[SOLVED] Probability of a Yarborough


by e(ho0n3
Tags: probability, solved, yarborough
e(ho0n3
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#1
Nov18-07, 09:56 PM
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1. The problem statement, all variables and given/known data
The second Earl of Yarborough is reported to have bet at odds of 1000 to 1 that a bridge hand of 13 cards would contain at least one card that is ten or higher. (By ten or higher we mean that it is either a ten, a jack, a queen, a king or an ace.) Nowadays, we call a hand that has no cards higher than 9 a Yarborough. What is the probability that a randomly selected bridge hand is a Yarborough?


2. Relevant equations
Axioms and basic theorems of probability.


3. The attempt at a solution
There are four players in bridge (and thus four hands) and since there are 32 cards which that are less than 10, there can be at most 2 Yarborough hands.

Define the following events.
A: the randomly selected hand is a Yarborough
E: there are two Yarborough hands
F: there is one Yarborough hand

Conditioning on E and F, then P(A) = P(A|E)P(E) + P(A|F)P(F) right? P(A|E) is 1/2 and P(A|F) is 1/4. P(E) is

[tex]\frac{\binom{32}{13}\binom{19}{13}\binom{26}{13}\binom{13}{13}}{\binom{ 52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}} = \frac{\binom{32}{13}\binom{19}{13}}{\binom{52}{13}\binom{39}{13}}[/tex]

P(F) should be

[tex]\frac{\binom{32}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13} - \binom{32}{13}\binom{19}{13}\binom{26}{13}\binom{13}{13}}{\binom{52}{13 }\binom{39}{13}\binom{26}{13}\binom{13}{13}} = \frac{\binom{32}{13}\binom{39}{13} - \binom{32}{13}\binom{19}{13} }{\binom{52}{13}\binom{39}{13}}[/tex]

Is this correct?
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Dick
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#2
Nov18-07, 10:19 PM
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I don't know why you are worrying about multiple hands. Isn't it just C(32,13)/C(52,13)?? Maybe I just don't see it because I don't play bridge.
e(ho0n3
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#3
Nov18-07, 10:25 PM
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What you have there is the probability that a hand of 13 cards is a Yarborough. I don't think this is equivalent to the probability that a random hand in a game of bridge is a Yarborough. (I've never played bridge either.)

Dick
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#4
Nov18-07, 10:43 PM
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[SOLVED] Probability of a Yarborough


I don't see the difference between a 'random hand' and a 'random hand played in a game of bridge'. I think you are over complicating again. How can how the rest of the cards are dealt make a difference? The problem doesn't ask anything about the probability of multiple Yarborourgh's in a game of bridge.
e(ho0n3
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#5
Nov19-07, 07:24 AM
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Quote Quote by Dick View Post
The problem doesn't ask anything about the probability of multiple Yarborourgh's in a game of bridge.
Right. I made use of it in order to calculate the probability sought. If the events "a hand of 13 cards is a Yarborough" and "a hand picked at random from a game of bridge is a Yarborough" are equivalent, then they should have the same probability. Unfortunately, they don't.
Dick
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#6
Nov19-07, 08:34 AM
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Quote Quote by e(ho0n3 View Post
Right. I made use of it in order to calculate the probability sought. If the events "a hand of 13 cards is a Yarborough" and "a hand picked at random from a game of bridge is a Yarborough" are equivalent, then they should have the same probability. Unfortunately, they don't.
You are going to have a hard time convincing me of that.
D H
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#7
Nov19-07, 09:41 AM
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You guys a nitpicking over an overloaded term. The 52 cards in a deck are randomized and dealt in an orderly fashion to four players. Each player ends up holding 13 cards, a "hand". The four hands together form a "deal". See http://www.worldbridge.org/departments/laws/Defin.htm.

Some people overload the term hand to mean the four hands together. This is what e(ho0n3 (eho for short) has done. Eho, I take it you play bridge. You are (1) reading too much into the problem, and (2) using an overloaded term. I think the author of the problem is getting at the simpler problem of the probability a specific player a getting a Yarborough hand as opposed to the probability of a deal coming up with a Yarborough.

BTW, you have not calculated the probability of a deal having a Yarborough correctly. As four hands are dealt out, there are four ways a deal can have exactly one Yarborough and six ways a deal can have exactly two Yarboroughs.
Dick
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#8
Nov19-07, 10:01 AM
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I don't think there is much confusion over the term 'hand'. If the OP follows his plan correctly and computes the odds of a game having exactly one Y and divides by 4, then the odds of having exactly 2 and divides by 2 and then adds them, he will get the correct answer. I'm trying to say this is unnecessary and that answer will be the same as what we both agree is the probability of a single hand being Y. Will confess to being lazy and not wanting to check the posted solution...
e(ho0n3
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#9
Nov19-07, 11:13 AM
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Quote Quote by D H View Post
Some people overload the term hand to mean the four hands together. This is what e(ho0n3 (eho for short) has done. Eho, I take it you play bridge. You are (1) reading too much into the problem, and (2) using an overloaded term.
You can call me echo1 for short.

BTW, you have not calculated the probability of a deal having a Yarborough correctly. As four hands are dealt out, there are four ways a deal can have exactly one Yarborough and six ways a deal can have exactly two Yarboroughs.
That is true if you're taking the order into account. In my calculations, I didn't.

Quote Quote by Dick View Post
I don't think there is much confusion over the term 'hand'. If the OP follows his plan correctly and computes the odds of a game having exactly one Y and divides by 4, then the odds of having exactly 2 and divides by 2 and then adds them, he will get the correct answer. I'm trying to say this is unnecessary and that answer will be the same as what we both agree is the probability of a single hand being Y.
Let's simplify the problem greatly and see if this is true. Suppose there are only two players and 4 cards, 1, 2, 3, 4, of the same suit. Each player gets two cards. A player has a Yarborough if he/she has a card greater than 2. The probability that a two-card hand has is a Yarborough is 5/6. The probability that a random hand is a Yarborough is 1 * 4/6 + 1/2 * 1/6 = 3/4. Unless I did something wrong, they're not equal.
Dick
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#10
Nov19-07, 06:12 PM
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Simplifying the problem is a great idea. But for your simplified problem, I get that a table must have either 1 (prob 1/3) Y or 2 (prob 2/3) Y's. So for the table odds, I get (1/2)*(1/3)+1*(2/3)=5/6. That said, I am still pulling my hair out trying to get a straight answer for the original problem using your counting argument. Grrrr....
e(ho0n3
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#11
Nov19-07, 11:00 PM
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Quote Quote by Dick View Post
Simplifying the problem is a great idea. But for your simplified problem, I get that a table must have either 1 (prob 1/3) Y or 2 (prob 2/3) Y's. So for the table odds, I get (1/2)*(1/3)+1*(2/3)=5/6.
I screwed up again. You are correct. They're are both equal.

That said, I am still pulling my hair out trying to get a straight answer for the original problem using your counting argument. Grrrr....
My counting argument is wrong. It must be because I got the probability of the simplified problem wrong. Don't use it. What is happening is that I'm counting one to many times. For example, I wrote in the denominator of the equations for P(E) and P(F) in the first post, that the number of possible bridge deals is:

[tex]\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}[/tex]

which is correct if I'm considering each of the four players distinctly. However, when I was doing the calculations, I was treating all four players the same, i.e. I didn't care who was N, S, E or W. It's no wonder I didn't get the right answer. D H was right all along.

So how can I fix this? Easy: Just consider each of the four players distinctly. The value of P(E) should actually be 6 times what I wrote in my first post (as D H mentioned). P(F) is a slightly trickier. I will fix my attention on N for now. There are

[tex]\binom{32}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}[/tex]

possible deals in which N has a Yarborough. But how many deals are there in which N has the only Yarborough? That's the above number minus the number of deals in which N and E, N and W, and N and S both have a Yarborough, i.e. minus

[tex]3 \cdot \binom{32}{13}\binom{19}{13}\binom{26}{13}\binom{13}{13}[/tex]

Dividing the difference by the total number of deals yields the probability that N has the only Yarborough. P(F) is just 4 times this amount (one for each of N, S, E, W).

Letting p = C(32,13)/C(52/13) and q = C(19,13)/C(39,13), you can see that P(E) = 6pq, P(F) = 4(p - 3pq) and P(A) = 1/2 * 6pq + 1/4 * 4(p - 3pq) = 3pq + p - 3pq = p.

Dick, you were right again.
Dick
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#12
Nov19-07, 11:10 PM
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Thanks, man. I was still ripping hair. Gotta confess, though, I still don't get it. Guess I'll get some sleep and think it over again in the morning with your hints. Let this be a warning against over complicating. You escalated the problem to the point I'm still baffled about reconciling the two approaches.
e(ho0n3
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#13
Nov19-07, 11:21 PM
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Hehe. Sorry about that. I will mark this post as solved. Thanks for all your help.
Dick
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#14
Nov19-07, 11:31 PM
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One final comment. I'm really impressed with how you dealt with this problem. Creating a simplified version of the problem to analyze your thinking is brilliant step. Kudos.


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