Prediction Streak

You're right; it's the mode that we're looking for, and modes and medians can differ by quite a bit depending on the distribution under analysis, and I think that this is such a distribution. I sense that the most economical way of calculating the mean is via a computer program such as that provided by Brock Brown the Python Wizard, but the mode is within our reach to determine by hand, since it is reasonable to assume that the mode will be somewhere between $1$ and $10$ inclusive.

I have concerns about the phrasing of and the answer to this question. First, it appears that we have to assume that there is a 50/50 chance of predicting the outcome of any match. Given this, there are then $2^{10} = 1024$ possible prediction sequences, each of which is equally likely.

It will be quite uncommon to get a high score, e.g., we can only get a score of $20$ three ways, namely

$WWWWLLWWWW, LWWWWLWWWW, WWWWLWWWWL.$

So we need only focus on scores of $10$ and $5.$ To get a score of $10$ , we have the following options:

• a subsequence of $4$ wins and the rest losses $\Longrightarrow 7$ sequences;

• separate subsequences of $3$ wins, $2$ wins and $1$ win and the rest losses $\Longrightarrow 60$ sequences;

• three separate subsequences of $2$ wins each and one subsequence of $1$ win and the rest losses $\Longrightarrow 1$ sequence.

This gives us the number of sequences yielding a score of $10$ as $S(10) = 68.$

Next, to get a score of $5$ , we have the following options:

• alternating wins and losses $\Longrightarrow 2$ sequences;

• one subsequence of $2$ wins and two subsequences of $1$ win each and the rest losses $\Longrightarrow 105$ sequences.

This gives us $S(5) = 107$ , which exceeds $S(10) = 68$ , which is why I chose $5$ as the answer. I haven't determined yet if $5$ is the actual mode of this distribution, but of the options given, $5$ is the most likely score.

@Janardhanan Sivaramakrishnan This is a great problem, but I have some concerns about the answer to this problem which I've outlined above. When you get a chance, could you tell me where I may have erred in my reasoning? Thanks.

@Trevor Arashiro Some unchecked calculations yield the following results:

$S(0) = 1, S(1) = 10, S(2) = 36, S(3) = 65, S(4) = 91, S(5) = 105,$

$S(6) = 89, S(7) = 85, S(8) = 84, S(9) = 75, S(10) = 68.$

So it would appear that there is a singular mode, which is $5$ . These ten values sum to $709$ , so there are $315$ sequences that yield scores from $11$ to $55$ which are progressively less common to achieve. (There would be another question; find the sum of all scores that cannot be achieved in any sequence. hehe) So Brock's mean of $~9$ seems reasonable.

Yup, $\boxed{5}$ is the mode.

Here are some more random simulations that count for each outcome of points to demonstrate:

Python 2.7:

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from time import time
import random
coin = (True, False)
count = {}
for i in xrange(56):
    count[i] = 0
def points():
    streak = 0
    output = 0
    for game in xrange(10):
        winner = random.choice(coin)
        prediction = random.choice(coin)
        if prediction == winner:
            streak += 1
            output += streak
        else:
            streak = 0
    return output
total_points = 0.0
trials = 0.0
seconds = 60*5
end = time() + seconds
while time() <= end:
    count[points()] += 1
biggest, answer = 0, 0
for key in count:
    if count[key] > biggest:
        biggest = count[key]
        answer = key
print "Mode:", answer