winner vs alternate break, computer simulated

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I've wasted an entire hour thinking about this. Easy to prove a race to 2 or 3, but to make a general proof for a race to n games is mighty tough...at least for me it is.

If you ever do work out a proof, let us know.

Someone responded earlier in this thread with a proof
 
An interesting stat might be the 'chance of the breaker winning if they broke and ran the previous game' =]

Here are some results for the streamed matches I watched for the 2015 and 2016 US Open 9-Ball Championships.

The overall stats (all tracked games) for the two years combined were:
  • • Successful breaks (broke legally, made at least one ball, and did not foul) -- 59.7% (593 of 994)
    • Breaker won game -- 53.2% (529 of 994)
    • B&R games -- 22.8% (227 of 994)

14 of those 227 B&R games were in the final game of a match, so there was no next game in those matches. Here are the stats for the "next game" after the other 213 B&R games:
  • • Successful breaks -- 64.8% (138 of 213)
    • Breaker won game -- 58.7% (125 of 213)
    • B&R games -- 23.5% (50 of 213)

So the results for the next game after a B&R game are a bit higher than the overall results.

Van Boening appeared in 6 of the matches tracked, and Shaw in 7. Their results (combined for both players for both years, and, even then, with caveats because of the small sample) were as follows -- first the overall results (all breaks), then results for just the games after B&R's. [4 of their overall 42 B&R's were in the last game of a match, so the second set of numbers is based on 38 B&R's.]
  • • Successful breaks -- 72% (91 of 126), 79% (30 of 38)
    • Breaker won game -- 56% (70 of 126), 61% (23 of 38)
    • B&R games -- 33% (42 of 126), 37% (14 of 38)

Again, the "next game" results are a bit higher than the overall results.
 
Here are some results for the streamed matches I watched for the 2015 and 2016 US Open 9-Ball Championships.

The overall stats (all tracked games) for the two years combined were:
  • • Successful breaks (broke legally, made at least one ball, and did not foul) -- 59.7% (593 of 994)
    • Breaker won game -- 53.2% (529 of 994)
    • B&R games -- 22.8% (227 of 994)

14 of those 227 B&R games were in the final game of a match, so there was no next game in those matches. Here are the stats for the "next game" after the other 213 B&R games:
  • • Successful breaks -- 64.8% (138 of 213)
    • Breaker won game -- 58.7% (125 of 213)
    • B&R games -- 23.5% (50 of 213)

So the results for the next game after a B&R game are a bit higher than the overall results.

Van Boening appeared in 6 of the matches tracked, and Shaw in 7. Their results (combined for both players for both years, and, even then, with caveats because of the small sample) were as follows -- first the overall results (all breaks), then results for just the games after B&R's. [4 of their overall 42 B&R's were in the last game of a match, so the second set of numbers is based on 38 B&R's.]
  • • Successful breaks -- 72% (91 of 126), 79% (30 of 38)
    • Breaker won game -- 56% (70 of 126), 61% (23 of 38)
    • B&R games -- 33% (42 of 126), 37% (14 of 38)

Again, the "next game" results are a bit higher than the overall results.

Cool findings. I'll modify my simulation to run these numbers and see what happens.
 
Here's a sketch of a proof. ... You could even have player A break n times in a row and then player B break n-1 times in a row, and you'd end up with the same probabilities of each player winning the set. Or you could play loser break, and the probabilities would also be the same....
After thinking about it for a while, I'm starting to agree with you, although I'm not comfortable with some of the details yet. You could even spot your opponent that he gets to pick who breaks each rack as long as he takes no more than n-1 breaks for himself and you flip if it gets to hill-hill. His choices do not change the match probabilities.

I don't remember seeing any similar probability problem that in effect discards a bunch of outcomes.
 
Here are some results for the streamed matches I watched for the 2015 and 2016 US Open 9-Ball Championships.

The overall stats (all tracked games) for the two years combined were:
  • • Successful breaks (broke legally, made at least one ball, and did not foul) -- 59.7% (593 of 994)
    • Breaker won game -- 53.2% (529 of 994)
    • B&R games -- 22.8% (227 of 994)

14 of those 227 B&R games were in the final game of a match, so there was no next game in those matches. Here are the stats for the "next game" after the other 213 B&R games:
  • • Successful breaks -- 64.8% (138 of 213)
    • Breaker won game -- 58.7% (125 of 213)
    • B&R games -- 23.5% (50 of 213)

So the results for the next game after a B&R game are a bit higher than the overall results.

Van Boening appeared in 6 of the matches tracked, and Shaw in 7. Their results (combined for both players for both years, and, even then, with caveats because of the small sample) were as follows -- first the overall results (all breaks), then results for just the games after B&R's. [4 of their overall 42 B&R's were in the last game of a match, so the second set of numbers is based on 38 B&R's.]
  • • Successful breaks -- 72% (91 of 126), 79% (30 of 38)
    • Breaker won game -- 56% (70 of 126), 61% (23 of 38)
    • B&R games -- 33% (42 of 126), 37% (14 of 38)

Again, the "next game" results are a bit higher than the overall results.

So I programmed into my simulation a "momentum factor" of 4% (if you break and run previous rack, then you're more likely to win the next). I used above data, comparing an average pro breaker to the van boening/shaw numbers. The better breaker wins about 0.3% more in winner break vs alternate break. Not a huge effect but it's there.

I also ran the same race to 9 simulations I did before (lesser player getting a 1 game spot). In that case adding a "momentum factor" just made the effect added around 1% more for alternate format wins.

Mathematically, alternate break = winner break in determining a winner, IF we look at the chance of winning as static. In reality, there is some sort relationship between a break and run and the next game. It's not like flipping a coin. Most good players know that when you get more table time and start running racks, you start playing better. When you sit in your chair for a long time, you can get cold.

However we don't know is if there is a "momentum factor" in alternate break. We do know there will be more games won in the race from the lesser player in alternate.

A good comparison would be looking at winner vs alternate break results, calculate two Fargo ratings for each format for the same player. I find it hard to believe they will match since alternate breaks clearly allows the lesser player to win more games, although the match outcome will probably be the same.

I've asked Mike Page about this and he went quiet when I mentioned these facts....
 
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