1. Probability

Probabilities ain't my favorites, but I liked this one:

Eight players P1, P2, P3,...., P8 play a knock-out tournament (lose and you're eliminated).

It's known that whenever the players Px & Py are paired, the player Px will win if x<y.

If the players are paired at random, what's the probability that P4 reaches the final round?

2. Originally Posted by JeffM
My venture: $\frac{3}{7}*\frac{1}{5}*\frac{1}{3} = \frac{1}{35} \approx 2.857\%.$
P(4) will win, first round, if he is paired with 5, 6, 7 or 8 - that would be 4 out of 7

then in the second round to survive he must be paired with one of 5, 6, 7 or 8 → 1, 2 or 3 (i.e. 1:2 or 1:3 or 2:3) must be paired among each other in the first round and 2 or 3 knocked out - oh my head hurts......

3. Originally Posted by Subhotosh Khan
...then in the second round to survive he must be paired with one of 5, 6, 7 or 8 - oh my head hurts......
To make it hurt more(!): "8" shouldn't be there: automatically loses in first round

Anyhoooo....I first got 4/21 as solution, then changed that to 4/35....not 100% sure it's correct....

Changed from (4/7) * (1/3) = 4/21 to (4/7) * (1/5) = 4/35
I initially wrongly assumed one of P1,P2,P3 lost in 1st round.

I see a "1/5" in Jeff's wrong(!) solution

Too lazy to check through a simulation program...

4. I, too, am getting 4/35 or 11.4% as the probability for P4 to make the final round.

Probability of winning Round 1 and going to Round 2: 4/7 (wins if he plays P5, P6, P7 or P8 of the 7 players)

Probability of winning Round 2 and going to Round 3: 2/7 [(4/7)*(1/2) wins if he plays P5, P6, or P7 of the 6 remaining players since P8 is out in Round 1]

Probability of winning Round 3 and going to The Finals: 4/35 [(2/7)*(2/5) wins if he plays P5, or P6 of the 5 remaining players since P7 is out in Round 2]

For what it's worth, I calculated the probability of P4 winning the tournament as 1/35: [(4/35)*(1/4) wins if he plays P5 of the 4 remaining players since P6 is out in Round 3]

I hope I'm right

5. What meanest thou with "round 3" then the finals, SR?

Round 3 is the finals...only 2 players left.

1: 4 games
2: 2 games
3: 1 game

6. Originally Posted by srmichael
I, too, am getting 4/35 or 11.4% as the probability for P4 to make the final round.

Probability of winning Round 1 and going to Round 2: 4/7 (wins if he plays P5, P6, P7 or P8 of the 7 players)

Probability of winning Round 2 and going to Round 3: 2/7 [(4/7)*(1/2) wins if he plays P5, P6, or P7 of the 6 remaining players since P8 is out in Round 1]

Probability of winning Round 3 and going to The Finals: 4/35 [(2/7)*(2/5) wins if he plays P5, or P6 of the 5 remaining players since P7 is out in Round 2]

For what it's worth, I calculated the probability of P4 winning the tournament as 1/35: [(4/35)*(1/4) wins if he plays P5 of the 4 remaining players since P6 is out in Round 3]

I hope I'm right
According to the law established by Sir Denis - P(x) wins against P(y) if x< y - P(1) will ALWAYS win the tournament and P(8) will lose in first round.

7. Originally Posted by Subhotosh Khan
According to the law established by Sir Denis - P(x) wins against P(y) if x< y - P(1) will ALWAYS win the tournament ....
Of course...and I'm P1; however, I'm still looking for 7 players

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