Rage Online

Apparently that's a thing.

Binary this week: unders and overs, point for each one correct.

(1) Oxford Possession: (U) 54% or less (O) 55% or more;
(2) Oxford Shots on Target: (U) Five or fewer; (O) Six or more;
(3) Oxford Fouls: (U) Ten or fewer; (O) 11 or more;
(4) Oxford Corners: (U) Six or fewer; (O) Seven or more;
(5) Oxford Yellow Cards*: (U) Three or fewer; (O) Four or more
(6) Oxford Goals: (U) One or Zero; (O) Two or more.

Bonus half mark for the first person to tell me what the probability is of getting exactly three correct, if the outcomes are random, independent and the probability of each (O) is the same as each (U).

TB: time of first goal in game, no repeats. If equidistant, prediction that is lowest wins.

* Yellow means yellow.

1) over
2) over
3) over
4) under
5) under
6) under

36.78%

TB 67

(1) U
(2) O
(3) U
(4) U
(5) U
(6) U

TB: 27

I suppose I should have said "show your workings" on the bonus ...

(1) O.
(2) U.
(3) U.
(4) U.
(5) U.
(6) O.

50%

TB: 38th minute.

1 O
2 O
3 O
4 U
5 O
6 O
Tb 8

OK, show your workings becomes germane here: OOUOUU gives Tom 2, Geoff 3, Radley and AColi 4 ...

Now Radley needs to justify his 36.78% ... which is very precise but not what I estimated. For me it's a binomial expansion so N! / (r! (N-r)!) . p^r . (1-p)^(N-r) with N = 6, r = 3 and p = (1-p) = 0.5 and 6! / (3! . 3!) .5^3 .5^3 = 0.3125 or 31.25% That's certainly what I'd teach for this type of problem. If Radley's got a viable and valid method for getting his % then he wins: otherwise, I'm afraid, I win on the tiebreaker ...

Zero yellow cards and only seven fouls? Are we soft, or what? I suppose 73% possession means less chance of fouling ...

You win then as I totally made it up and am delighted to be within 5% of the answer.