Poker Math Question

Hello,

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

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

In a heads up game 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?

Would love to know the mathematical odds of this happening.

Thanks in advance.

Southern Belle

3*4 cards of 52 (12/52) ,
then 11 of 51 (12/52*11/51)
happening 20 times (12/52*11/51)^20
one of 20 ways 20*(12/52*11/51)^20

About 1.8*10^-25. A very small number.

One, not both.

I think A,Q would qualify as valid. I get 1 in 85,685,896,462

Maybe OP means Q-high or better? As in, Q-2 would also count. Then the chance is more reasonable

One in 85 billion is not reasonable!

I had to recalculate but it's still one in 71.6 billion.

The odds of getting at least one Queen or higher are 12/52 plus 11/51 = 0.446 455 505 279
The odds of that happening 20 times in a row are 0.000 000 098 984 or 1 in 10,102,674.

For what it's worth, the odds of getting pocket deuces thru pocket jacks are 40/52 * 3/51 = 0.045 248 868 778
The totoal odds of getting at least one queen, or pocket pairs, is almost 50%: 0.491 704 374 057
The odds of it happening 20 times in a row are 0.000 000 682 467 or 1 in 1,465,273.

For the record, I'm not sure if any of these are exactly what is being asked in the original post.

Oops. My thinking was wrong. 12/52 + 11/51 is only part of a complex formula.

It's so complex, that it's actually easier to work it backwards. I.E. Take the odds of NOT getting any card higher than a Jack, and subtract that from 1. Therefore the odds of getting at least one Queen is 1 - ( 40/52 * 39/51 ) = 0.411 764 705 882

The odds of that happening 20 times is 0.000 000 019 633 or 1 in 50,935,154.

That makes the odds of getting a queen or better or a pocket pair 0.457 013 574 661

The odds of that happening 20 times is 0.000 000 157 973 or 1 in 6,330,208.



By comparison, the odds of getting Red on a double zero Roulette table is 0.473 684 210 526

The odds of getting 20 Reds in a row is 0.000 000 323 428 or 1 in 3,091,874.

Quote: s2dbaker

One in 85 billion is not reasonable!

I had to recalculate but it's still one in 71.6 billion.

I wasn't including queens, D'oh!

Here's my math:
Chance of not pulling a Q or better with the first cad is .76923....
Chance of not pulling a Q or better on the second card is. 76471....
Combined = .58824...
Inverse to get remaining (desired result) = .41176...
20 times in a row = 1 in 50,935,154

Better than a lottery ticket.

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.