[MickyB]

In an article on the Italian trials, the author stated that, if the best three pairs each had an 80% chance of finishing in the top three places, the probability of them all doing so was 51%. Presumably he reached this number by cubing 0.8, but this ignores the fact that Pair A qualifying decreases the chance of pair B qualifying.

Trivially, if there are only four pairs in this model, the probability of the best three qualifying is 40%.

If there are more pairs playing, this number is higher, as sometimes more than one of the top three will miss out, so we need to adjust for double-counting.

At this point, I got stuck.

What is the probability of the best three qualifying if there are five pairs in the event, with probability 80%/80%/80%/30%/30%?

What is the probability of the best three qualifying as the number of pairs in the event tends towards infinity?

[helene_t]

There is no simple answer to this because it depends on the source of the uncertainty. If the uncertainty is exclusively related to the performance of the three seeded teams, their ranking (relative to the non-seeded teams) is probably independent. OTOH if the uncertainty is (partly) related to the possible appearance of a dark horse, they will be positively correlated because the absence of a dark horse is good for all three seeded teams.

But let's assume that the uncertainty relates to the performance of the seeded teams. Suppose each seeded team has a probability of p of ranking top among the non-seeded teams, a probability q of ranking second among the non-seeded teams, and a probability r=1-p-q of ranking third among the ns teams.

We can now calculate the probabilities of the position of the three seeded teams in the top-five ladder:

Three seeded teams qualifying:
SSSxx: p^3

Two seeded teams qualifying:
SSxSx: 3p^2 * q
SxSSx: 3p * q^2
xSSSx: q^3
SSxxS: 3p^2 r
SxSxS: 6pqr
xSSxS: 3q^2r

One seeded team qualifying:
SxxSS: 3pr^2
xSxSS: 3qr^2
xxSSS: r^3

We have one constraint (that the probabilities for each of the seeded teams qualifying add up to 240%) so there is still one degree of freedom. Anybody has an idea of a further constraint on the relation between p and q?

[manudude03]

Looking at the opposite view (checking if a position is occupied by one of the top 3) the calculation looks a bit simpler. 1st is taken by one of the top 3 80% (240/300) of the time. If it is taken by one of the top 3, then 2nd position is taken by one of the top 3 72.7% (160/220) of the time and if that is taken, then the 3rd position is only taken by one of the top 3 57.1% (80/140) of the time. Multiplying these 3 figures together gives me a value of 33.25%. If the probability of a non-top-3 pair finishing in the top 3 is given as 60%/(n-3) where n is the number of pairs, then this number is unaffected by the number of pairs entering, but in practice, that 80% figure would be dropping.

edit: This is probably really wrong.

[kenberg]

The following occurs to me: Consider two extreme cases for five teams.

1. One team is really awful, no chance in hell that they are finishing anywhere but last. Then we really have the four team situation and the probability of all three of the 80% teams teams finishing in the top three is 40%, just as if the fifth team went home.

2. (Dump this, as Cher points out this is impossible) There is room for one team, Meckwell or whatever, to have an essentially 100% probability of finishing in the top three without destroying the condition that three other teams each have an 80% chance of finishing in the top three. In this case the probability of the three 80% teams all finishing in the top three is 0, since Meckwell already grabbed one of those spots.


Of course you could say that there was an unstated condition that the threee 80% teams were presumed to be the threee best teams so that case 2 should be knocked out. Ok, but there are still variations on case 1. Maybe the worst tem is ot quite so bad, and as we change our assessment, the other numbers will change as well.



We could look at the highly implausible formulation:
3. Suppose that there are three good teams, all interchangeable in ability, and several (maybe just two) lesser teams, also interchangeable in ability. So, if a,b,c are the better teams and d,e are the lesser teams, we suppose that a final ranking of, say, a>d>b>c>e is just as likely as c>e>b>a>d since we have (only) permuted equivalent teams a,b,c and permuted equivalent teams d,e. We allow that b>a>e>d>c might have a different likelihood of occurring. since in this ordering a lesser team came in third instead of second and a better team came in last instead of fourth.


So we look at ways to assign probabilities to these rankings subject to the condition that each of the better teams has its 80% shot of finishing in the top three and subject to our interchangeability assumptions. We could see if this leads unambiguously to a probability for the three better teams getting all three top places. My guess is that it doesn't.


I think that what I am saying is not exactly the same as what Helene is saying, although it appears to be close.

Note: If we lookk at how likely it is that our model is realistic, it seems to me that having one of the five teams being almost w/o hope of finishing in the top three is more realistic than assuming that the three top teams are interchangeable and the two lower teams are interchangeable. If we really try for realism, this will get tough, I think.

[aguahombre]

If 3 teams each have 80% chance of being in top 3, isn't 1st place taken closer to 90% of the time?

[kenberg]

aguahombre, on 2013-December-24, 12:31, said:

If 3 teams each have 80% chance of being in top 3, isn't 1st place taken closer to 90% of the time?

I'm not sure what you mean. But here is one thing that could happen with four teams, call them a,b,c,d where a,b,c are the good teams. It could be that the probability of d coming in either second or third is 0.6 and the probability of d coming in fourth is 0.4. Assume also that when d does not come in last, the teams a,b and c are equally likely to land in fourth place. This gives a,b,c each a probability of 0.8 of being in the top three and a probability of 1 that one of a,b,c will be in first place.

I'm not sure what you mean by " 1st place taken" so I am not sure what I said is relevant. Taken by whom?

The probabilities I suggest are not crazy. Maybe d is slow and steady, just not flamboyant enough to reach the top, but a,b,c all want to try for top place and swing a bit. In so doing they sometimes crash and land in fourth.Any precise scenario is open to challenge, the world is not that predictable, but I think something like I say is not crazy on the face of it and mathematically it fits.

[cherdano]

kenberg, on 2013-December-24, 10:38, said:

2. There is room for one team, Meckwell or whatever, to have an essentially 100% probability of finishing in the top three without destroying the condition that three other teams each have an 80% chance of finishing in the top three. In this case the probability of the three 80% teams all finishing in the top three is 0, since Meckwell already grabbed one of those spots.

Can't resist catching a fellow mathematician in flagranti of making a mathematically impossible assumption
I can prove that no other pairs has a chance higher than 60% of qualifying. I think so can you

[cherdano]

The BBF idiot's guide "How not to look like an idiot on BBF" says that there isn't anything to add to a play problem after rhm has posted, nor to a statistics question after Helene has posted. Let me try anyway...

Let's simplify: the probably that none of the top three qualifies is negligeable. The problem of computing the probability we are interested in is then equivalent to the figuring out the probability that two of the non-top pairs qualify.
(If x, y, z are the probabilities of 3, 2 or 1 top pair qualifying, then we are interested in x, and we know that 3x + 2y + z = 2.4, and that x + y + z = 1.)
It's easy to see that this depends very much on whether the strength distribution of the remainder of the field. In the extreme case there is just one other strong pair, in which case z = 0, and thus x = 0.4. If there are two other strong pairs, then z becomes non-zero, and x becomes bigger: x = 0.4 + z.

[kenberg]

cherdano, on 2013-December-24, 18:13, said:

Can't resist catching a fellow mathematician in flagranti of making a mathematically impossible assumption
I can prove that no other pairs has a chance higher than 60% of qualifying. I think so can you

Oops. You mean I have to think these things through before posting?

The eggnog is good, Merry Christmas.

[aguahombre]

cherdano, on 2013-December-24, 18:21, said:

Let's simplify: the probably that none of the top three qualifies is negligeable.

To this idiot (me), this is the same as saying that one of the top three will have taken up the #1 spot at least 90% of the time, from which the other calculations began.

[cherdano]

We have already had one extreme example: just one other strong pair, no other pair will ever qualify. The probability we look for is 40%.
The other extreme is: if the boards are flat/reward superior play, then the three top pairs will qualify. If the boards are wild/have all bad slams making/good slams going down, then none of the top 3 pairs will qualify.
That would make the probability 80%. The actually probability will be between 40% and 80%.

[gwnn]

We just got told that there are only 5 pairs so the probability of none of the top 3 to qualify is 0, no?

Anyway, I made a very simple model on an Excel sheet. I think it's what Helene is doing but simplifying it even more.

1. Bridge performance is modelled with just a binary variable: playing well and playing badly. This metric is the same for the good pairs and the bad pairs. The only difference is that the good pairs play well more often than the bad pairs. The good pairs (A-pairs) play well a% of the time and the bad pairs (B-pairs) play well b% of the time.
2. A pair playing well will always place above a pair not playing well.
3. The relative rankings of pairs who play well/bad are random.

Of course this is a problem with 2 unknowns (a and b) and only 1 known: that a and b combine for an 80% individual success probability after summing up all the relative probabilities. A bit tedious, so this is where meddling with Excel comes in, both to sum up the probabilities quickly and to let me play with a and b and see what kind of numbers I get for all three qualifying. My first guess was
a=78%, b=20% (refined from 80-20) => there is an 80.05% chance for any particular good team to qualify and a 44.2% chance for all three
a=80%, b=22.5% => 80.11%, 44.17%
As per helene_t, we may want to set b=0 because no random joe beats Meckwell on their day.

Then I obtain
a=64%, b=0% => 80%, 45.57%
At the other end of the spectrum, putting the onus on the underdogs to prove themselves (a=100%, b=?):
a=100%, b=43.8% => 80.02%, 45.8%

So I always get a probability between 44-46%.

My Excel didn't like some division by 0 stuff so I put in 0.0001 and 0.9999 instead.

edit: And I just noticed that there was a second question in the OP. That will need something more than Excel presumably

[gwnn]

I just realised that my two-state model applied to the infinite case with b=0 (b is necessarily 0 or else clearly only B-pairs will qualify) and a=0.8 is the exact equivalent for the interview calculation. Unless the A-pairs bring their A-game, there will be infinitely many B-pairs that will challenge them for a qualification spot, therefore the A-pairs have 0 chance of beating them. You could reformulate the model to have more states but if in the end it will always reduce to whether an A pair is capable of killing all of the B-pairs and it will have nothing to do whether an A-pair can beat another A-pair, when there are finite A pairs (and the number of A pairs is smaller or equal to the number of qualification spots) and infinite B pairs.

[helene_t]

cherdano, on 2013-December-25, 05:21, said:

We have already had one extreme example: just one other strong pair, no other pair will ever qualify. The probability we look for is 40%.
The other extreme is: if the boards are flat/reward superior play, then the three top pairs will qualify. If the boards are wild/have all bad slams making/good slams going down, then none of the top 3 pairs will qualify.
That would make the probability 80%. The actually probability will be between 40% and 80%.

Now you make me look like an idiot but maybe that was what you meant

Seriously, I like your explanation better than mine. In particular, what I said about the dark horse is misleading because it would take the appearance of two or three dark horses simultaneously to create positive correlation between the seeded pairs' qualifying probabilities. So your example with the evil boards is better.

[helene_t]

I just simulated the experiment 100000 times, assuming each of the five teams has a performance that is N(mu,1) where mu=0 for the seeded teams and some handicap for the nonseeded teams which apparently has to be -1.233935 to give the seeded teams a 80% chance of qualifying. It happened 44372 times that all the seeded teams qualified. So this is consistent with Csaba's results.

Edit: with six teams the handicap is -1.39945 and all seeded teams qualified 46831 times.

This post has been edited by helene_t: 2013-December-26, 09:58

[gwnn]

I applied my two-state model to a bunch of pair numbers, assuming b=0 and finding a for each case.

pairs	a (%)	prob. 3 qualify (%)
4	38.93	40.00
5	63.98	45.55
6	69.88	47.19
7	72.59	48.07
8	74.15	48.61
9	75.16	48.99
10	75.88	49.28
11	76.41	49.49
12	76.81	49.65
13	77.14	49.80
14	77.4	49.91
15	77.62	50.01
20	78.33	50.34
25	78.71	50.52
30	78.95	50.64
35	79.12	50.74
40	79.23	50.78
45	79.33	50.84
50	79.4	50.88
60	79.5	50.92
70	79.58	50.97
80	79.63	50.99
90	79.67	51.01
100	79.71	51.04
(inf)   80.00   51.20

The plot is a bit boring. Anyway, this model slightly overestimates the probability for small numbers as per helene_t's more accurate numbers, but they would also converge to 80% and 51.2%.