Comment Re:Another solution.. (Score 1) 101
Obviously I was talking about the general probability of a (weighted) fair coin toss. (8 of them)
Actually, you're totally right about the 4 games being all that matters. (assuming he doesn't alter the moves, and just plays them off each other).
For some reason I automatically assumed that the question was only talking about games that were either won or lost.
But, let's assume you're right. Then the con-artist only won his bet 4.1% of the time. This is not a very good con if the con-artist loses 95.9% of the time.
You suggested there was a 4.1% chance of winning the wager. In fact you have answered, "What is the probability that the next round of 4 games played will be all non-draws?" But, a figure of 4.1% wins is nonsensical for a bet.
The real wager must be, "I cannot be beaten more than 50% of the time." That's the point of playing the opponents off each other.. Because the claim will always be true.
However, as the AC who replied to me pointed out, I failed to notice the brain teaser here which is that "winning 4 or more out of 8 fair coin tosses is greater than 50%". In 8 fair coin tosses, on average there will be an even distribution of x heads per 8 tosses. So, 50-50. But you must add the probability of ((exactly 4 heads out of 8 tosses)/2) to 50% to get the answer. In other words whenever you get 4 heads, there are 4 tails, so it's a draw. The trick is, you're just deciding that all of those are wins.
Then the probability of getting at least x heads in n fair coin tosses is sum(from i = x to n, ((n choose x)/2^n)_i).
Taking your example: (where x = 2 and n = 4) we get: 0.687500 = ~69%
So, while you may have been correct on the literal interpretation of the given problem, showing I was wrong in that regard, you are arguing a moot point, and have provided only useless results. I would offer some constructive criticism, but unfortunately for you, they do not teach common sense in high-school.
oooo burn
Hey, j/k man, I'm an idiot.
Actually, you're totally right about the 4 games being all that matters. (assuming he doesn't alter the moves, and just plays them off each other).
For some reason I automatically assumed that the question was only talking about games that were either won or lost.
But, let's assume you're right. Then the con-artist only won his bet 4.1% of the time. This is not a very good con if the con-artist loses 95.9% of the time.
You suggested there was a 4.1% chance of winning the wager. In fact you have answered, "What is the probability that the next round of 4 games played will be all non-draws?" But, a figure of 4.1% wins is nonsensical for a bet.
The real wager must be, "I cannot be beaten more than 50% of the time." That's the point of playing the opponents off each other.. Because the claim will always be true.
However, as the AC who replied to me pointed out, I failed to notice the brain teaser here which is that "winning 4 or more out of 8 fair coin tosses is greater than 50%". In 8 fair coin tosses, on average there will be an even distribution of x heads per 8 tosses. So, 50-50. But you must add the probability of ((exactly 4 heads out of 8 tosses)/2) to 50% to get the answer. In other words whenever you get 4 heads, there are 4 tails, so it's a draw. The trick is, you're just deciding that all of those are wins.
Then the probability of getting at least x heads in n fair coin tosses is sum(from i = x to n, ((n choose x)/2^n)_i).
Taking your example: (where x = 2 and n = 4) we get: 0.687500 = ~69%
So, while you may have been correct on the literal interpretation of the given problem, showing I was wrong in that regard, you are arguing a moot point, and have provided only useless results. I would offer some constructive criticism, but unfortunately for you, they do not teach common sense in high-school.
oooo burn
Hey, j/k man, I'm an idiot.
