vpFREE2 Forums

Here we go again...

#1

Thanks, sounds good!

I neglected to divide my number (137,747,462) by Q^44 (.884901525). Doing so
gives your result. I think my number gives the chances of six OR MORE quads,
while yours gives the chances of EXACTLY six quads, which was what I was
after.

So, it's the product of--the probability of hitting the hand six times, the
probability of NOT hitting the hand 44 times, and the number of ways you can
arrange the six hits among the 50 hands. Sounds right to me!

No wonder I've only hit six quads once!

Brian

ยทยทยท

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In a message dated 1/16/2008 6:57:05 PM Pacific Standard Time,
jeff-cole@comcast.net writes:

bjaygold@... wrote:
My biggest 50-play longshot was holding a pair of 2s and getting SIX
quads. I could use some help with this one, as the number I come up
with is too high to be correct. I think.

and also wrote:
Here's how I calculated it.
Chances of hitting a quad from a pair is one in 360.333.
So, (360.333 ^ 6) / [50! / (44! * 6!)] = one in 137,747,462.
Did I calculate this correctly?

Brian, if we let P = 45/C(47, 3) [that's for the 360.333]
and let Q = 1 - P,
then C(50, 6) * P^6 * Q^44 gives us the probability of 6 hits, which
is 1/155,664,171. How does that sound?
Jeff

**************Start the year off right. Easy ways to stay in shape.
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[Non-text portions of this message have been removed]

#2

bjaygold@ wrote:

My biggest 50-play longshot was holding a pair of 2s and getting SIX
quads. I could use some help with this one, as the number I come up
with is too high to be correct. I think ...
Here's how I calculated it.
Chances of hitting a quad from a pair is one in 360.333.
So, (360.333 ^ 6) / [50! / (44! * 6!)] = one in 137,747,462.
Did I calculate this correctly?

jeff-cole wrote:

Brian, if we let P = 45/C(47, 3) [that's for the 360.333]
and let Q = 1 - P,
then C(50, 6) * P^6 * Q^44 gives us the probability of 6 hits, which
is 1/155,664,171. How does that sound?

bjaygold wrote:

Thanks, sounds good!
I neglected to divide my number (137,747,462) by Q^44 (.884901525).
Doing so gives your result. I think my number gives the chances of
six OR MORE quads, while yours gives the chances of EXACTLY six
quads, which was what I was after.

So, it's the product of--the probability of hitting the hand six
times, the probability of NOT hitting the hand 44 times, and the
number of ways you can arrange the six hits among the 50 hands.
Sounds right to me!

Your last phrasing of the problem is accurate, right on, Brian.
However, your calculation was not the probability of six hits or mroe.

The only way to calculate a cumulative probability such as that is to
subtract from 1 the individual probabilities of hitting once, twice,
... and five times -- using the same method as that used to calculate
P(6 hits).

Excel provides a function that can do calculations for you: BINOMDIST
-- where you specify the number of suceesses (hits), trials (hands
played), probability of a success, and whether you want a cumulative
result (specified as a 1 or 0).

Using this function non-cumulatively yields Jeff's result of
1/155,664,171 for 6 hits. To determine the probability of 6 or more
hits, subtract the cumulative (0 to 5) hit result from 1 and you end
up with a probability of 1/152,948,203. (The odds of getting more
than 6 hits are exceedingly small.)

- Harry