more math

[Tom239]

A puzzle in probability.

A deck is shuffled and bridge hands (13 cards each) are dealt.
You choose one player and ask if his hand has at least one ace; he says yes.

The deck is shuffled and hands are dealt again. You again choose one
player but this time ask if the ace of spades is in his hand; he says yes.

Which hand was more likely to contain more than one ace,
the first hand you asked about or the second?

 

[RJD1022]

The hand with the ace of spades is more likely to have another ace.

 

[sahtt]

I'll let the audience play first. Probability problems are my favorite.

Tip - read the problem out loud and very slowly.

 

[Tom239]

I'll let the audience play first. Probability problems are my favorite.

Tip - read the problem out loud and very slowly.

While I like this tip I also want to assure everyone I tried to word
the question as clearly as I could. I wanted getting the answer to
be about probability rather than having to read between the lines.
E.g., I meant that when the people answered that the aces were
in their hands, they were telling the truth.

 

[RJD1022]

While I like this tip I also want to assure everyone I tried to word
the question as clearly as I could. I wanted getting the answer to
be about probability rather than having to read between the lines.
E.g., I meant that when the people answered that the aces were
in their hands, they were telling the truth.

I agree. I'm not sure why you have to read it out loud or slowly.
There is no trick. It is just math.

 

[ENT]

Neither...

 

[sahtt]

While I like this tip I also want to assure everyone I tried to word
the question as clearly as I could. I wanted getting the answer to
be about probability rather than having to read between the lines.
E.g., I meant that when the people answered that the aces were
in their hands, they were telling the truth.

Words like at least one is what I was referring to, not trying to suggest any trickery on your end.

 

[Tom239]

I'll let the audience play first. Probability problems are my favorite..

Looks like the audience has finished playing :smile:
Woulda been nice if we had people giving all three possible answers, but at least we got two.

 

[nicholas421]

I agree with rjd. I simplified the problem by making it a deck of 4 cards (2 aces and 2 others) and 2players

 

[Mirroredshades]

The first hand.

 

[Tom239]

I agree with rjd. I simplified the problem by making it a deck of 4 cards (2 aces and 2 others) and 2players

If you calculate as conditional probability, P(A|B) = P(A∩B) / P(B)
the 4-card deck example throws some of the baby out with the bath water.
P(A∩B) is the same in both cases in the simplified example and it isn't
the same in both cases of the original question about bridge hands.
That doesn't mean it gets the wrong answer, just that it isn't all that
convincing an argument.

 

[nsxtasy]

I simplified the problem by making it a deck of 4 cards (2 aces and 2 others) and 2players

If you really want to simplify the problem, make it a deck of 4 cards, all aces.

 

[CB72]

A puzzle in probability.

A deck is shuffled and bridge hands (13 cards each) are dealt.
You choose one player and ask if his hand has at least one ace; he says yes.

The deck is shuffled and hands are dealt again. You again choose one
player but this time ask if the ace of spades is in his hand; he says yes.

Which hand was more likely to contain more than one ace,
the first hand you asked about or the second?

I didn't go back to my books on probability but I believe neither
In the first hand the guy had 4*13/52 chances of having an ace.
In the second hand the guy only had 1*13/52 chances to get an ace of spade.
To get a second ace, they chances are equal for both guys 3*13/52.
I'm not sure about the maths:frown:
This question is kind of similar to the one where you ask a guy if he would rather travel with an airline that just had a crash to one that hasn't had one in a long time.
The answer is: it doesn't matter as there is no "memory" in probability:smile:

 

[RJD1022]

If your really ineterested in knowing why the hand with the ace of spades is more likely to have another ace.

Check out problem 4 on the page below:
http://www.datagenetics.com/blog/may12012/index.html

It is a detailed explanation.

 

[Carguy!]

If your really ineterested in knowing why the hand with the ace of spades is more likely to have another ace.

Check out problem 4 on the page below:
http://www.datagenetics.com/blog/may12012/index.html

It is a detailed explanation.

I read their explanation and I still don't understand why simply mentioning that you have the Ace of Spades increases the chances of a second ace. In the first explanation they stated that they had at least one ace, if they would of mentioned its suit how would that increase the chances of having another ace?

Normally I love probability questions and I can't remember one I didn't get, on the explanation page they mention the one about having two kids and at least one of them is a boy, what are the odds of the 2nd one being a boy. I figured this one out right away as it makes complete sense to me.

I have written software many times to test probability questions and plan on doing that if I can't get my mind around why these two don't have the same chance of having two or more aces.

I have now written a simulation program and when one of the hands has an ace I am seeing another ace in the same hand about 58.6% of the time. So the percentages that are in the explanation don't seem to be correct.

I also don't understand this line in the explanation:
"Well, this puzzle is more about a confidence problem than a probability problem. It’s about the disclosure of information."

 

[RJD1022]

If you understand the bonus question about why specifying which of the children is a boy, then it is the same concept applied to a much larger set.
The concept being that when you have more information you can change the equation used to determine the possibility of other events.

Using the boy problem as an example.
In the first case, “at least one is a boy”, there are three possible cases: MM, MF, FM.
There is only one chance in three that the other child is also a boy.
Calculated as:
# solutions with 2 boys / # total solutions
1/3 = .3333

When you specify which child is the boy (let’s call it the first child), your solution set reduces to: MM, MF and so the chance that the other child is a boy is 50:50.
Calculated as:
# solutions with 2 boys / # total solutions
1/2 = .50

Related back to the ace problem:
If you know that the hand has at least one ace, then the solution is:
# hands with at least two Aces / # hands with at least one Ace
I didn't do the math but from the website: 36.27%

If you know that the hand has the ace of spades, the solution becomes:
# hands with at least two Aces (one of which is the Ace of Spades) / # hands with the Ace of Spades
I didn't do the math but from the website: 56.12%

What language did you write your simulation in. I would love to see it if you would be willing to send it to me.

 

[Tom239]

I read their explanation and I still don't understand why simply mentioning that you have the Ace of Spades increases the chances of a second ace. In the first explanation they stated that they had at least one ace, if they would of mentioned its suit how would that increase the chances of having another ace?

Although a smaller number of potential hands contain the ace of spades,
within those hands a greater proportion contain a second ace.

I have now written a simulation program and when one of the hands has an ace I am seeing another ace in the same hand about 58.6% of the time. So the percentages that are in the explanation don't seem to be correct.

I have confirmed the probability 5359/14498 = 0.369637... by doing the math.
After you said you wrote a simulation I did that too and my
program gives a result close to the calculated probability.
If you post your program I would be happy to review it.

I also don't understand this line in the explanation:
"Well, this puzzle is more about a confidence problem than a probability problem. It’s about the disclosure of information."

I wouldn't have phrased it that way.
It's a straightforward problem in conditional probability.

C program:

#include <stdio.h>
#include <stdint.h>

static uint32_t  rand_52()
{
   /*  uses a primitive rng from _Numerical Recipes in C_   */
   static uint32_t   r = 19;
   r = 1664525 * r + 1013904223;
   return (r ^ (r >> 11)) % 52;
}

static void shuffle(uint32_t* deck)
{
   uint32_t   hold;
   uint32_t   i,j;
   for (i = 0; i < 52; i++) {
      j = rand_52();
      hold = deck[j];
      deck[j] = deck[i];
      deck[i] = hold;
   }
}

int main(int argc,char** argv)
{
   uint32_t   deck[52];
   uint32_t   i,n,ace_count;
   uint32_t   denominator = 0;
   uint32_t   numerator   = 0;
   uint32_t   hand_count  = 10000000;

   for (i = 0; i < 52; i++) deck[i] = i;

   for (n = 0; n < hand_count; n++) {
      shuffle(deck);
      ace_count = 0;
      for (i = 0; i < 13; i++) {
         if (deck[i] < 4) ace_count++;
      }
      if (ace_count > 0) {
         denominator++;
         if (ace_count > 1) {
            numerator++;
         }
      }
   }

   printf("%u hands;  %u have >=1 ace;  %u have >=2 aces;  quotient=%.6f\n",
          hand_count, denominator, numerator,
          ((float)numerator) / ((float)(denominator)));

   return 0;
}

results:

10000000 hands; 6963043 have >=1 ace; 2572922 have >=2 aces; quotient=0.369511

 

[sahtt]

Here are a couple I'd encourage you all to watch:

http://www.youtube.com/watch?v=9vRUxbzJZ9Y&feature=relmfu (<3 minutes)

http://www.youtube.com/watch?v=WZ6PoqGvSbk&feature=related (<6 minutes)

And a bonus:

Three prisoners are informed by their jailer that one of them has been chosen at random to be executed, and the other two are to be freed. Prisoner A asks the jailer to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information, since he already knows that at least one will go free. The jailer refuses to answer this question, pointing out that if A knew which of his fellows were to be set free, then his own probability of being executed would rise from 1/3 to 1/2, since he would then be one of two prisoners. What do you think of the jailer's reasoning?

 

[RJD1022]

Here are a couple I'd encourage you all to watch:

And a bonus:
Three prisoners are informed by their jailer that one of them has been chosen at random to be executed, and the other two are to be freed. Prisoner A asks the jailer to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information, since he already knows that at least one will go free. The jailer refuses to answer this question, pointing out that if A knew which of his fellows were to be set free, then his own probability of being executed would rise from 1/3 to 1/2, since he would then be one of two prisoners. What do you think of the jailer's reasoning?

I'm not going to answer because I know the answer, but If you watched the first video you should be able to get this one.

 

[sahtt]

For Tom's problem (which I like), here is an all words discription -

Probability of having at least 2 aces in my 13 cards given I have one of four aces but do not know which ace I have.

versus

Probability of having at least 2 aces in my 13 cards given I have an ace and know which one I have, in this case an ace of spades

The odds are the same whether the ace you know you have is any of the 4 suits.

Once you assign each of the odds, the fraction is 11686 / 20825 you have at least one more ace if you already know you have an ace of spades, diamonds, clubs, or hearts.

The fraction is only 5359 / 14498 you have at least one more ace if all you know is you have an ace.

Definitely fits in the category of confusing if not counter intuitive for some.

 

[sahtt]

I'm not going to answer because I know the answer, but If you watched the first video you should be able to get this one.

My applied probability professor had a good explanation for this one I'll share in a couple days.