Probability of finding a three.

[BMAD]

Suppose there is a hat that contains the numbers 1,2,3,4, and 5.  You seek to find the three. You blindly reach into the hat pulling out a number. If it is wrong, then without replacement you reach for another. When you pull out the three you stop. Each pick before three is subtracted from 0 with your final pick (the three) being added to that total.  What is your expected value? 

[bonanova]

A quick and dirty analysis
 

Spoiler

On average two numbers will be drawn before the three. The average of the other numbers is (1+2+4+5)/4 = 3.

So the average score will be -3 + -3 + 3 = -3

 

[bonanova]

Nice puzzle.
 

Spoiler

There are 5! = 120 ways to draw the numbers.
The 3 will be drawn first, second, third, fourth and fifth 1/5 of the time (24 draws each.)

1. When drawn first, nothing is deducted from the score of 3 per draw.
   Score = 3x24 = 72.
2. When drawn second, each of the other cards appears 1/4 of the time (6 draws each), once.
   Score = 72 - 1x6x(1+2+4+5) = 0.
3. When drawn third, each of the other cards appears 1/4 of the time, twice.
   Score = 72 - 2x6x(1+2+4+5) = -72.
4. When drawn fourth, each of the other cards appears 1/4 of the time, three times.
   Score = 72 - 3x6x(1+2+4+5) = -144.
5. When drawn fifth, each of the other cards appears 1/4 of the time, four times.
   Score = 72 - 4x6x(1+2+4+5) = -216.

Total score for all 120 cases = 72 + 0 - 72 - 144 - 216 = -360

Expected score = -360/120 = -3

 

[FUZZY]

On ‎10‎/‎13‎/‎2016 at 2:30 AM, BMAD said:

Suppose there is a hat that contains the numbers 1,2,3,4, and 5.  You seek to find the three. You blindly reach into the hat pulling out a number. If it is wrong, then without replacement you reach for another. When you pull out the three you stop. Each pick before three is subtracted from 0 with your final pick (the three) being added to that total.  What is your expected value? 

You could get 3 in the first, second, third, fourth or fifth draws.

 

3 in the first draw:  One number (3). 1 way

 

In  the second draw means _ 3 : for the number drawn before 3, it is ⁴C₁ ways_.  Then, for 3, it is 1 way. So, 4*1= 4 numbers have 3 in second place. (13,23,43,53)

 

In the third draw means _ _ 3 : for the two numbers drawn before 3, it is ⁴C_2. The two numbers drawn can arrange themselves in 2! ways. Next, 3 is drawn (1 way).

 ⁴C₂*2!*1= 12 numbers have 3 in third place.

In the fourth draw means _ _ _ 3 : for the three numbers drawn before 3, it is ⁴C_3. The 3 numbers can arrange themselves in 3! Ways.  Finally, 3 is drawn.

⁴C₃* 3!* 1= 24 numbers have 3 in fourth place.

In the fifth draw means _ _ _ _ 3: for the four numbers drawn before 3, it is ⁴C_4. The 4 numbers can arrange themselves in 4! Ways.  Finally, 3 is drawn.

⁴C₄* 4!*1= 24 numbers have 3 in fifth place.

Totally, 1+4+12+24+24 = 65 numbers have 3 in them.

First draw value = (1/65)*3 = 3/65

Second draw (13 or 23 or 43 or 53) value for the four numbers

=  (1/65)*(0-1+3) + (1/65)*(0-2+3) +  (1/65)*(0-4+3) + (1/65)*(0-5+3)

= 1/65 [–(1+2+4+5)+(3*4)]= 0

Third draw value for the twelve numbers =-36/65

Fourth draw for the 24 numbers = -144/65

Fifth draw = -216/65

Total expected value= (3/65)+0+(-36/65)+(-144/65)+(-216/65)= -393/65

[bonanova]

5 hours ago, FUZZY said:

You could get 3 in the first, second, third, fourth or fifth draws.

 

3 in the first draw:  One number (3). 1 way

 

....

 

In the fifth draw means _ _ _ _ 3: for the four numbers drawn before 3, it is ⁴C_4. The 4 numbers can arrange themselves in 4! Ways. 

 

 

Yikes! Really? It's 24 times more likely for a 3 to be drawn 5th than to be drawn 1st?

Spoiler

But 3 is not special, so that would be true for all the numbers.

In other words, it is highly unlikely that any number will be drawn 1st.
On average, they all will be drawn maybe 4th or 5th.

So you could make money this way. Give someone 2 to 1 odds that you will draw the number 3 on the 4th or fifth draw. You'll get takers who think your winning chances are 40%. They will expect to double their money 60% of the time. You, however, will expect to win 48 times out of 65.    (Better yet, don't try it.)

A more intuitive way to see what's wrong with saying there is only one way to draw a certain number 1st but multiple ways to draw it 2nd, and so on, is to write down all 120 equally probable ways to draw the five numbers without replacement. Then pick one at random.

You'll see that every number has a 1/5 chance of being drawn first, 1/5 chance of being drawn second, ... or fifth, just as symmetry demands.

I recall another puzzle where a community prohibited any family from having more than a single male child: A couple could keep having female children as long as they liked, but they had to be sterilized after the birth of their first son. So after a long time there were many more females than males in the population, right?

No. Every birth has a 50% chance of producing a son. End of story. The population, blissfully unaware of the local laws, maintained its gender balance.

 

[BMAD]

On 10/27/2016 at 9:37 AM, bonanova said:

Yikes! Really? It's 24 times more likely for a 3 to be drawn 5th than to be drawn 1st?

  Hide contents

But 3 is not special, so that would be true for all the numbers.

In other words, it is highly unlikely that any number will be drawn 1st.
On average, they all will be drawn maybe 4th or 5th.

So you could make money this way. Give someone 2 to 1 odds that you will draw the number 3 on the 4th or fifth draw. You'll get takers who think your winning chances are 40%. They will expect to double their money 60% of the time. You, however, will expect to win 48 times out of 65.    (Better yet, don't try it.)

A more intuitive way to see what's wrong with saying there is only one way to draw a certain number 1st but multiple ways to draw it 2nd, and so on, is to write down all 120 equally probable ways to draw the five numbers without replacement. Then pick one at random.

You'll see that every number has a 1/5 chance of being drawn first, 1/5 chance of being drawn second, ... or fifth, just as symmetry demands.

I recall another puzzle where a community prohibited any family from having more than a single male child: A couple could keep having female children as long as they liked, but they had to be sterilized after the birth of their first son. So after a long time there were many more females than males in the population, right?

No. Every birth has a 50% chance of producing a son. End of story. The population, blissfully unaware of the local laws, maintained its gender balance.

 

your last comment was actually the inspiration of this question.