Basic maths question about +/- when pushing all-in over time

[jimmer]

I played some HU tourneys last night. Over the course of 4 or 5 tourneys, I ended up with all my chips in the middle and being called PF about 8 times. By my reconning i was favorite 7 times but only won 2 of them.

My question;
If i added up the total percentage chances of me winning each hand and then divided it by the total amount of times i pushed all-in, would this give me an 'overall' statistic for whether I'm making the right plays?

ie; if i pushed all-in 5 times and i worked out my chances of winning were; 60%, 40%, 80%, 50%, 60%; if i add this up and divide it by 5, it gives me 58%. Long term, this means I'm onto a winner right?

[HalfSugar]

Correct. You only need to apply a number to each scenario to see why:

Let's say each example you play is hand #1 of a HU game where you both started with 500 chips. Total pot each time therefore = 1,000 chips.

Example 1 - 1,000 x 60% = 600 chips
Example 2 - 1,000 x 40% = 400 chips
Example 3 - 1,000 x 80% = 800 chips
Example 4 - 1,000 x 50% = 500 chips
Example 5 - 1,000 x 60% = 600 chips
Total = 5,000 chips / 2,900 chips = 58%

Of course, each spot needs to be identical for the math to work like that so really I'd be looking at the spots where the money went in as a less than a statistical favourite (that does not just mean getting the chips in as a 50% favourite but allowing for pot odds in each scenario) and see if there are any obviously reasons (read: leaks) for this occurring. Not just the all-in spot either but the lead up to it.