Simple gambling problem

Originally posted by LemonJello
Let K denoted drawing a king; Let FC denote drawing a face card.

Bayes tells us that

P(K|FC) = [P(FC|K)*P(K)]/P(FC) = [(1)*(4/52)]/(12/52) = 1/3.

that just does not make sense, the odds of drawing a king next are 4/51.

oops nevermind

Oops nevermind.

Originally posted by eldragonfly
yeah i understand that, that is the solution that is given but the odds of drawing a king next is 4/51. You cannot eliminate the rest of the deck minus the remaing face cards, as you propose. 😕

The odds for what you draw next is a different problem.

That would be an interesting question to pose though.

You have a standard deck of 52 cards that has been thoroughly shuffled, so that any card is equally likely to be in any location in the deck. The person across from you draws a card and keeps it, and informs you that he drew a face card, but does not show you what the card is. You then draw a card randomly from the deck.

Assuming he was being honest, what are the odds that the card you pick is a king?

Originally posted by eldragonfly
that just does not make sense, the odds of drawing a king next are 4/51.

Oh, then I think you don't understand the question that was put forth in whatever book you got the question from. The question is: given that you draw a face card, what is the probability that said card also happens to be a king? This is the question I answered because I thought that is what you were asking (maybe my bad, but more likely you are misinterpreting the original question).

You seem to be interpreting the question differently as: suppose you have a deck of cards and then you draw a face card and remove it from the deck. What is the probability that on your next draw (now from the 51 remaining cards) you happen to draw a king?

Those are two very different questions.

Originally posted by eldragonfly
i will have a look at that, but here's one i clearly do not understand.

What is the probability that you will draw a king given that you have drawn a face card. "Statistics the Easy Way" gives the answer as p=1/3, using the conditional probability formula P(A|B).

4/51 seems a much more reasonable number, or 12/52 * 4/51.

The question here involves one drawing only, not two.

Given that the card you draw is a face card, what are the chances that it is also a king?

Originally posted by LemonJello
Oh, then I think you don't understand the question that was put forth in whatever book you got the question from. The question is: given that you draw a face card, what is the probability that said card also happens to be a king? This is the question I answered.

You seem to be interpreting the question differently as: suppose you have a deck of car ...[text shortened]... m the 51 remaining cards) you happen to draw a king?

Those are two very different questions.

that was not the question lemmonjello. I have already provided a problem description.

1) a face card is drawn p= 12/52
2) what are the odds that a king will be drawn next.
3) the solution is given as 1/3 using P(A|B)

Originally posted by eldragonfly
that was not the question lemmonjello. I have already provided a problem description.

1) a face card is drawn p= 12/52
2) what are the odds that a king will be drawn next.
3) the solution is given as 1/3 using P(A|B)

Sigh. Re-read my post. You simply don't understand the actual intended content of the question that is being asked in your book.

In other words, their answer of 1/3 is for the question as I laid it out, and it is correct. The other, very different question in your head (which clearly isn't a faithful interpretation of the original question relative to what the authors intended) has a different answer.

Originally posted by eldragonfly
that was not the question lemmonjello. I have already provided a problem description.

1) a face card is drawn p= 12/52
2) what are the odds that a king will be drawn next.
3) the solution is given as 1/3 using P(A|B)

The question in the book may have been poorly worded and confusing, but it only involves drawing the one card.

The question it is asking, in essence is this..
If you draw a face card from a fair deck of cards, what are the odds that it is also a king?

The question you thought it was asking was different, although I can see how it could be interpreted in that manner given the wording.

nevermind

nevermind, you are right.

Originally posted by geepamoogle
That would be an interesting question to pose though.

You have a standard deck of 52 cards that has been thoroughly shuffled, so that any card is equally likely to be in any location in the deck. The person across from you draws a card and keeps it, and informs you that he drew a face card, but does not show you what the card is. You then draw a car ...[text shortened]... m the deck.

Assuming he was being honest, what are the odds that the card you pick is a king?

You have to consider the cases
1. He drew a K
2. He drew a different face card

P(You draw K | He draws face card)
= P(You draw K & He draws face card)/P(He draws face card)
= (8/52*4/51 + 4/52*3/51)/(12/52)
= 11/153

Originally posted by eldragonfly
lemonjello stop playing out like a dummy. 🙁 In fact here is the problem description verbatim from the text, page 65:

What is the probability that you will draw a king, given that you have drawn a face card?

The text gives the solution as (4/52)/(12/52) = 1/3

Look, I can understand how you would interpret that question in the manner you have. But, clearly based on the proposed answer, you are not interpreting the question in the manner they intend.

Jesus, you're a dense mofo. For the last time, this is how that question was intended to be interpreted:

Given that you draw a face card, what is the probability that this same card also happens to be a king? The answer is 1/3.

If you want to just insist on interpreting the other way, then the answer will be different.

Originally posted by LemonJello
Sigh. Re-read my post. You simply don't understand the actual intended content of the question that is being asked in your book.

In other words, their answer of 1/3 is for the question as I laid it out, and it is correct. The other, very different question in your head (which clearly isn't a faithful interpretation of the original question) has a different answer (the one you quote, 4/51).

eldragonfly's question, as I understand it, is as follows.

If a random face card is removed from a fair and complete deck, and you draw a card randomly afterwards, what are the chances it will be a king?

My proposed answer is 11/153 with the following reasoning.

The card removes has a 1-in-3 chance of being a king, in which case the next card as a 3-in-51 chance of being a king. So the chances that a king is removed, then a king is drawn is 3-in-153.

If the card removed was a jack or queen, then the next card has a 4-in-51 chance of being a king. So the chances of this possibility are 2/3 * 4*51 or 8-in-153.

Since these two possibilities have no union (the card removed was either a king, or else it wasn't), they combine to give you 11-in-153 odds the next card is a king.

The other two possibilities both have cards other than kings being picked, and aren't a part of the calculation.

You'll note here that I weighted the possibilities when calculating odds, which is important with some problems. (I initially forgot to do this with the Sports problem earlier.)

Originally posted by geepamoogle
eldragonfly's question, as I understand it, is as follows.

If a random face card is removed from a fair and complete deck, and you draw a card randomly afterwards, what are the chances it will be a king?

My proposed answer is 11/153 with the following reasoning.

The card removes has a 1-in-3 chance of being a king, in which case the next card as ...[text shortened]... ortant with some problems. (I initially forgot to do this with the Sports problem earlier.)

Oops, of course you are right that it is not as simple as 4/51. Thanks. I was just trying to get eldragon to understand the question in his book as the authors intended for it to be interpreted (which is not the question you are solving here).

Originally posted by LemonJello
Oops, of course you are right that it is not as simple as 4/51. Thanks. I was just trying to get eldragon to understand the question in his book as the authors intended for it to be interpreted (which is not the question you are solving here).

i went and looked it up, it was a one draw one card calculation.

Originally posted by LemonJello
Look, I can understand how you would interpret that question in the manner you have. But, clearly based on the proposed answer, you are not interpreting the question in the manner they intend.

Jesus, you're a dense mofo. For the last time, this is how that question was intended to be interpreted:

Given that you draw a face card, what is t ...[text shortened]...

If you want to just insist on interpreting the other way, then the answer will be different.

My mistake, still stop being rude, it serves no purpose. 😉