The only time the extra run matters is when a team lost by one or went to extra innings.

When a team lost by one the extra run brings them to extra innings where they have a 50% chance of winning. Since 22.5% of games are decided by one run, 11.3% of the time the extra run goes to the team that lost and that converts to an a win 5.6 of the time - or 0.56 wins.

9% of games go to extra innings. If you add a run to one of those teams then they win the game - but they would have won half of them anyway. So the extra run matters 4.5% of the time and that converts to 0.45 wins.

0.56 wins + 0.45wins = 1.01 wins (Remember your denominonsense).

A more recent example that I have seen of good critical thinking is from the book 'Fifty Challenging Problems in Probability with Solutions' by Frederick Mosteller. In question 6, Chuck-a-Luck, he poses the following question:

How would you approach it?

I approached it with ugly brute force:

- In 5*5*5 = 125 cases we lose one unit
- In 1 case we win 3
- In 1*1*5*3 we win 2
- In 1*5*5*3 we win 1

This gives us 3+30+75 =108 wins and 125 losses.

By approaching the problem from the carnival operator's point of view Mr. Mosteller came to a more elegant way of figuring it out - even if it did involve the ugly maths at the end.

Imagine six players have each bet 1 unit on a different number. Start with the case where three different numbers are rolled. Your return the money to the players whose numbers have come up and pay their profit from the other three players bets.

Now do the situation where a double number has come up. Now you only have to return money to two people and that means you have four units to pay three units of profit.

Similarly when a triple number comes up, you only have to return money to one person and so you are left with five units to pay three units of profit.

What I enjoy about his view is that it figures out where the profit is coming from - when double or triple numbers are rolled - whereas I had no idea when my calculations were complete.