Re: Martingale in the field



On 3 Mar 2007 14:09:49 -0800, "Doc Dice"
<steve_johnson8@xxxxxxxxxxxxx> wrote:

{snip} > Lets talk very simple probabilities (not statistics) ...
The chances of flipping a "fair" coin and getting a specific result (H or T) is
.5 or 50% or 1 in 2 or 1 to 1. All these express the same probability (I will
stick with the first). The chances of getting *any* specific fair coin flipping
result in sequence is calculated by the chance of getting one specific result
times the number of trials in the sequence.

I agree with you Mason, however, we both know that the odds are in
play each time the dice are rolled. My small point is that when one
goes to the table those odds are hardly ever seen as we are at the
table for a such a short period of time.

You are confused, Doc. The odds are seen EVERY single roll, EVERY
single bet, EVERY single session, and over your entire lifetime of
playing craps. What is often not seen when you go to the table for a
sessions is a set of outcomes (rolls, bets, or the whole session) that
follows the expectation very closely. That is to be expected, and
FLOWS directly from the odds being "in play each time the dice are
rolled", as you say.

If you roll a pair of dice 36 times, what are the odds that you will
actually see 1 2, 1 12, 2 11s, 2 3s, 3 4s, 3 10s, 4 9s, 4 5s, 5 6s, 5
8s and 6 7s? I think we figured this out once, and it's pretty damn
unlikely. However, it is MORE likely than any other distinct set of
rolls, just much less likely than ALL the other sets of rolls put
together.

This same principle can be illustrated by examining the probability
distribution of 10 coin flips. What are the chances of getting exactly
5 head and 5 tails, which is the "expected value"? The answer is
..2461. So the probability of NOT getting 5 and 5 is over .75. However,
the probability of getting 4 heads is .20508 and 6 heads is also
..20508. In fact, the distribution forms a pyramid, just like the
"perfect 36".

The odds are ALWAYS there; they are built into the fabric of the game,
like pi and e are built into the fabric of the Universe. See below:

Probability of exactly 0 successes in 10 trials is 0.000976562
cumulative: 0.000976562
Probability of exactly 1 successes in 10 trials is 0.009765625
cumulative: 0.010742188
Probability of exactly 2 successes in 10 trials is 0.043945312
cumulative: 0.054687500
Probability of exactly 3 successes in 10 trials is 0.117187500
cumulative: 0.171875000
Probability of exactly 4 successes in 10 trials is 0.205078125
cumulative: 0.376953125
Probability of exactly 5 successes in 10 trials is 0.246093750
cumulative: 0.623046875
Probability of exactly 6 successes in 10 trials is 0.205078125
cumulative: 0.828125000
Probability of exactly 7 successes in 10 trials is 0.117187500
cumulative: 0.945312500
Probability of exactly 8 successes in 10 trials is 0.043945312
cumulative: 0.989257812
Probability of exactly 9 successes in 10 trials is 0.009765625
cumulative: 0.999023438
Probability of exactly 10 successes in 10 trials is 0.000976562
cumulative: 1.000000000

Cheers,
Alan Shank

Therefore each player looks
for variants that occur at the table. For example, how many times have
you seen a shooter hit two, three or even fourteen numbers, as stated
earlier, in a row. Thats the variant that occurs and on occasion turns
into a super roll. Each one of us has stories of super rolls that
happen to themselves or others playing at a table.

i enjoy reading your, Shanks & Gregs stuff

DD
.



Relevant Pages

  • Re: Multiplicative Chernoff bound
    ... and the average number of failures and successes m per sample. ... the chernoff bound to find e such that with probability c, ... relative error of my estimation m of p is as small as I want (with ... and the confidence intervals from this will be asymptotically ...
    (sci.stat.math)
  • Re: please check my homework
    ... > /* return the probability of x successes in n events, ... > given the probability p for success in a single event, ... In the above comment you should specify that n must be a positive integer, ...
    (comp.programming)
  • Re: Simple confidence interval
    ... Say I run 10 trials and I get 7 successes. ... Is there a way to calculate a probability range (such ... You can obtain confidence intervals for the binomial parameter by using the relationship between the binomial distribution and the beta distribution and the relationship between the beta distribution and the F distribution. ... Let x be the observed number of successes, n be the number of trials, and 1-alpha be the confidence level. ...
    (sci.stat.math)
  • Re: Incoherent Odds?
    ... Probability is a measurement of the likelihood of an event happenning. ... the limit of the ratio s/n, ... odds are the ratio of failures to successes (where ... you get four failures for every one success; ...
    (sci.math)
  • Re: Why is the
    ... I neither claim nor disclaim membership in the math commune. ... To come out even or better after 30 Any 7 bets the player must see at ... probability of having this much "good luck" is zero to a few hundred (or ... rolls does the math say the ratio of sevens to rolls must become 1:6; ...
    (rec.gambling.craps)