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Wednesday, December 03, 2008
What would happen if the shootout period was 10 minutes, not 5?
This is going to be a math-heavy post. Be forewarned.
Java Geek answers it most correctly, in this blog entry by James Mirtle, with data supplied by Gabriel Desjardins. Gabe says that the number of goals scored in OT (4-on-4 hockey) is 7.14 per 3600 seconds, and that it’s fairly uniform. In these kind of scenarios, it’s always best to answer the question: what are the chances of it NOT happening. The chance of not scoring in each second is 1 minus 7.14/3600 (or .998). If you have a 5-minute OT (300 seconds), then you take that .998 figure we just got, raise it to the power of 300, and that tells you the chance of the game still being tied. That figure is 55% (meaning 45% of the time you have a winner). If it was a 10-minute OT (600 seconds), then you do the same thing, but raise to the power of 600, not 300. That figure is 30%, meaning that 70% of the time, you have a winner.
Another way would be to realize that with 62 goals and 54 shootouts in the 5-minute OT, then this means that you were tied 54 out of 116 times, or 47% of the time. If you had a 10-minute OT, you square that number (that is, if the chance of not scoring per 5 minutes is 47%, then the chance of not scoring per 10 minutes is .47*.47), and you get 22%. So, 78% of the time, you have a winner.
The short of it is, that if you double the amount of overtime, then you chop in half the chance of going into a shooutout. It works out this clean, because per 5 minutes of OT, half the time the game ends still tied.