Probability question
This is a variation of a famous probability question (okay, questions) that often drives people nuts.
Assume the deck of cards is well shuffled, and that the statements made about the hole cards are truthful. This is not meant to be a trick question.
1. Suppose someone is dealt two hole cards, looks only at one of them, and tells you that it is a spade. What is the probability that the other card is also a spade?
2. Suppose someone is dealt two hole cards, looks at both hole cards, and tells you that one of the cards is a spade. What is the probability that the other card is also a spade?
ScottyZ
P.S. Super-Duper Lucky Bonus Question: Does this mean that the chances of you being dealt s0000ted hole cards go up if you only look at one hole card instead of both?
Assume the deck of cards is well shuffled, and that the statements made about the hole cards are truthful. This is not meant to be a trick question.
1. Suppose someone is dealt two hole cards, looks only at one of them, and tells you that it is a spade. What is the probability that the other card is also a spade?
2. Suppose someone is dealt two hole cards, looks at both hole cards, and tells you that one of the cards is a spade. What is the probability that the other card is also a spade?
ScottyZ
P.S. Super-Duper Lucky Bonus Question: Does this mean that the chances of you being dealt s0000ted hole cards go up if you only look at one hole card instead of both?
Comments
(simple enough)
2. the other card is a spade exactly 156/770 of the time... ~20.26%
(there are 1482 (39*38) hands that contain no spade of 2652 (52*51)... therefore 770 hands contain atleast one spade allowing the claim..."one card is a spade"... 156 (13*12) of those are sooooted)
Super-Duper Lucky Bonus Question. NO! (but don't you wish this were true)
What I would like to know is, who are the people that had a hard time figuring that one out. But more importantly where do they play cards? lmao hehe
Many people get hung up on the idea that the probabilities "must" be the same for both questions. No pulling a fast one on Mickey though...
ScottyZ
Thus,
Probability of 2 spades given at least one is: 156/1170 = 13.33%
Please correct me if I'm wrong...
ScottyZ
770 indeed should have been 1170... we crap players use 7-11 interchangably!:D
Gide
I assume anyone holding cards could be lying, so this should be ignored. I would only believe this, if the hole card was turned over. Hole card unexposed, must assume that player could be lying. Same good question, slightly different answer
1) 3 to 1 odds or 25% probability that one card is spade
2) 52(2) / (13(2) * 4) = 1326 / 312 = 4.25 to 1 odds or 19% probability (See note below)
Edit -- (above is probability of two suited cards) For suited spades == 52(2) / 13(2) = 1326 / 78 = 17 to 1 or 5.5%
BONUS
Of course the odds go up if you only look at one card, allowing a non-George Costanza to really believe the other card is of the same suit. If you are George Costanza, then it doesn't matter as you can believe your cards are any two cards you want them to be. Hey, maybe Larry David/Jerry Seinfeld modelled George from Doyle Brunson.
NOTE
The total number of 2 cards hands is 52 choose 2 as order is not important -- total of 1326 hands. The formulas below assume that order is important and are double counting.
In these sorts of questions, you can assume either that order does or does not matter, as long as you are consistent in chosing one option and sticking with it.
The double counting you are concerned about will simply result in a bunch of 2's cancelling out in the computations if they are done in the "order does matter" style.
Close, but the 7 card stud fanatics will tell you that one of your spades is dead when you are drawing for the second one. It's
12/51
since 12 spades are left from a deck of 51 cards because our friend already looked at one card from the deck, which is known to be a spade.
I disagree with this. The number
13(2) * 4 / 52(2)
is the unconditional probability of being dealt two suited cards (of whatever suit). We're looking for the conditional probability of both cards being spades, given that we know one of the cards is a spade. Aside from the arithmetic error, MickeyHoldem's calculation is right IMO.
ScottyZ
Yes, it may work out properly for this calcuation, but It's by accident. IMO, your question helps illustrate some important concepts and understanding that order is not important is an important part of the process. If you use permutations instead of combinations, you'll often arrive at the incorrect answer.
Yes, you are correct if you can believe that you were told the truth about a spade being one of the hole cards. Who believes anyone holding cards, without being able to see them? I was being a smart ass -- so, 13/52 if you have no information. other than being a smart ass, many beginners make the mistake of assuming that the cards dealt are not part of the odds calculation because they are already out of the deck and therefore cannot come out. However, the are incorrect because they don't know the cards that are out, so the total sample is 52 - seen cards. So, there's a method to my madness.
Yes, I went back and edited my note to highlight the fact that I had computed the probability of any two suited cards, not spades. I went back and updated and the probability of two spades, given no information removes the multiplication by four.
That's a good point. All unknown cards, whether out in hands or still in the deck, need to be counted in these kinds of odds/probability calculations.
ScottyZ