Poker Math Question

I will also assume that the original post meant "at least one of the two cards is a Queen or better".

The probability of it happening once is 1 minus the probability that both cards are 2 through Jack.
There are (52 * 51) / 2 different sets of pocket cards, of which (40 * 39) / 2 are both 2-J, so the probability of getting at least one A/K/Q is 1 - 780/1326 = 0.4117647.

The probability of it happening exactly 20 times out of 21 is (the probability of it happening once)^20 * (the probability of it not happening, once) * 21 (since any of the 21 hands can be the "bad" one) = 1 / 4,123,322.

If one of the two cards is AKQ and the other is 2-J, there are 12 * 40 = 480 sets of these pocket cards, so the probability of getting exactly one A/K/Q is 480/1326 = 0.362, and the probability of this happening exactly 20 times out of 21 = 0.362^20 * 0.638 * 21 = about 1 / 50,000,000.

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-0.

I agree with this calculation.. Once the p = .411764706 is established, the next calculation is a simple binomial probability of 20 successes in 21 trials, except the binomial uses combin(21,20) instead of combin(20,1) and both give the same result.

Quote: dwheatley

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

Yes...this is what I meant.

In 20 out of 21 hands straight ONE of the players cards was never lower than a queen. However remember that I am speaking of the same player as well. The other was any card from 2 thru Ace.

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


Recalculations anyone?

Thanks in advance.

Quote: ThatDonGuy

The probability of it happening exactly 20 times out of 21 is (the probability of it happening once)^20 * (the probability of it not happening, once) * 21 (since any of the 21 hands can be the "bad" one) = 1 / 4,123,322.

Technically, the Non-Queen-or-better hand could not be the first or last hand since that would make it 20 in a row.

Granted, the chances are remote (Mike - please help this thread...)
Since we're talking about poker - did we consider the flop here, and how many hands were won/lost?
You hold 9-8 of diamonds, and the flop is 9h,8s, and 3c, you might want to stay in...
You play your hand, not just your hole cards...

Quote: s2dbaker

Technically, the Non-Queen-or-better hand could not be the first or last hand since that would make it 20 in a row.

20 in a row would still be 20 out of 21, whatever the denominator (number of trials) is specified.

Quote: s2dbaker

with the first cad....

Boston pokah playah: These caads ah maaked! LOL.

Quote: NowTheSerpent

20 in a row would still be 20 out of 21, whatever the denominator (number of trials) is specified.

OP was using a real world example. If it was 20 in a row, he would have said it.

Quote: NowTheSerpent

Quote: s2dbaker

with the first cad....

Boston pokah playah: These caads ah maaked! LOL.

Many of my missives are written on a phone on a bouncy Long Island Rail Road train. I think I do pretty well considering the circumstances.