Friday Puzzle -- best of seven

Re: Friday Puzzle -- best of seven

John Veerkamp

probabilities on the outcome of any game do not change due to the outcome of a prior game.

Re: Friday Puzzle -- best of seven

Alex Y

Correct, each game is assumed to be independent. No "momentum", so in my example, the probability of the Celtics winning game 4 after already winning the first three was still 50%.

Re: Friday Puzzle -- best of seven

David Weaver

This is a good example of the red balls and blue balls picked from an urn thing, but with a twist, I think.

It's made easier than the base case (no experience thus far, seven game series) by the fact that four of the games are put away.

The chance that the series ends in five games is zero. The chance that it ends in six is 2*(1/2)^2 = 1/2

The chance that it ends in seven is the complement to that, which is also 1/2.

A more interesting question might be from the beginning, how likely is the series to go to 5, 6 or 7 games.

Re: Friday Puzzle -- best of seven

David Weaver

Some day, I'll learn to read the whole question. Bbiaf with an answer.

Re: Friday Puzzle -- best of seven

David Weaver

Sort of working this out as we go along...

In order to have six games to begin with, we have to assume that there are two possible outcomes of 1-5, and that is 3 wins for boston and 2 for the lakers or 3 for the lakers and 2 for boston. 4 and 1 and 5 and 0 scenarios are out, because the series is over in either case or not possible to begin with.

Either has an equally likely chance of occurring (each 3-2 scenario). All other scenarios can be disregarded.

Given that we're already to that state, the chance of the game going to seven is 1/2 chance for boston and 1/2 chance for the lakers in each respective situation where they lead.

given that they each had 1/2 chance of being the conditioned status, we have2(1/2)(1/2)

Wait...that's the same answer as before. 1/2 chance that the series goes to 6 and 1/2 chance that it goes to 7.

This could be done by brute force - there are only 128 possible combinations, as I see it (2^7), even without specifying that the series is over after four wins.

(i'm not sure yet that I'm comfortable with this answer, still trying to prove to myself that it's OK - I think it is, but I'm way too lazy to do brute force)

  Glad to see I'm not the only one who does that!

I guess the question is whether it's correct?

David Weaver

I haven't been here long enough to know if it needs to be a wink for it to be the right answer.

Alex Y

That's the answer, and the "trick", as you found, is to ignore my lead toward a brute strength solution.

If it helps you to consider it "proved", look at it this way:

In a best of seven series, the only way to get to a game six is for one team to have a 3-2 lead. But we have said that each team has a 50% chance of winning each game. If the team with a lead wins game six (50% chance), the series ends with game six. If the team with a lead loses game six (50% chance), the series goes to a seventh game, which decides the victor.

Re: I guess the question is whether it's correct?

Alex Y

That was a reply to your post about reading the question before answering . See other post about your answer, which was correct.

BTW, since you seem to have enjoyed this one, I pass on another that I posted here some time ago, but which generated little or no response:

You and a friend want to bet $1,000 on the outcome of a best-of-seven series, but are only allowed to place bets on individual games. Is it possible devise a betting schedule such that whenever the tournament ends, the person whose team won the tournament had won exactly $1,000?

Re: I guess the question is whether it's correct?

David Weaver

To start, I think that you have to be betting *on* the winning team when they win to get 1000. If you start by betting on the team you want to win, then if they win in 4, there's no way to be only 1k ahead.

So, I'd say, bet for the other team until your team has won 3, and then bet for your team.

That's my quick answer, gotta go to a meeting. I'll come back later to see if I need to think harder.

Jeez...I already see a problem with this one. I'll have to get back to it later. if I used my strategy, if your team one in five, then you'd be down pretty significantly.

I'll get back to this one later.

Re: I guess the question is whether it's correct?

David Weaver

by the way, I'm assuming that the person who does the betting must bet all of the games in the series?

If so, I'm not sure how to be something other than up 2000 or at zero at the end of the series if your team wins the first four.

Re: I guess the question is whether it's correct?

Alex Y

To start, I think that you have to be betting *on* the winning team when they win to get 1000. If you start by betting on the team you want to win, then if they win in 4, there's no way to be only 1k ahead.

If you bet $250 on each game, and your team won in four, you would be $1,000 ahead. But this strategy might not work too well if you couldn't guarantee a sweep!

That assumption not needed

Re: I guess the question is whether it's correct?

David Weaver

So, must the individual bets never be greater than a thousand per?

If not, the answer is easy.

What about the cumulative total of the bets, is there any limit on it other than the sum of individual limits?

My knee-jerk reaction with no limits at all is to just bet whatever you want and when your team has won three games, bet $1000 more than you owe each time they play.

That doesn't work well if there is a $1000 limit.

Alex Y

No such limit. But I don't think that makes it easy. How does you scheme work if they win the first three games, then lose the next four?

Re: I guess the question is whether it's correct?

Larry Barrett

Alex, can you please clarify how this game works?

I assume that each player starts with $1000; (an alternative might be that there is a "pot" of $1000, and each player starts with $0). What is not clear is the amount that can be bet on any game.

It could be the minimum of what either player is willing to bet on that game. For instance, player A might be willing to be $1000 on first game, player B $0. In this case the agreed to bet would be $0 and after the game they would each have $1000 regardless of outcome of game.

Alternatively, a player could bet up to the max that the other player had at that time, regardless of what the other player wanted to bet. So player A could bet up to $1000 on game 1, player B could still bet $0. If team A won that game, player A would have $2000 (if he had bet $1000), player B $0; but if team B won that game, player A would have $0, player B $2000. Maybe there are other rules.

I don't have a strategy for either case so far.

starting a new thread on this one, Larry and David

Alex Y

The betting puzzle is one I have posted here before, but it is getting confusing in this thread. I'll post it anew.