Probability question

[Alchemist]

I made a post similar to this in the Probability forum but it didn't get any responses so I'll give it a shot here.

I'm trying to come up with a general formula to do this:

Suppose you play 200 hands in a session. Let's say you're dealt at least one Nine in 55 of your starting hands. What's the likelihood of that and how many SD is that from the norm?

About as far as I've gotten was the probability of getting dealt at least one Nine:
1-[(48/52)(47/51)(46/50)] ~= 0.21738

Would it be correct to say that the expected number of starting hands with a Nine would be: p*H ? In this case: 0.21735 * 200 = 43.5? Then what?

[Alchemist]

Ok I think I found the solution to my own problem. There was a similar question asked at the Wizard of Odds website. So I adapted the equation to my purposes.

let p = probability of an event; in this case, the probability of getting dealt at least one 9 in stud
which shown above is ~=0.21738

h = # hands in a given session

Thus hp = expected number of hands dealt in that session with at least one 9

Using the Wizard's formula:

SD = sqrt (hp - (1-p))

= sqrt (200 x 0.21738 - (1-0.21738))

= sqrt (43.5 - .78262)

= 5.83

So in 200 hands within 1 std deviation, you should see at least one Nine in ~37-49 hands.

Not especially useful I guess, especially since this forum is pretty devoid of any 'poker is rigged' posts.

[beta1607]

I understand exactly none of this.

[Alex/Mugaaz]

[ QUOTE ]
I understand exactly none of this.

[/ QUOTE ]

[Nick_Foxx]

All I know is the probability of catching an offsuit 9 when you start with 3 babies in Stud/8 is about 80-90%.

[BeerMoney]

P(At least one nine) = 1 - P(No nines) = 1 - .78262^55

[jon_1van]

[ QUOTE ]
Ok I think I found the solution to my own problem. There was a similar question asked at the Wizard of Odds website. So I adapted the equation to my purposes.

let p = probability of an event; in this case, the probability of getting dealt at least one 9 in stud
which shown above is ~=0.21738

h = # hands in a given session

Thus hp = expected number of hands dealt in that session with at least one 9

Using the Wizard's formula:

SD = sqrt (hp - (1-p))

= sqrt (200 x 0.21738 - (1-0.21738))

= sqrt (43.5 - .78262)

= 5.83

So in 200 hands within 1 std deviation, you should see at least one Nine in ~37-49 hands.

Not especially useful I guess, especially since this forum is pretty devoid of any 'poker is rigged' posts.

[/ QUOTE ]

Did you get your forumla right?

I don't think you did.

I think the appropriate formula is

SD = sqrt(npq) where n = number of hands, p = prob of getting dealt a 9 in a single hand, q = prob of not get dealt a 9 (or 1-p)

E[x] = p*n = 43.5

SD = 1.86

Your results of 55 hands with a 9 was 6 standard devs away from the average.

[Alchemist]

[ QUOTE ]

Did you get your formula right?

I don't think you did.

I think the appropriate formula is

SD = sqrt(npq) where n = number of hands, p = prob of getting dealt a 9 in a single hand, q = prob of not get dealt a 9 (or 1-p)

E[x] = p*n = 43.5

SD = 1.86

Your results of 55 hands with a 9 was 6 standard devs away from the average.

[/ QUOTE ]
I copied it from the sheet of paper I messily wrote it out on. It should be what you have written but the math I did above should be correct:

SD = sqrt (npq); where q = (1-p)

= sqrt (200 x .21738 x .78272)
= sqrt 34.014
= 5.83

So 55 hands out of 200 with a Nine is
(55-43.5)/5.83 = 1.97
Just under 2 SD from the norm, right?

[jon_1van]

oops. I said you only did 20 hands inside the SQRT.

thats good....6 SD is scary.

[Alchemist]

[ QUOTE ]
P(At least one nine) = 1 - P(No nines) = 1 - .78262^55

[/ QUOTE ]

This would be the probability of getting at least one hand with a Nine in any of 55 deals.
1 - (.78262)^55 = 99.99986%