# It's a serious game when six-packs are at stake

I'm playing a small-stakes game of a casino poker variant called Okay You Can Stop Holding 'Em. The rules are simple: the player and the house are both dealt two cards face down, there is a round of betting, then one community card is dealt face up on the table, followed by another round of betting. Three of a kind is the best hand, then a pair, then high card. There are no straights or flushes in this game. A single 52-card standard deck is used.

Now, the house/dealer has an enormous tell that everyone in the room is aware of. He always makes a big raise on the first betting round if he is holding exactly a pair of aces, kings or queens, and only with these hands.

So, cards are dealt. I bet, the house makes a big raise. Alarm bells go ding ding ding. I call.

The community card is a king. I have one king in my hand for a pair of kings, meaning that I lose if the house holds three kings or a pair of aces, and win if he holds a pair of queens.

I show the king to my friend sitting next to me and give him a wink. He says: "I bet you a six-pack that you have the best hand".

I love getting six-packs and hate giving them away. Should I take this bet?

• What's the second card in your hand? – ffao Sep 7 '18 at 21:32
• @ffao I think that’s something you have to include in the probability calculations, which frankly was a pain. I’d be interested for the community to find my mistakes though, since I think I’ve made a couple... – El-Guest Sep 7 '18 at 22:01

I believe it depends on what your second card is. If it's an Ace, the chances of the dealer having 2 Queens is higher than the chance of him having 2 Kings or 2 Aces. Otherwise, The chance of him having 2 Queens is the same as him having 2 aces, and then add the chance to have 2 kings means it's more likely the dealer wins.

• This is the intended answer, but I don't think it's immediately clear from the post why the second statement is true. – jafe Sep 8 '18 at 13:45

Okay so you know that

The dealer has either AA, KK, or QQ. You also have a K, and there’s a K on the table. This means that there are 48 cards which your second card might be.

We can break things up into cases:

Case 1: There’s a $\frac{40}{52} = \frac{10}{13}$ chance that you have a 2-J. In this case, there are $C_2^4 = 6$ ways that the dealer could have QQ, times 6 ways that you and the community card could be KK; 6 ways that the dealer could have KK times two ways that you and the community card could be KK; and 6 ways that the dealer could have AA times 6 ways that you and the community card could be KK. You lose $\frac{36 + 12}{36+12+36} = \frac{48}{84} = \frac{4}{7}$ of the time.

Case 2: There is a $\frac{4}{52}=\frac{1}{13}$ chance that you have a Q. In this case, there are 6 ways the dealer could have QQ times 2 ways that you could have the remaining Q times 4 ways for the community K. There are 6 ways the dealer could have KK times 2 ways for the community K times 4 ways for your Q. There are 6 ways the dealer could have AA times 4 ways for the community K times 4 ways for your Q. You lose $\frac{3}{5}$ of the time.

Case 3: There is a $\frac{4}{52}=\frac{1}{13}$ chance that you have a K. In this case, there are 6 ways the dealer could have QQ times 4 ways that you could have a K times 3 ways for the community K. There are 6 ways the dealer could have KK times 2 ways for the community K times 1 way for your K. There are 6 ways the dealer could have AA times 4 ways for the community K times 3 ways for your K. You lose $\frac{14}{26}=\frac{7}{13}$ of the time.

Case 4: There is a $\frac{4}{52}=\frac{1}{13}$ chance that you have a A. In this case, there are 6 ways the dealer could have QQ times 4 ways that you could have the remaining A times 4 ways for the community K. There are 6 ways the dealer could have KK times 4 ways for the community K times 4 ways for your A. There are 6 ways the dealer could have AA times 4 ways for the community K times 2 ways for your A. You lose $\frac{2}{5}$ of the time.

When you put this all together, you will lose

$\frac{10}{13}\frac{4}{7} + \frac{1}{13}\frac{3}{5} + \frac{1}{13}\frac{7}{13} + \frac{1}{13}\frac{2}{5} = \frac{1}{13} + \frac{40}{91} + \frac{7}{169} \approx 55.8\%$ of the time.

We conclude that

You should take the bet, since your preference is that you’d rather win the beer than win the hand.

• The chance you have a K can't be the same as the chance you have a Q given that you already know where two of the Ks are. There are lots of similar mistakes all around. The exact value is kind of hard to compute, but for the yes/no answer I think Chris's argument suffices. (I misread your answer to have the opposite conclusion earlier, sorry). – ffao Sep 7 '18 at 23:17
• @ffao fair enough, thanks for the feedback. I need to stay away from the stats questions, they get me nothing but downvotes! I figured I had covered it the other way round — since the community card is pulled after the second card K I thought I had taken care of it by imagining a new deck of 52 being dealt out. – El-Guest Sep 7 '18 at 23:42

Yes, as stated, there are 3 situations, in 2 of them you'll lose, the other you win. 1/3 vs 2/3 is not good odds. So your friend is likely to be wrong, and you win the six pack, but lose the hand.

• This assumes the outcomes are equally likely, which may or may not be true. (Haven't confirmed yet) – Quintec Sep 7 '18 at 20:57
• okay but as the card on table isn't A or Q, the chances of them having A pair or Q pair is equal. Then there is the additional chance they have a K pair, so the overall value may not be 2/3 but it will still be skewed in their favour – AHKieran Sep 7 '18 at 22:35