# Probability question

Discussion in 'Odds & Ends' started by Steady Eddy, Feb 6, 2011.

1. ### Steady EddyHall of Fame

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Really, two questions.

You deal 13 cards from a well shuffled deck to a friend. Then you ask, "Do you have an ace?" If he answers yes, what is the probability that he has another ace?

Second. You deal 13 cards to a friend. You ask, "Do you have the Ace of Spades?" If he answers yes, what is the probability he has another ace?

2. ### mikelerModerator

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If all 13 cards are aces, 100% probability.

3. ### sureshsBionic Poster

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Probability of the first one is 0.3827.

Probability of second is 0.3623.

Both are probably wrong LOL.

Tough one this.

4. ### tennisnoob3Professional

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1. 1/13(simplified from 4/52 x 3/51

2. 1/52 x 3/51

edit: i'm probably wrong too. i think i solved for drawing one card at a time.

Last edited: Feb 6, 2011
5. ### Steady EddyHall of Fame

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That first one is close enough for government work! To four digits it's .3696.

Suppose we abbreviate the # of ways of getting 1 ace as N(1).

Then the fraction is: [ N(2) + N(3) + N(4) ] / [ N(1) + N(2) + N(3) + N(4) ]. The numerator is the total # of ways to have more than one ace, the denominator is the given condition, i.e. that of having at least an ace.

If you have only one ace, the chance is only .25 that it is the ace of spades. If you have two aces, the chance that one of them is the ace of spades is .50. Realizing this the fraction for the 2nd part is:

[(.50)N(2) + (.75)N(3) + N(4) ] / [ (.25)N(1) + (.50)N(2) + (.75)N(3) + N(4) ], which really changes the outcome. If you know you have an ace, the chance of having a 2nd ace is UNDER 50%. But if you know it's the ace of spades, (or any suit, for that matter), the chance of having a 2nd ace is OVER 50%.

6. ### mikelerModerator

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I came so close to writing that out in my first answer.

7. ### Steady EddyHall of Fame

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You have not made an error yet! Are you this consistent in tennis as well?

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^^ didn't want to have to look it up in my stats book.

9. ### mikelerModerator

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Most days.

Something tells me there is some serious dust on that book! Mine got sold a long time ago for beer money.

10. ### sureshsBionic Poster

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By first I meant the very first event (1 ace) and by second I meant getting the second ace.

I did not even attempt what you call the second one.

11. ### lonuxHall of Fame

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... Too complex.

Actually, not all to hard if you use a chance tree, but writing it out is painful.

12. ### sureshsBionic Poster

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You deal 13 cards from a well shuffled deck to a friend. Then you ask, "Do you have an ace?" If he answers yes, what is the probability that he has another ace?

First part: p = 4/52 = 1/13. q = 12/13, n = 13, r = 1

For the second part, since 1 ace is gone, p = 3/51, q = 48/51, n = 12, r = 1. Again, nCr*p^r*q^(1-r) is the answer.

Are either of these above correct?

13. ### Steady EddyHall of Fame

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If each card dealt was an independent event this is on track. That is, if we wrote down each card, put it back in the deck and reshuffled, then dealt another, and so on, (BTW, the exponent for q would be n - r).

But each card changes the probabilities of the following cards, so instead think this way: in how many ways can a hand be dealt containing 1 ace? 4 cards are aces, 48 cards are non aces. Of the 4 aces, we need 1, so 4C1, of the 48 non aces we need 12, so 48C12. The product 4C1 * 48C12 yields the # of ways of constructing hands w/ exactly one ace. A similar process works for the other choices too.

14. ### AndresG.O.A.T.

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Geeeeeeeeks!!!!

Go outside and play some tennis!

:mrgreen:

15. ### sureshsBionic Poster

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Right, it is n-r.

I find probability questions to be very difficult. One of the toughest branches of math, because there are very few things to study, yet very difficult problems to face.