My Math Forum Fun with World Cup Group Play

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July 5th, 2018, 02:44 PM   #1
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Math Focus: Electrical Engineering Applications
Fun with World Cup Group Play

Attached are the point totals of this year's World Cup group (round robin) play.

Just for fun, and looking only at the number of points for each team, in which groups may:

1) The outcomes of all 6 matches be determined?
2) The outcomes of 2 matches be determined?
3) The outcome of 1 match be determined?
4) The outcome of no matches be determined?

To clarify: Suppose that 3 different results of the 6 matches produces the given point total for a group. If the outcome of only 1 of the matches is the same for all 3, then we can say that only 1 outcome may be determined.

Some may find this very easy, but I have to admit that I bungled one of them. To check, I wrote a script.

Edit: I forgot to mention (for those that do not know), the winner of a match gets 3 points, the loser 0 points, and 1 point to each team for a tie.
Attached Images
 wc1.JPG (52.3 KB, 3 views) wc2.JPG (55.5 KB, 2 views)

Last edited by jks; July 5th, 2018 at 03:26 PM.

 July 15th, 2018, 03:43 PM #2 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 635 Thanks: 96 Math Focus: Electrical Engineering Applications Well the World Cup is over and congratulations to France. I thought I would go ahead and post my results to the question that I asked above (these results my need further checking, but I did check them against the actual results for at least one level of a sanity check). Note that I use T for a tie (draw) instead of D because of notation in my program (not shown). First, for groups A and G where the points are 9,6,3,0. Let's just take group A: Code: U - 9 3W R - 6 2W SA - 3 1W (or 3T, which is shown to be wrong - remember we are looking only at points, not the W, D (T), or L E - 0 SA must have 1W since there cannot be an odd number of ties. Obviously, there is only one possible outcome as U won all 3 matches, R lost to U and beat SA and E. SA lost to U and R and beat E. And E lost all of its matches. So we can determine the outcomes for groups A and G. ---------- Next, let's take groups C and E where the points are 7,5,3,1. Taking group C: Code: F - 7 2W 1T D - 5 1W 2T P - 3 1W (or 3T, which is shown to be wrong) A - 1 1T P must have 1W since 3T would give an odd number of ties. The matches are: FD - this must be a tie since D never loses. So F must beat P and A. FP - F wins FA - F wins DP - D must win since D never loses and P never ties. DA - tie since that is all that is left for D PA - P wins since this is all that is left for both. So we can determine all of the outcomes for groups C and E. ---------- For group D we have: Code: C - 9 3W A - 4 1W 1T N - 3 1W (or 3T, which is shown to be wrong) I - 1 1T N must have 1W or there would be an odd number of ties. The matches are: CA - C wins since C wins all 3 matches. CN - C wins CI - C wins AN - A wins, due to the outcomes of CA and AI AI - tie, since these are the only 2 teams with a tie. NI - N wins, which is all that is left for N. So we can determine all of the outcomes for group D. ---------- For group B we have: Code: S - 5 1W 2T P - 5 1W 2T I - 4 1W 1T M - 1 1T The matches are: SP - must be a tie since neither loses. SI SM PI PM IM - I wins as I cannot beat S or P since they do not lose. Only 2 match outcomes may be determined as both of the following results end up in the points being correct: Code: SP_T SI_T SM_S PI_P PM_T IM_I SP_T SI_S SM_T PI_T PM_P IM_I (a T after the underscore denotes a tie, otherwise the winning side is after the underscore). ---------- For group H we have: Code: C - 6 2W J - 4 1W 1T S - 4 1W 1T P - 3 1W (or 3T, which is shown to be wrong) P must have 1W since 3T would give an odd number of ties. Here we can only say that 1 match may be determined, which is that the JS match must have been a tie since these are the only teams with a tie. There are 3 outcomes that produce the points: Code: CJ_C CS_C CP_P JS_T JP_J SP_S CJ_C CS_S CP_C JS_T JP_J SP_P CJ_J CS_C CP_C JS_T JP_P SP_S ---------- For group F we have: Code: S - 6 2W M - 6 2W K - 3 1W (or 3T, which is shown to be wrong) G - 3 1W (or 3T, which is shown to be wrong) With 18 total points for the 6 matches, all must have had winners so there can be no ties. There are 4 outcomes that produce the points: Code: SM_S SK_S SG_G MK_M MG_M KG_K SM_S SK_K SG_S MK_M MG_M KG_G SM_M SK_S SG_S MK_M MG_G KG_K SM_M SK_S SG_S MK_K MG_M KG_G Therefore, no matches can be determined.

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