A POKER MATH QUESTION...I CAN'T FIGURE IT OUT.

Hello,

I am brand new here so (mods) please forgive me if I have placed this question in the wrong section.

I am re-posting this math question once more because I don’t believe that my previous post explained it properly and didn’t give it justice. Thank you to those who answered previously. Perhaps this will make the question a bit clearer.

So here it is again:
For any of you mathematically minded folks out there here is a Texas hold Em math question.

In a heads up game (2 players only) what are the odds of one particular player (not combining both players...one player only) getting nothing less than a Queen (that is A, K, Q only) as one of their pocket cards in 20 out of 21 hands?

EXAMPLE: Heads up poker. 2 players only.
Pocket Cards:

Player 1 - A-6 /Hand 1
Player 1 - A-9 /Hand 2
Player 1 - Q-7 /Hand 3
Player 1 - K-10 /Hand 4
Player 1 - Q-J /Hand 5
Player 1 - A-J /Hand 6
Player 1 - K-8 /Hand 7
Player 1 - K-Q /Hand 8
Player 1 - Q-6 /Hand 9
Player 1 - A-10 /Hand 10
Player 1 - K-7 /Hand 11
Player 1 - A-J /Hand 12
Player 1 - A-10 /Hand 13
Player 1 - K-4 /Hand 14
Player 1 - Q-5 /Hand 15
Player 1 - 2-3 /Hand 16 (The only hand that does NOt have an A-Q in the pocket)
Player 1 - K-J /Hand 17
Player 1 - A-9 /Hand 18
Player 1 - K-4 /Hand 19
Player 1 - A-Q /Hand 20
Player 1 - Q-3 /Hand 21

Would love to know the mathematical odds of this happening.

Thanks in advance.

Southern Belle

>I am brand new here so (mods) please forgive me if I have placed this question in the wrong section.
That's okay. Anyone named Southern Belle can do no wrong!

>I am re-posting this math question
Ah... math question. Well, that lets me out. As I can only count to twenty usually but only to ten in your case since I would never remove my shoes in the presence of a Southern Belle.

I do have one question though. This question that you pose: does it have something to do with a sequence of events that actually took place, perhaps at an online game or something?

Yes...the scenario I have described did indeed take place at a live poker tournament in Las Vegas.

Edit-I screwed up - here is corrected calculation

P(no AKQ) = 40/52 1st card
P(no AKQ given no AKQ 1st card) = 39/51 2nd card

P( no AKQ both cards) = 40/52 x 39/51 = .5882352
P(at least 1 AKQ) = 1- .5882352 = .4117648

20 times out of 21 hands is Binomial P=.4117648, N = 21, x = 20. P(x=20) = 2.42524 exp -07

I think that in any hand, the odds of having at least one card queen or greater is (12/52)*2=46.154%. So the odds of doing it 20 times is .46154^20=.00000019239, or 1 in about 5,200,000. I don't know how to factor in the one failed hand (or even if this is right for the other 20...)

This is what I calculated on the previous thread. This answers your question. This is for any given 21 hands and includes the cases where the single hand without the queen+ happens on the first or last hands. It also assumes that you got to see the player's cards for every hand.

Quote: CrystalMath

I calculate that the probability of getting any Queen or higher in two cards is 1-combin(40,2)/combin(52,2). Actually, I'm calculating the probability of getting two cards that are jacks or lower (which is a simpler calculation) and then subtracting that from 1 to get the odds of getting any queen or higher.

p(queens+ in one hand) = 0.411764706

The original question said what is the probability of this happening in 20 out of 21 hands.

p(queens+ in 20 of 21 hands) = 0.411764706^20 * (1-0.411764706)^1 * combin(21,1) = 2.42523E-07

Therefore, the odds of this happening is 1 : 4,123,322.026, which is four times better than hitting the top award on most slot machines.

So, who won? How many all-ins? How long did it last (ending at 21 hands)?

Quote: matilda

Edit-I screwed up - here is corrected calculation

P(no AKQ) = 40/52 1st card
P(no AKQ given no AKQ 1st card) = 39/51 2nd card

P( no AKQ both cards) = 40/52 x 39/51 = .5882352
P(at least 1 AKQ) = 1- .5882352 = .4117648

20 times out of 21 hands is Binomial P=.4117648, N = 21, x = 20. P(x=20) = 2.42524 exp -07

Thank you matilda and all for your input.

CrystalMath...are you stating that the odds of this event are 1 in approx. 4.1 million?

I am not really a math whiz so for the sake of us folks who are not able to decipher mathematical equations what do your figures boil down to in layman's terms?

Like 1 in 1,000,000 ...

Thanks in advance matilda and all.

They both agree, 1 in 4.1 million. About the same odds of winning some lotteries.

Thank you everyone...I really appreciate your time and answers.