vpFREE2 Forums

Here we go again...

#1

Earlier, I 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.

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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?

Thanks,
Brian

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#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.

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?

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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