Two Aces Puzzle

If a poker player who always tells the truth announces to the table that he has an ace, what are the odds that his other hole card is also an ace?

It is against the rules at virtually all poker tables in casinos and tournaments to orally reveal any part of your hand to anyone and/or everyone at the table before the action has concluded.

This is different than the two dice problem, because the first ace has been dealt from a card deck which represents a finite pool of cards. I believe the odds of the other card being an Ace is 3/51 or 1/17.

Quote: JackSpade

If a poker player who always tells the truth announces to the table that he has an ace, what are the odds that his other hole card is also an ace?
link to original post

JAckSpade,
Intentionally or unintentionally, you have just posted a variation of the 'Two Dice Puzzle...
https://wizardofvegas.com/forum/questions-and-answers/math/15508-two-dice-puzzle/

That puzzle has been argued here for over NINE YEARS. It is one of the bitterest threads, actually sets of threads, in forum history and has seen more member suspensions than any other.

To answer the question.....

There is not enough information:
Does he only declare anything if he holds an ace, or does he always declare the value of one of his cards.
Without the calling rules being explicit, the probability could be anywhere between 0 and 100%

And on that note, I'm outta this thread.
Send in the clowns.

Quote: gordonm888

It is against the rules at virtually all poker tables in casinos and tournaments to orally reveal any part of your hand to anyone and/or everyone at the table before the action has concluded.

This is different than the two dice problem, because the first ace has been dealt from a card deck which represents a finite pool of cards. I believe the odds of the other card being an Ace is 3/51 or 1/17.
link to original post

Can we fairly say that this is a prop bet outside the course of a regular hand, where the always honest poker player has two (and only two) cards face down?

Quote: gordonm888

I believe the odds of the other card being an Ace is 3/51 or 1/17.
link to original post

I like this answer.

We could say that one player who is holding two cards that aren't aces is trying to figure out the answer before deciding whether to go all-in. At many casinos, if there are only two players left in the hand, there are no rules against one of them telling the other what he has or even turning a card face up.

But first, you'd have to decide whether a person who always told the truth would ever say he had "an" ace if actually he had two aces.

Imagine if this truth teller said "I have a daughter." Could he have two daughters? Maybe, depending on the context of the conversation.

What if the player was asked specifically before announcing his hand, "Do you have pocket aces?" THEN he responded, "I have an ace." That statement in that context would seem to specifically negate the possibility that he has two aces.

But if he did have two aces, wouldn't he be lying if he said "I don't have an ace"?

I agree with your statement

Yeah, this is a lot like the two dice problem. One of the biggest stumbling blocks of those threads was whether the peeker was always reporting on a specific value or merely either of the values he sees.

For this, let's assume he is always reporting on the presence of at least one ace.

Also, let's assume we're talking about a two card hand, and ignoring the rule about revealing hand values. If you want, we can say he has a 'tell' where everyone knows when he has at least one ace.


The odds of getting an ace on the first card and not caring about the other card is:
( 4 / 52 ) * ( 51 / 51 ) = 0.076923
Note that that is also the odds of getting an ace on only one card.

The odds of not getting an ace on the first card but getting it on the second card is:
( 48 / 52 ) * ( 4 / 51 ) = 0.072398

Combined, that means the odds of getting at least one ace is:
0.076923 + 0.072398 = 0.149321
Note that that is slightly less than double the odds of getting an ace in one card, which makes sense.


For what it's worth, the odds of not getting an ace in two cards is:
( 48 / 52 ) * ( 47 / 51 ) = .850679

Note that if you add those two numbers, you get 100%.


Next, the odds of getting two aces is:
( 4 / 52 ) * ( 3 / 51 ) = 0.004525

So, the odds of getting two aces, after revealing you have at least one ace, is:
( Two aces ) / ( At least one ace ) = 0.030303


Here's an entirely different way to calculate it:

There are 52 * 51 = 2652 unique 2 card hands

There are 4 * 51 unique hands where the first card is an ace.

There are 4 * 51 - 6 = 198 unique hands where at least one card is an ace. Had to eliminate the 6 hands that got counted twice.

There are 6 / 198 = 0.030303 hands where at least one card is an ace and the other card is also an ace.


For the record, even though I came up with the same number two different ways, I think it should be higher, closer to the 3/51=0.058824 that was already suggested. Closer, but not exactly.

Can someone confirm or contradict my math, please?


-----


On a completely separate side note, there was an episode of High Stakes Poker several years ago where Doyle was short stacked and under the gun. He looks at the first card, and goes all in before the dealer gives him his second card. Someone (Daniel?) remarks, "I guess we all know what that card is..."

At the showdown, he won the main pot when both cards were aces!

As has been stated, the wording of the original question is vague.

I think what you mean to ask is something like then following.

There is a poker player who always answers all questions truthfully. After checking his two hole cards he is asked, "Do you have at least one ace?" He responds with a 'yes.' What is the probability he has two aces?

Quote: DJTeddyBear

Can someone confirm or contradict my math, please?

link to original post

For what it's worth I got to the same result with nearly the same reasoning, as copied raw from my notepad before reading your answer:

52*51 combinations

4*51 has the first card as an ace (EDIT: and possibly also the second card as ace)
48*4 has the second card as an ace ((EDIT: but the first card is definitely not an ace)

of these

4*3 has the first and second as an ace

chance of 2 aces is 4*3/(4*51+48*4) == 3,03%

Wow, that 2 dice thread must be a doosey, 3 mods and the wiz, all on the first page. Sheeesh.

Quote: JackSpade

If a poker player who always tells the truth announces to the table that he has an ace, what are the odds that his other hole card is also an ace?
link to original post

Zero, assuming that, in addition to "always telling the truth," he is speaking "proper" English. "An ace" means that he has just one.