Hockey Shootout Proof
 

Over at my hockey message board, I've gotten in somewhat of an argument with another poster who insists that you are more likely to win if you put your best shooter in a shootout first.

Here are the parameters, and I need a proof more or less of why it doesn't matter.

Two teams in a shootout. The only two possible out comes are WIN or LOSE. It does NOT matter if you win 2-0 or 2-1 or 3-1 or 2-3 et cetera. Furthermore, if a team is up 2-0 going into the third round, the last shooter does not shoot. (They seem to be getting that since you are more likely to lose 2-0 if you put your best shooter last, then you are more likely to lose period.)

Here's the link to the thread if anyone cares.

http://www.letsgopens.com/scripts/ph...pic.php?t=9839

Re: Hockey Shootout Proof
 

The only reason why the best players may shoot last is to deal with the pressures of close situations . That being said , I do agree that it's probably best to have your best snipers shoot first , since this would give your team a confidence boost if you happen to find yourself ahead . This also asserts that your best snipers will in fact shoot before the game terminates . Having your best players shoot first will maximize your expectation of winning .

Re: Hockey Shootout Proof
 

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The only reason why the best players may shoot last is to deal with the pressures of close situations . That being said , I do agree that it's probably best to have your best snipers shoot first , since this would give your team a confidence boost if you happen to find yourself ahead . This also asserts that your best snipers will in fact shoot before the game terminates . Having your best players shoot first will maximize your expectation of winning .

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How? IF you ignore the intangibles is there any concrete mathematical proof that matters?

The only reason your last shooter doesn't shoot is when your team is going to lose anyway.

Re: Hockey Shootout Proof
 

You are right. The order does not matter as long as the players score independently.

That the third players sometimes don't shoot is precisely balanced by the fact that when they do shoot, more equity is at stake.

Suppose you force the third players to shoot, even if it doesn't matter (just as you deal the river in hold'em even when a player is all-in and drawing dead). The final score doesn't change if you reorder the players (addition is commutative), so the winning team doesn't change if you reorder the players.

Will this convince someone who has publicly stated the opposite, and is unwilling to take back anything? Possibly not. However, you could tell him that a mathematician is eager to bet against him if he wants to bet $51 against $49 that a line-up of 0.7,0.5,0.3 will beat a line-up of 0.3,0.5,0.7 for at least 100 trials. If he doesn't think it's worth at least $1, but still provides some numerical advantage, we can work out something with more trials. If the offer to bet doesn't end the discussion, you might at least get to see an amusing excuse.

In that situation, by the way, there are 33.41% ties, and each side wins 33.295% of the time, regardless of which team goes first, and regardless of whether you force the third players to shoot when the result is already decided.

Re: Hockey Shootout Proof
 

Pzhon , if you have a 5 player shoot-out , then it's possible for a team to win by the sixth shot . It wouldn't matter as long as each shot is independent and that both teams will take 5 shots each . In a hockey shootout , the game ends if another team cannot equal the number of goals scored by the other team . This means that the game can end in as few as 6 shots and at most 10 shots .

Re: Hockey Shootout Proof
 

[ QUOTE ]
You are right. The order does not matter as long as the players score independently.

That the third players sometimes don't shoot is precisely balanced by the fact that when they do shoot, more equity is at stake.

Suppose you force the third players to shoot, even if it doesn't matter (just as you deal the river in hold'em even when a player is all-in and drawing dead). The final score doesn't change if you reorder the players (addition is commutative), so the winning team doesn't change if you reorder the players.

Will this convince someone who has publicly stated the opposite, and is unwilling to take back anything? Possibly not. However, you could tell him that a mathematician is eager to bet against him if he wants to bet $51 against $49 that a line-up of 0.7,0.5,0.3 will beat a line-up of 0.3,0.5,0.7 for at least 100 trials. If he doesn't think it's worth at least $1, but still provides some numerical advantage, we can work out something with more trials. If the offer to bet doesn't end the discussion, you might at least get to see an amusing excuse.

In that situation, by the way, there are 33.41% ties, and each side wins 33.295% of the time, regardless of which team goes first, and regardless of whether you force the third players to shoot when the result is already decided.

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Thanks. I quoted the post on the other site. I'll keep you updated with responses.

Re: Hockey Shootout Proof
 

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Thanks. I quoted the post on the other site. I'll keep you updated with responses.

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There is some really silly math going on over on that site

Re: Hockey Shootout Proof
 

[ QUOTE ]
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Thanks. I quoted the post on the other site. I'll keep you updated with responses.

[/ QUOTE ]

There is some really silly math going on over on that site

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You're telling me.

But wait, I am just being a "hindsight coach" right?

Re: Hockey Shootout Proof
 

best player first, unless you have two pretty solid others in front of him who probably will score, in which case maybe put him 3rd.

Otherwise put him first.