In an NCAA basketball tournament, are 64 teams who play single elimination tournament, hence 6 rounds, and you have to predict all the winners in all 63 games. Your score is then computed as follows: 32 points for correctly predicting the final winner, 16 points for each correct finalist, and so on, down to 1 point for every correctly predicted winner for the first round. (The maximum number of points you can get is thus 192.) Knowing nothing about any team, you flip fair coins to decide every one of your 63 bets. Compute the expected number of points.

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Let's start with terms: "expected value" means the average amount that you would win or lose over a large number of plays. The expected value of a given outcome is (probability of that outcome occurring)`*`(loss or payout of that outcome). The expected value of the entire game is the sum...

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Let's start with terms: "expected value" means the average amount that you would win or lose over a large number of plays. The expected value of a given outcome is (probability of that outcome occurring)`*`(loss or payout of that outcome). The expected value of the entire game is the sum of the expected values of all possible outcomes of the game.` `

For first round games, this is straightforward. You get 0 points for a wrong prediction and 1 point for a valid prediction. Since you're deciding by coin flip, you have a 50/50 chance of winning or losing. Thus, your expected value for a single first-round game is 

`(0.5)(0)+(0.5)(1)=0.5`

So, we expect to get half a point on average for each round one game.

What about for the second round? Well, for the right team to win that game, it needs to win the first and second rounds--that's two coin flips. When you stack probabilities on top of one another, they multiply, so your odds of correctly guessing a second-round match are:

`(0.5)(0.5)=0.25`

The right team wins 1/4 of the time, and they do not win the other 3/4 (sometimes they lose the second game, and sometimes they lose out in round one, but we get zero points either way, so there's no point in distinguishing those cases). Conversely, the points for a second-round win double. Multiplying the new outcomes by the new payout, we get:

`(0.75)(0)+(0.25)(2)=0.5`

Half a point again. Curious!

Conveniently, the pattern holds to the third round. We're adding another coin flip, meaning it's now a 1/8 chance of the right team winning, but we're doubling the points one more time. Plugging that in to the formula once more, we find the expected value for a game in the third round is:

`(0.875)(0)+(0.125)(4)=0.5`

At this point, there's no reason to keep going. We've shown that for each round, the payout doubles and the likelihood of winning halves, canceling out. Every game in every round has the exact same expected value of half a point. All that's left is to find the total expected value of the whole bracket. And since the bracket contains 63 games, each of which is worth half a point on average, our expected score is

`(63)(0.5)=31.5`

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