Probability Question?

Question

What are the odds against drawing two face cards from a standard deck? The answer is 11:13.

Thanks

Yes random deck, equal chance of drawing any card, etc. Also by face cards I mean J, Q, K.

Oops the answer is 210:11. Sorry about that!

Answer 1

There are 12 face cards in a standard 52-card deck. Assuming the 1st card is not replaced the a priori probability of drawing two face cards is
p = (12/52)(11/51)
The probability of not drawing two face cards is
q = 1 - (12/52)(11/51)
The odds against drawing two face cards are
o = q/p = (1 - (12/52)(11/51)) / ((12/52)(11/51))
o = (52)(51) - (12)(11)) / ((12)(11)
o = (13)(17) - (1)(11)) / (1)(11)
o = 210:11
not 11:13

Answer 2

I suspect my argument is not going to be well received. I say the probability is 50% Let a ∈ℝ, Let b ∈ℝ and randomly select the values for a and b. As already noted, for a ≤ 0, P( a < b²) = 1, this is trivial. Only slightly less trivial is the idea that P(a < 0 ) = 1/2 and thus P( a < b² | a ≤ 0) = 1 and P( a < b² ) ≥ 1/2 Now consider what happens when a > 0 For a > 0, while it is easy to show there is a non zero probability for a finite b, the limit, the probability is zero. a < b² is equivalent to saying 0 < a < b², remember we are only looking at a > 0. If this a finite interval on an infinite line. The probability that a is an element of this interval is zero. P( a < b² | a > 0) = 0 As such we have a total probability P( a < b² ) = P( a < b² | a ≤ 0) * P(a ≤ 0) + P( a < b² | a > 0) * P(a > 0) = 1 * 1/2 + 0 * 1/2 = 1/2 Remember, this is because of the infinite sets. No matter what type of interval you draw on paper or on a computer you will find a finite probability that appears to approach 1. But this is due to the finite random number generators on the computer and if we had this question asked with finite values there would be a a solution greater than 50%. I don't mean to be condescending, but please explain why using the Gaussian to approximate a uniform distribution is a good idea? Aren't infinite numbers fun. Cantor when mad working with them! :)

Answer 3

A face card is Jack, Queen & King. There are 4 x 3 =12 face cars in a deck of 52 cars.
Let the odds of drawing two face cards = p. Then
p = 12/52 x 11/51 = 33/221
so odds against drawing two face cars = 1 - p
= 1 - 33/221 = (221-33)/221 = 188/221

Answer 4

It comes down to defining odds.

There are 12 face cards, so the probability of getting 2 face cards
= 12/52 * 11/51 = 11/221

So that means 210/221 = prob of not getting 2 face cards.
So if odds = prob of not getting / prob of getting
then it's 210:11

Answer 5

You didn't specify whether you're replacing the card back in the deck before drawing the 2nd card, so I'll give you both fractions and approximate percentages.

Assuming you do NOT replace:

((4 suits) * (3 face cards per suit) / (52 total cards)) * (((4 suits) * (3 cards per suit)) - 1 (already drawn face card)) * (51 total cards remaining) = 132/2652 or in simplest form, 11/221 or approximately a 4.977% chance of drawing two consecutive face cards without putting the first one back in the deck.

If you DO replace:

(((4 suits) * (3 face cards per suit))/(52 total cards)) ^2 = 144/2704 = 72/1352 or in simplest form, 9/169 or approximately a 5.325% chance of drawing two consecutive face cards, putting the first one back in the deck before selecting the second card.

Answer 6

I assume you mean 1) in a fully randomized deck 2) random draw without looking 3) drawing only two cards...
3 fc per suit 4 suits = 12 out of 52 (excluding jokers, right?)
1st 12/52 after 1st fc there's 11/51
P=12*11/(52*51)

Answer 7

One card is selected at random from a standard deck, not replaced, and then a second card is drawn. Find the probability of selecting two face cards.