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Hosayif
07-14-2005, 01:14 AM
what is the probability of NOT being dealt aces in 600 hands?
uuDevil
07-14-2005, 01:24 AM
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what is the probability of NOT being dealt aces in 600 hands?

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In Excel,

=BINOMDIST(0,600,1/221,FALSE) gives a result of 0.065802786
BruceZ
07-14-2005, 02:07 AM
[ QUOTE ]
[ QUOTE ]
what is the probability of NOT being dealt aces in 600 hands?

[/ QUOTE ]

In Excel,

=BINOMDIST(0,600,1/221,FALSE) gives a result of 0.065802786

[/ QUOTE ]

(220/221)^600 works too.
Hosayif
07-14-2005, 08:54 PM
thanks both of you, but is it .06 percent or 6 percent (the running count is at 790 now btw)
Hosayif
07-14-2005, 09:01 PM
i finally got them!!! 795 hands and it ended up making the nut flush
Paxosmotic
07-14-2005, 09:09 PM
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
what is the probability of NOT being dealt aces in 600 hands?

[/ QUOTE ]

In Excel,

=BINOMDIST(0,600,1/221,FALSE) gives a result of 0.065802786

[/ QUOTE ]

(220/221)^600 works too.

[/ QUOTE ]
Hey now, that's a pretty good trick, I'm going to use that one.
uuDevil
07-14-2005, 10:40 PM
[ QUOTE ]
i finally got them!!! 795 hands and it ended up making the nut flush

[/ QUOTE ]

(220/221)^600=.065803 is ~6.58%, or odds of 14.2:1 against

(220/221)^794=.027299 is ~2.73% or odds of 35.6:1 against

Spreadsheets are handy. The windows calculator works ok too.
Hosayif
07-14-2005, 11:44 PM
ok so i got them again on hand 881, so does that mean the chances of that happening are (1-((220/221)^(881-795)=67.7%)=32.3%?

I got them again in hand 911, though they were called and sucked out on by some moron with 93s. Dont play at bugsys club
uuDevil
07-15-2005, 02:02 AM
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ok so i got them again on hand 881, so does that mean the chances of that happening are (1-((220/221)^(881-795)=67.7%)=32.3%?

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I don't understand that expression.

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I got them again in hand 911,....

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So 3 times in 911 hands? If you want to understand this you may want to read up on the binomial probability distribution. The calculation is easy in Excel:

=binomdist(3,1/221,911,true) gives .409492, i.e. a 40.9% probability of getting AA 3x or fewer in 911 hands

=binomdist(3,1/221,911,false) gives .189411, i.e. an 18.9% probability of getting AA exactly 3x in 911 hands

The most likely number of times you would get AA in 911 hands is 4, and that will happen 19.5% of the time.

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....though they were called and sucked out on by some moron with 93s. Dont play at bugsys club

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You've got this wrong. Even if you never learn to calculate probabilities, you should understand this.