Re: No result is likely!!!(I could be confused)
- From: Alan Shank <goatcabin@xxxxxxxxxxxx>
- Date: Mon, 05 Dec 2005 12:30:17 -0800
On Mon, 5 Dec 2005 12:11:10 -0600, jbkbub@xxxxxxxxx (John Kerr) wrote:
>I understand that the mode is, to be even for any even money odds bet
>set where being even is a possible oucome.
>
>I also understand that after the game is started, that the payer can
>expect to remain ahead, or behind, depending on the outcome of the 1st.
>decision. The most probable number of times a player can expect to be
>even after the game begins, is zero, followed by one time, two times,
>etc.
>
>These two statements seem to be in conflict to me, when thinking only of
>one game set...and one player.
>
>If one player is expected to be behind or ahead, depending on the 1st.
>decision, then how can they also expect to be even once the game begins?
>
>I understand that expectation, prior to the game beging....but it must
>begin, or everything remains "theoretical".....and Alan wouldn't allow
>me to use theoretical when I said "After you win your first bet in a 20
>coin flip, you can expect to win 9.5 of the remaining 19 flips, giving
>you a theoretical expectation of being better than even."
>
>I certainly have no quarrel with breaking-even being the theoretical
>mode in an even odds set of wagers.
>
>It just seems to me, that in reality, the first decision changes things
>from theoretical to actual....in a 20 decision set, with the first
>decision being predetermined, we are now actually left with 19 as the
>set, which does not have a break even mode.
Yes, it is actual, and so you have to use the ACTUAL possibilities for
the remaining 19 flips. I demonstrated that, whichever way the first
toss went, the sum of the probabilities of
heads first toss + 9 more heads next 19 AND
tails first toss + 9 more tails next 19
is still exactly the same as the probability of 10 of each before you
start.
You are mixing the theoretical with the actual, because you start with
an actual result, then use the theoretical 9.5 expectation to add on.
They cannot be mixed.
>As any set unfolds, and
>after each decision, inclding the first, a new calculation is in order!
Yes, and I have shown the new calculation after one decision for coin
flips and odds bets on 4/10, and they support the position that, for
any number of trials divisible by the number of "ways," the most
likely single outcome is the "expected" one. The more trials, the less
likely that outcome is, but it is still the MOST likely. What is
happening is that, as more trials or bets are finished, the number of
different distinguishable outcomes grows to a huge number, of which
being even is only one.
>
>I believe the problem lies in you all, holding out that the set is 20,
>when I am seeing it as 19, because of the first decision being
>predetermined, thus removing it from the original theoretical set of 20.
No, the problem is you mixing the actual and theoretical. The first
decision is not "predetermined." You have to consider both branches of
the decision tree.
>
>You are looking at it before the happening, and I am looking at it one
>decision into the happening.
Then you are talking about 19 trials, a set where it's impossible to
be even.
Cheers,
Alan Shank
.
- References:
- Re: No result is likely!!!
- From: Cat_in_awe
- Re: No result is likely!!!(I could be confused)
- From: John Kerr
- Re: No result is likely!!!
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