An Algorithm Challenge (finding the five fastest horses out of a large group)

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Here's the result of another simulation to calculate the probability of which is the slowest horse (after the WOW race):

A5 is the slowest horse: 0.1675
B5 is the slowest horse: 0.1753
C5 is the slowest horse: 0.1879
D5 is the slowest horse: 0.2086
E5 is the slowest horse: 0.2608

Note that he probability that A5 is the slowest horse is ##\frac 1 6## as previously calculated precisely.
 
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On the extra extra credit, it shows B4 is faster than A5, which I expected. Both lose to 4 other horses, but we know B4 is faster than one other horse - we don't know that for A5.

I was surprised that C3 was not 13.00.
 
Been playing with a much more subtle problem: identify the median-speed horse.

One reason this is subtle is the condition "fewest races" - is this on average? Or the smallest maximum guranteed to find the answer? Or something else?
 
Vanadium 50 said:
Been playing with a much more subtle problem: identify the median-speed horse.

One reason this is subtle is the condition "fewest races" - is this on average? Or the smallest maximum guranteed to find the answer? Or something else?
Dunno. I'll have to check with the sheik.
 
PeroK said:
If I understand the question, you can model this by considering the numbers 1-25 put at random into five groups of five. The numbers in each group are then sorted in ascending sequence. We then choose group A as the group with the number 1 in it. A1 is the number 1 and A5 is the highest number in that group. Group B is the group with the next lowest first number. E.g., if the number 2 is not in group A, then group B will be the group with the number 2 in it. Group E is the group with the highest first number.

It's clear than the distribution for E5 has a higher mean than the distribution for A5.

This could be simulated using a Python (or C++) script, or possibly calculated - if it wasn't so late in the day.

PS the lower the number, the faster the horse it represents.
Though I no longer play the game of bridge, I frequent a group of bridge players. They are confronted with such situations all the time. The also have the constraint of solving the problem without any aids and doing so in a timely manner. The problem is quite similar as the number of cards in a hand is fixed and a higher rank in one hand means lower ranks in the other hands.

The way they would approach this is called "empty spaces." In the A group we know that one of the horses isn't the slowest. This leaves 4 empty spaces in group A. The chance that A contains the slowest is 4/24 = 1/6. In the E group we know that none of the horses is in the top four. The chance that E contains the slowest is about 5/(25-4). That's 24%, as opposed to the simulation value of 26%. Close enough for bridge.
 
Hornbein said:
Dunno. I'll have to check with the sheik.
Sheik al-Abdullah said, "What may I ask is this "average" of which you are speaking?" When I tried to explain he replied, "I hear your words only as they rattle about in the deep well of time. The horses win or not win by the will of Allah, peace be upon him. So it is written." I elected not to pursue the matter further.
 
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