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# Poker Deal Probability PuzzlerPrev TopicNext Topic

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• Sugar Land
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You are an observer of a poker deal experiment.

The experiment administrator has two decks of playing cards: one deck is the standard 52 cards (no Jokers), and the second deck is the standard 52 cards plus 2 Jokers (54 total).

The administrator chooses one of the two decks at random with 50% probability of each, but you are not informed of which one is chosen.

Next the administrator deals two five-card poker hands from the (same, but unknown) chosen deck to two players. The result is both players have drawn a Royal Flush. Furthermore, no Jokers appear in either hand (the Royal Flushes are not completed with any Jokers).

Given the result of the experiment, what is the probability that the administrator chose the standard 52 card deck for the deal?

• United States
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Nice puzzle. I suppose it doesn't really matter that the hands are royal flushes, it's only important that 10 cards were dealt and none was a joker. In that case the probability the 52-card deck was chosen is 1431/2377 ≈ 0.60202

I worked it out like this: The probability of getting no jokers in 10 cards with a 52-card deck is 1. And the probability of the same with a 54-card joker deck is 946/1431 ≈ 0.66108. And since both decks are chosen with equal likelihood, the conditional probability is 1/(1 + (946/1431)) = 1431/2377.

I also checked it with simulations and got the same. I hope I'm not missing a subtlety where the fact that the hands are royal flushes turns out to be important.

• Sugar Land
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Quote: Originally posted by cottoneyedjoe on Aug 8, 2022

Nice puzzle. I suppose it doesn't really matter that the hands are royal flushes, it's only important that 10 cards were dealt and none was a joker. In that case the probability the 52-card deck was chosen is 1431/2377 ≈ 0.60202

I worked it out like this: The probability of getting no jokers in 10 cards with a 52-card deck is 1. And the probability of the same with a 54-card joker deck is 946/1431 ≈ 0.66108. And since both decks are chosen with equal likelihood, the conditional probability is 1/(1 + (946/1431)) = 1431/2377.

I also checked it with simulations and got the same. I hope I'm not missing a subtlety where the fact that the hands are royal flushes turns out to be important.

You're correct. It doesn't matter what the two hands are, other than the fact that no Joker appears. That was the "trick" part of the question. The exact answer is 1431/2377, as you correctly reported.

• Sugar Land
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Question 2: Suppose after the deal in the first experiment, the decks are gathered up and reshuffled. However, the deck used on the first experiment is still not revealed to you. The experiment is then repeated for a second time, except for the fact that the same (unknown) deck from the first experiment is used, not a randomly chosen deck. Two five-card hands are dealt, neither of which contains a Joker. What is the updated probability that the deck dealt (in both experiments) was the standard 52 card deck with no Joker?

Question 3: Suppose the particulars of the first experiment in the original question, except that the first deck is now the standard 52 cards plus ONE Joker (53 cards total) The second deck is still 52 standard cards plus TWO Jokers (54 cards total). The result is no Jokers in the two-five card hands dealt. Given this result, what is the updated probability that the deck chosen for the deal was the one with 53 cards (one Joker)?

• Sugar Land
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Question 4: Same particulars as Question 3 except that the result of the two five-card hands dealt is that one of the two players is dealt a single Joker (the other player was not dealt a Joker). With this result, what is the probability the deck used for the deal was the 53 card deck with one Joker?

• Sugar Land
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Quote: Originally posted by Orange71 on Aug 8, 2022

Question 2: Suppose after the deal in the first experiment, the decks are gathered up and reshuffled. However, the deck used on the first experiment is still not revealed to you. The experiment is then repeated for a second time, except for the fact that the same (unknown) deck from the first experiment is used, not a randomly chosen deck. Two five-card hands are dealt, neither of which contains a Joker. What is the updated probability that the deck dealt (in both experiments) was the standard 52 card deck with no Joker?

Question 3: Suppose the particulars of the first experiment in the original question, except that the first deck is now the standard 52 cards plus ONE Joker (53 cards total) The second deck is still 52 standard cards plus TWO Jokers (54 cards total). The result is no Jokers in the two-five card hands dealt. Given this result, what is the updated probability that the deck chosen for the deal was the one with 53 cards (one Joker)?

You are the one in a million fascinated by these puzzles, and a glutton for punishment. We aim to please, so let's propose a spin on Question 2. Call it "Question 2b".

Question 2b: Instead of two rounds of deals as in Question 2, we have a single round of deals, but with two additional players. So, there is a total of four hands of five cards dealt to four players, 20 cards total, and the result of the experiment is no player has a Joker. What is the probability that the deck chosen was the standard 52 card deck with no Jokers? If you wish, comment on the magnitude of this result compared to the result of the previous Question 2.

• United States
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Quote: Originally posted by Orange71 on Aug 8, 2022

Question 2: Suppose after the deal in the first experiment, the decks are gathered up and reshuffled. However, the deck used on the first experiment is still not revealed to you. The experiment is then repeated for a second time, except for the fact that the same (unknown) deck from the first experiment is used, not a randomly chosen deck. Two five-card hands are dealt, neither of which contains a Joker. What is the updated probability that the deck dealt (in both experiments) was the standard 52 card deck with no Joker?

Question 3: Suppose the particulars of the first experiment in the original question, except that the first deck is now the standard 52 cards plus ONE Joker (53 cards total) The second deck is still 52 standard cards plus TWO Jokers (54 cards total). The result is no Jokers in the two-five card hands dealt. Given this result, what is the updated probability that the deck chosen for the deal was the one with 53 cards (one Joker)?

Q2) In this one you square all the intermediate probabilities because it's two independent trials with the same outcome back to back. Probability the 52-card deck was used, given that two deals contained no jokers is 1^2/(1^2 + (946/1431)^2) = 2047761/2942677 ≈ 0.69588.

Q3) The fractions become simpler (and closer to 0.5) as the difference between the number of jokers in the decks decreases. This one should be 27/49 ≈ 0.55102.

While I was doing the first problem, I inadvertently rediscovered a nice combinatorial identity. (I worked it out two different ways to make sure the answer was coherent.)

(N choose K)*(N+M choose M) = (N+M-K choose M)*(N+M choose K)

I didn't have a chance to work out the other two questions yet, but here is a meta-question. How could one minimally alter the original question so that the fact of being dealt two royal flushes is not a red herring but actually a pertinent detail? One possibility is that the 54-card deck contains two extra cards that are part of royal flushes, for example, K♠ and A♠, or A♠ and A♥. Then the probability of the deck being the 52-card deck would be slightly less than half. Working out that question seems harder, with a lot of cases to analyze separately.

• Sugar Land
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Cottoneyedjoe, agree/correct on both of your solutions. Try Q2b and 4. They are fun.

I always loved trick questions when I was in school, as they test true knowledge of the concept. (OK, I'm lying a bit on loving them.) However, in "real life" we're confronted with dealing with lots of confounding/irrelevant information, so, in that sense, it's good to throw them into problems for good measure.

• Sugar Land
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To take up your meta-question challenge, let's try a spin on Q4. You have 53 card (1 Joker) and 54 card (2 Joker) decks. One is chosen at random (50% probability of each). Two hands are dealt (5 cards each) from the same deck. You're told that both hands are Royal Flushes, but one of them is a natural Royal Flush (no Jokers), and the second Royal Flush hand is completed with exactly one Joker. However, you are not shown the hands, so you don't know which missing natural Royal Flush card is replaced by the Joker. Given the experiment results, what is the probability the deck from which the hands were dealt is the 53-card / 1 Joker deck?

Note: I haven't yet worked the answer myself. I think this will be harder than the others.

• United States
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Quote: Originally posted by Orange71 on Aug 8, 2022

To take up your meta-question challenge, let's try a spin on Q4. You have 53 card (1 Joker) and 54 card (2 Joker) decks. One is chosen at random (50% probability of each). Two hands are dealt (5 cards each) from the same deck. You're told that both hands are Royal Flushes, but one of them is a natural Royal Flush (no Jokers), and the second Royal Flush hand is completed with exactly one Joker. However, you are not shown the hands, so you don't know which missing natural Royal Flush card is replaced by the Joker. Given the experiment results, what is the probability the deck from which the hands were dealt is the 53-card / 1 Joker deck?

Note: I haven't yet worked the answer myself. I think this will be harder than the others.

This new spin still gives the same answer as Q4, which is 27/71.

In the first deck, there's 4 ways to choose the suit of the natural r.f., 3 ways the choose the suit of the unnatrual r.f., and 5 ways to choose the card that's being replaced by the joker, and 1 way to choose the joker. So the probability is 4*3*5/(53 choose 10)

In the second deck, there's again 4 ways to choose the suit of the natural r.f., 3 ways the choose the suit of the unnatrual r.f., and 5 ways to choose the card that's being replaced by the joker, but now 2 ways to choose the joker. So the probability is 4*3*5*2/(54 choose 10)

Any kind of permutation factor for how the cards are dealt would just get factored out of both. The conditional probability is

(4*3*5/(53 choose 10))/(4*3*5/(53 choose 10) + 4*3*5*2/(54 choose 10)) = 27/71.

Perhaps the way to keep the jokers and make the r.f. relevant is to not reveal whether or not the r.f.s are completed with jokers. The condition is simply that two hands are dealt that count as royal flushes.

I have an idea about another deck puzzle and comparing poker hands, but I'll start a new thread.

• Sugar Land
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Cottoneyedjoe, when the problem is posed in the way you suggested, two Royal Flushes with one of them possibly being "Unnatural" (completed w/ Joker), the fact that the hands are Royal Flushes ("Natural" or "Unnatural") doesn't alter the posterior probability of 52-card deck (no Joker) or 53-card deck (with Joker). With any random deal of 2 hands of 5 cards (doesn't matter if they are Royal Flushes or not), if the prior probabilities were "p" for 52-card and "1-p" for 53-card decks, then so are the posterior probabilities as long as (this is the key point) we are not given any information on the whether a Joker showed up in the experiments.

With my meta-question, there is information imparted about the chosen deck because you're told that one of the hands has a Joker. So the posterior probabilities are altered. However, it doesn't matter whether they are Royal Flushes or any random hands, given that a Joker replaces any specific card that is missing in the deal.

• Sugar Land
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I worked out the answer to my meta-question is p = 1161/1601 ≈ 0.7252.

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