> That's what I did with my abacus

>
> (8/104)*(7/103)*(6/102)*(96/101)*(95/100)*(94/99)*
>
> (93/98)*(92/97)*(91/96)*(90/95)*(89/94)*(88/93)*
> (87/92)*(86/91)*(85/90)*455=

> =0.07332209549650789440

> Sincerely,
> Cacarulo

That's right, Sir. I've got also:

0.073321885

So probably Don is right and the final probability
is 100/7.33 = 13.64 or one in 13.64 odds.

This figure makes a lot of sense, btw it doesn't
mean that your approach using the RC tricks or
mine, with the binomial distribution for an ace
count typically with sampling without replacement,
are both wrong or producing erroneous data. It's
just that we are reading the results under the
Normal curve wrong.

My z =1.92 is yours (1.96 + 1.88)/2 don't forget.

For z=1.9245 you get Qz = 0.0271 but that's to the
right in other words, that's the probability for
more than 3 aces, so is not the correct answer and I was correct not to 'gamble' on it. Too much
speed for a not so easy problem.

Have a look at your figure 0.0051. That will mean
you'll have 3 aces in the first 15 cards only once every 196 deals. Doesn't make sense, does it?

Are we both wrong Cac? I guess probably not, maybe just blind under the Gaussian curve.

I need more time to think about this. Let's
improve the normal approach.

Regards
Z

Regards