View Full Version : Probability question

Steady Eddy

02-06-2011, 07:44 AM

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?

mikeler

02-06-2011, 08:40 AM

If all 13 cards are aces, 100% probability.

sureshs

02-06-2011, 08:57 AM

Probability of the first one is 0.3827.

Probability of second is 0.3623.

Both are probably wrong LOL.

Tough one this.

tennisnoob3

02-06-2011, 09:04 AM

Really, two questions.

1. 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?

2. 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?

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.

Steady Eddy

02-06-2011, 09:46 AM

Probability of the first one is 0.3827.

Probability of second is 0.3623.

Both are probably wrong LOL.

Tough one this.

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%.

mikeler

02-06-2011, 09:58 AM

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%.

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

Steady Eddy

02-06-2011, 10:12 AM

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

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

bad_call

02-06-2011, 10:18 AM

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

mikeler

02-06-2011, 10:41 AM

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

Most days.

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

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

sureshs

02-06-2011, 11:03 AM

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%.

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.

lonux

02-06-2011, 11:05 AM

... Too complex.

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

sureshs

02-06-2011, 11:08 AM

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

nCr*p^r*q^(1-r) is the answer

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?

Steady Eddy

02-06-2011, 12:10 PM

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

nCr*p^r*q^(1-r) is the answer

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.

Andres

02-06-2011, 07:16 PM

Geeeeeeeeks!!!!

Go outside and play some tennis!

:mrgreen:

sureshs

02-07-2011, 07:47 AM

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.

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.

vBulletin® v3.6.9, Copyright ©2000-2014, Jelsoft Enterprises Ltd.