# Matchplay, what are the chances of...

For fun if you enjoy math.

I’ve never seen this happen in random pairings. Swiss style, yes.

Given:

Typical 4 strike format (not progressive) bottom 2 get one strike
28 players
4 player groupings
Player groupings set on random
Assume nobody out after 4 rounds, ie 28 players play in the first 5 rounds.

Find the probability of having one particular opponent in your group in rounds 1-5 inclusive

Isnt it just 1 in 6561?

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It’s 27 / 9^5 (ish) which is 1 in 2187.

There are 27 possible opponents with whom this can happen and there’s a 1 in 9 shot of drawing them each round.

If it’s meant to be any two opponents, there are 351 possible pairs of opponents which drops the rarity down to about 1 in 168 tournaments.

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I’ll add the extra step of how you get 1 in 9, so that we can eliminate the ish. So you are matched against player B in all 5 rounds. There are “26 choose 2” (26!/(2!24!)) ways of choosing the remaining two opponents besides you and player B from the remaining 26 players, which is 325. This is out of the “27 choose 3” ways of choosing your other 3 opponents out of the remaining 27 opponents which is 2925. 325 over 2925 is 1/9. For a specific player, there is a (1/9)^5 chance you would encounter this player in all 5 rounds. The other solution of 27(1/9)^5 has already been given if the question is interpreted as the chance of this happening against any of your 27 opponents. I think both interpretations of the question have now been covered and thanks to bkerins for the heavy lifting.

Ah yes, the three. I forgot to multiply by 3. I am an idiot.

Pedantic time … everyone ignore

It is “ish” for a few reasons. The 1/9^5 is correct for one opponent. But: If I’m not matched up with one opponent it makes being matched up with someone else slightly more likely. Also it is possible but very unlikely to have this happen with two different people in the same event. Multiplying 1/9^5 by 27 is not legal in this case in probability land, but the real answer is close enough for jazz.

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1 in 9 each round over five rounds.

9x9x9x9x9 or 1 in 59,049

First round doesnt matter and there are 3 possibilities from the first round hence 3/9^4

First round still matters; this rearranges the question into multiple steps. .

It now reads: What is the probability of having one of 3 specific opponents 4 times in a row after round 1?

In order to have someone repeat in rounds 2-5 they would have to be chosen in round 1.

So the first step is to determine the probability of having any one of those opponents in the group.

Then the next step is to calculate the probability of encountering any one of those 3 opponents 4 more times in a row.

So it is still in essence an event occurring 5 times in a row or P(event)^5.

Similar to a question - what are the odds I’ll flip a coin the same way 3 times in a row - the first flip only sets the mark - the next two must match it. But the first flip doesn’t ‘matter’ in the sense that it must be heads or tails, but it matters in that it sets the mark. So I was being quick with my answer - in that it doesn’t matter who is in the first group, other than that one of those 3 will join you 4 more times.

Now for a related question - what are the odds that in 4 strike tourney with 28 people nobody is eliminated after round 4?

Are we assuming Swiss Pairing for this?

If so, I think it’s (1/6)^3, or about 0.463%.

The goal is to get into Round 4 where there you’ve minimized the number of people with three strikes, and I believe the key is to look at the matchups where there are different amounts of strikes and assume that those who have a greater amount of strikes win. I think the best you can get to is two players with 3 strikes if a few key matches in prior rounds go a specific way.

Round 1 - Everyone at 0

Round 2 - 14 at 0; 14 at 1
In Round 2, there’s one matchup where it’s 0011. If the two people with 1 Strike win, then the strikes are more spread out. I think it’s a 1/6 chance (4 of 24 permutations) where that scenario happens.

Round 3 (assuming Round 2 played out as above) - 6 at 0; 14 at 1; 6 at 2
In Round 3, there’s one matchup where it’s 1122. There’s also a 0011 matchup but it doesn’t matter as none of them could be eliminated in Round 4. If the two people with 2 Strikes win, then the strikes are more spread out. Assuming same (1/6) chance.

Round 4 (assuming Round 2 & 3 played out as above) - 26 at either 0/1/2; 2 at 3
In Round 4, there’s one matchup where it’s 2233. If the two people with 3 Strikes win, no one is eliminated. Assuming same (1/6) chance.

You need the same situation to play out over three specific games. Assuming random winners in each, I think it would be (1/6)^3.

In the original problem I took the liberty of stating nobody was eliminated in rounds 1-4 to keep the question easier.

In reality there had been 3 people eliminated after round 4. So there were groups of 3 and groups of 4 to consider which I felt would needlessly complicate the analysis.

@coreyhulse in the original problem it was not Swiss.

You’ve changed the question.

You originally said:

The only other relevant details were each round being 7 x 4 player groups (28) - the rest is just background filler.

So let’s say you are player A, the ‘particular person’ is player B, everyone else is player X

Probability of you getting player B in your group is 1/9, as it is for the next 4 rounds.

So the answer is 1 in 59,049.

This bit I’m confident I’m right probably

If however you meant “What’s the probability of ANY player being in the same group as you for the 5 rounds”, that’s where the ish comes in, without

First round is irrelevant as you are guaranteed 3 other players in your group (B, C & D).
2nd round.
Possibilities are
ABXX, = 3 * (1/27 * 24/26 * 23/25)
ACXX, = 3 * (1/27 * 24/26 * 23/25)
ADXX, = 3 * (1/27 * 24/26 * 23/25)
ABCX, = 6 * (2/27 * 1/26 * 23/25)
ABDX, = 6 * (2/27 * 1/26 * 23/25)
ACDX, = 6 * (2/27 * 1/26 * 23/25)
ABCD = 6 * (3/27* 2*26 * 1/25)

etc. etc.

I’d say this is worded the way you say, but could also mean what everyone else is saying, like he didn’t know the particular person until he looked at his group for the first 5 rounds, and said - that’s crazy - what are the odds?

I can see the potential for confusion. Certainly was not the intent.

In one interpretation, it can be considered as: what is the probability of players A and B being in the same 4 player group 5 times in a row where B is known ahead of time.

However I see how it could be interpreted that a group of 4 has been formed with A as one of the players and what is the probability of playing any of those remaining 3 players 4 more times in a row.

Either way, I feel it is still a pretty amazing occurrence

Given the number of tournaments I think it would be more amazing if it had never happened to anyone.

Of course. That is a much larger universe than any one player’s universe.