Re: True odds bets? Probability says........

On Thu, 5 Jul 2007 13:41:54 -0500, jbkbub@xxxxxxxxx (John B) wrote:


Re: True odds bets? Probability says........

In a fair, even-money game, the most probable single outcome is to be
even after 100 decisions. The probability is .079589. The probabilities
of every other distinct outcome fall away on either side of 50/50, like
this:

Probability of exactly 47 successes 100 trials is 0.066590500
Probability of exactly 48 successes 100 trials is 0.073527010
Probability of exactly 49 successes 100 trials is 0.078028664
Probability of exactly 50 successes 100 trials is 0.079589237
Probability of exactly 51 successes 100 trials is 0.078028664
Probability of exactly 52 successes 100 trials is 0.073527010
Probability of exactly 53 successes 100 trials is 0.066590500

At every odd number of trials, it is impossible to be even. At every
even number of trials, the most likely outcome is to be even. The
probability of being even declines with every additional pair of trials,
because the more trials, the more possible different outcomes. Being
even, however is ALWAYS the most probable for any even number of trials.

Cheers,
Alan Shank
=========
Alan, I appreciate your expertise in the area of probability, and
expectation. I also acknowledge that your expertise is far superior to
mine in these areas.

Lately I have broke out some old probability books, to see if I could
learn anything :).

The math is sometimes beyond me, but I like to think I grasp a concept
now and then.

Maybe I am mistaken on the following concept, I would appreciate your
analysis of it.

"And individual player's expected distance from "0" after "n" number of
trials." (in a fair game)

As I understand the concept in the books:

after one trial....expected distance is 1

after two trials...expected distance is 1

after three trials...expected distance is 1.5

expected distance after (n) trials, (about .8 times the square root of
the number of decisions)

This is only good for even-money, fair bets. The "expected" distance
is the weighted average of the absolute values of the distances from
even for each possible outcome. This "distance from even" is not based
on the mathematical expectation, however, which, for one trial, is .5.
It recognizes that you cannot win half a bet. So, 1-0 and 0-1 are both
distance of 1 away from even.

I have verified up to 20 decisions. For 20, I get 3.52 as the expected
distance from even, which is .787 times square root of 20.

I concluded from this that an individual player is more likely to be
ahead or behind than even after any number of trials....expectation of
an individuals "position" in a random walk. Am I mistaken in that
conclusion?

No, more likely to be ahead or behind than be even, but more likely to
be even than any other single W-L record.

I further concluded that an individual player that loses their first
bet, is more likely to be behind than ahead after (n) number of
decisions. Am I mistaken in that conclusion?

No, we demonstrated that. I sim'd an odds bet on point 4 with a
"handicap" of one loss. The more trials, the less significant the
initial handicap; after 1026 trials, including the initial loss we
consider "as read", the probability of being even or ahead is now
greater than the probability of being behind.

I read expectation as the "expected" distance from zero after (n) number
of decisions.

As I said, it's the weighted (by probability and number of
combinations) average of the distance from even of all the possible
outcomes.

I understand your statement that the "single" most likely position is
zero, but that is not the "expected" position as I read it.

Correct. But that is not the expected value of the bet. The expected
value of a single bet is .5 * bet, just like the expected value of a
passline bet is .01414 * bet. So the difference from expected value is
not the same as difference from an even W-L record. In my program, for
distance from even I used (trials - wins), but for expected value I
would use (expected wins - wins), so for one trial it would be (.5 -
wins), either .5 or -.5.

I understand that I have a long way to go in this discipline.....but I'm
trying :). I also know I can be opinionated when I believe I am right,
it's my nature....but I am working on that too :).

You are making an error, in my view, when you say:

After one point-4-odds bet, you are more likely to be ahead than
behind. (This is a W-L type of expectation). Therefore, since the
expectation from now on is zero (an expected value, not a W-L record),
more players will be ahead.

This flaw is demonstrated because I can argue:

after two bets, more players will be ahead than behind, and since the
expectation from then on is zero, more players will be ahead.

p(0-2) = .444444
p(1-1) = .22222 * 2 ways = .4444444
p(2-0) = .111111
So, p(being ahead) = .5555556.

It's the same argument, with the same flaw - mixing W-L expectation
with bet expectation.

It doesn't wash because this is a fair bet, but not an even-money one.

If you talk about a fair, even-money bet, then the W-L expectation is
almost the same thing as the expected value.

Cheers,
Alan Shank
.

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