Math: Kid's Car Tournament Odds

[QuikSand]

So, my (now 5yo) son and I have taken to playing with his burgeoning set of hot wheels and similar cars. We have set up a game as a competition, and in my mind, it seems to present a possible math puzzle. Let's see...

For related reference, this is just how my mind works...

The game is simple. He grabs a bunch of cars, we set up a track, insert some hills and stuff, and he pushes the cars down the track. If they make it all the way, they advance, if they don't, they are eliminated. Those that advance get another try, and the field dwindles from there. When there's a round where exactly one car succeeds, it is declared the champion.

So, at its heart, the question is: What is the probability of a given car being declared the champion?

[QuikSand]

Musing... in a set of n=2, this is simple, right?

P1 = 50%
P2 = 75%

chance of both succeeding = .375
chance of both failing = .125
(both of these result in a null result and a start-over)

chance of P1 yes and P2 no = .125
chance of P2 yes and P1 no = .375

so Bayes tells us that P(P2 wins) = .375 / .500 = 75%

For n=2, expressing this as a generic is simple, too.

How much harder does it get when n is unknown? Or if it's 50? Or, say, as it approaches infinity?

[QuikSand]

Expanding to n=3 explodes the work a little, but it remains manageable.

I'll reference P1,P2,P3 and refer to wins of each in a binary way, respectively...

So 101 means #1 succeeds, #2 fails, #3 succeeds
and the chance of that is (P1)*(1-P2)*(P3)

So, we have 8 possible outcomes:
000 (null, re-start with n=3 again)
100 (resolves)
010 (resolves)
001 (resolves)
110 (continues with n=2)
101 (continues with n=2)
011 (continues with n=2)
111 (null, re-start with n=3 again)

So, for the P(#1 wins) we'd need to look at that one outcome 100, then dig one level deeper into the two cases 110 and 101 where the tournament continues with n=2 an #1 remains in play. Sorting out those other two is just the generic form of what I ran out above, replacing the given percentages with dummy variables.

So you'd end up with:

P(#1 winning)=
sum of
P(100 outcome)
+
P(110 outcome)*P(#1 beats #2 when n=2)
+
P(101 outcome)*P(#1 beats #3 when n=2)
all divided by
P(result is neither 000 nor 111)

Logically that makes sense, but this is getting away from being manageable quickly.

I feel like there's a pattern that would emerge here if we carried it out far enough...and that as n approaches infinity we might have some way to express this. But it isn't sitting in front of me, even after thinking it through this far.

[EagleFan]

2/3

[QuikSand]

Basically the same system as a spelling bee, but I can't find any place where someone noodled through that, either.

[hollmt]

i dont care about the math, but hot wheels racing tournaments/competitions are fun!

and lets be honest here....it was you that wanted to 'play' the most, right?
over christmas my girlfriends nephew got a nice hot wheel track and i spent a good couple hours having a hot wheel tournament w/ him. because, why not? i started organizing it and just basically took control of the entire thing.