10, Blue 10, Yellow 4 counts as a different score combination to Red 4, Green 10, Blue 10, Yellow 20

[Dinos Sfyris]

Nobody?

[Edwin Mead]

I got as far as working out the number of combinations after the first round (I reckon this is 94, but am ready to be corrected), but then my head started to hurt when I began thinking about round 2.

[Bob De Caux]

Had a quick go and got 260

[Gavin Chipper]

I got 256. Not entirely confident about it though.

[Gavin Chipper]

No - actually I agree with Bob.

[Dinos Sfyris]

Had a quick go and got 260

Oh dear I got 248 when I worked it out which makes me think I've missed a score with 12 permutations Care to explain your reasoning?

[Bob De Caux]

Care to explain your reasoning?

Sure. I worked out permutations of the 3 players who haven't got 20, and then just multiplied the final answer by 4 (to recognised that the leader could be R,G,B or Y).. I've labelled it by total points (which can be a max of 36 and a min of 24).

36
12,12,12 - 1 perm
32
12,12,8 - 3 perms
12,10,10 - 3 perms
30
12,10,8 - 6 perms
28
12,12,4 - 3 perms
12,10,6 - 6 perms
12,8,8 - 3 perms
10,10,8 - 3 perms
26
12,10,4 - 6 perms
12,8,6 - 6 perms
10,10,6 - 3 perms
10,8,8 - 3 perms
24
12,8,4 - 6 perms
12,6,6 - 3 perms
10,10,4 - 3 perms
10,8,6 - 6 perms
8,8,8 - 1 perm

Total perms = 65

Therefore total possible scores = 65*4 = 260

[Dinos Sfyris]

Cheers Bob. Had another fiddle with this at work last night and realised I'd missed 20, 12, 6, 6. Will have to think of some more.

[Ian Fitzpatrick]

PUZZLE D1: GOLF

There's a 9-hole golf course near my house where I like to play, but my golf swing is very inconsistent. One day while totting up the scores after playing with my friend, he points out to me that my difference in scores on subsequent holes are all different prime numbers. What is the lowest possible 9-hole total score I could have achieved?

Dinos, your golf is slightly worse than mine if my calculations are correct your score was 254

[Dinos Sfyris]

That puzzle was answered a while back, Ian. The optimum score is a LOT lower than 254. Don't forget the prime differences can go up OR down eg. 1 (+19) 20 (-17) 3 etc. For the solution see here

[Howard Somerset]

Cricket Puzzle

This isn't one of mine; I saw it in today's Sunday Times. And this seems like an appropriate thread for it. I've not tried it yet, so I don't have a solution, and don't yet know how difficult or easy it is.

George and Martha were watching cricket. After the visitors' innings, George said: "How odd! Every time two batsmen were together, the partnership was broken when the team's total was the product of their two batting-order numbers. But one partnership created a stand of over 50." "And," said Martha, "each time an odd-numbered batsman was out, the next batsman out was even-numbered. This continued until the dismissal of batsman 11." The home side's innings saw a similar pattern, but their biggest stand was higher.
What was a) the home team's biggest stand; b) the result of the match?

[Howard Somerset]

I've not tried it yet, so I don't have a solution, and don't yet know how difficult or easy it is.

I had a go last night, before dropping off to sleep. Turned out reasonably straight forward, and not up to the standard of some of the puzzles in this thread. I'l leave it here though, as some might like to try it.

[Matt Bayfield]

I think I've worked this out:

(a) the home team's largest stand was 81 (and the away team's largest stand was 63)
(b) the match was tied, both teams finishing on 99 all out

I can provide my working on request!

[Howard Somerset]

I think I've worked this out:

(a) the home team's largest stand was 81 (and the away team's largest stand was 63)
(b) the match was tied, both teams finishing on 99 all out

I can provide my working on request!

I agree on both counts