Poker Deal Probability Puzzler

Reply #11

Orange71 — Tue, 09 Aug 2022 22:12:31 GMT

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

Reply #10

Orange71 — Tue, 09 Aug 2022 21:32:57 GMT

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 post... [ More ]

Reply #9

cottoneyedjoe — Tue, 09 Aug 2022 14:54:39 GMT

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 tha... [ More ]

Reply #8

Orange71 — Tue, 09 Aug 2022 01:03:26 GMT

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 repla... [ More ]

Reply #7

Orange71 — Tue, 09 Aug 2022 00:17:26 GMT

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.

Reply #6

cottoneyedjoe — Mon, 08 Aug 2022 22:30:26 GMT

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 rediscov... [ More ]

Reply #5

Orange71 — Mon, 08 Aug 2022 20:54:02 GMT

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 st... [ More ]

Reply #4

Orange71 — Mon, 08 Aug 2022 18:35:57 GMT

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

Reply #3

Orange71 — Mon, 08 Aug 2022 17:07:55 GMT

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 5... [ More ]

Reply #2

Orange71 — Mon, 08 Aug 2022 16:55:04 GMT

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.

Reply #1

cottoneyedjoe — Mon, 08 Aug 2022 15:37:09 GMT

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... [ More ]

Poker Deal Probability Puzzler

Orange71 — Mon, 08 Aug 2022 03:02:23 GMT

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 bo... [ More ]