0
   

The chances of a higher power.

 
 
Reply Tue 10 Jul, 2018 12:33 am
Farmerman and Leadfoot decide to settle the question about the existence of a higher power using a set of ESP cards. Farmerman randomly chose the deck containing six cards with a square on each of them and one card with a triangle on it. By design Leadfoot got the other deck containing five cards with a square on each of them and one card with a triangle on it. The existence of a higher power is proven if Leadfoot draws the first triangle. The non-existence of a higher power is proved if Farmerman draws the first triangle. If they both draw a triangle from their decks at the same time they redraw. They draw randomly, simultaneously with replacement.

What is the probability that a higher power is proven to exist by Leadfoot drawing the first card with a triangle on it?

What is the probability that Farmerman is proven correct if he graciously allows the chances of a draw to be added to Leadfoot's probability?
 
View best answer, chosen by ascribbler
Setanta
 
  3  
Reply Tue 10 Jul, 2018 12:54 am
Math question, huh?
chai2
  Selected Answer
 
  3  
Reply Tue 10 Jul, 2018 01:21 am
@Setanta,
My head hurts
0 Replies
 
fresco
 
  3  
Reply Tue 10 Jul, 2018 07:11 am
@ascribbler,
Zero, since the general assumption of a 'higher power' or otherwise has nothing to do with a two party experiment even if the participants thinks it does.
ascribbler
 
  2  
Reply Wed 11 Jul, 2018 09:54 pm
@fresco,
Now, now fresco. The question has a mathematical answer and you need to show your working for any glorious credit.
fresco
 
  0  
Reply Wed 11 Jul, 2018 11:54 pm
@ascribbler,
No its not even that! A probability question involving card drawing is elementary high school stuff BUT the phrase '..generously allowing the chances of a draw..etc' is mathematically meaningless.
0 Replies
 
fresco
 
  1  
Reply Thu 12 Jul, 2018 12:25 am
@ascribbler,
If you mean what are the chances of Leadfoot winning or drawing
the answer is : 1/5 x 5/6 +1/5 +1/6.
0 Replies
 
fresco
 
  1  
Reply Thu 12 Jul, 2018 06:21 am
@fresco,
slight mistype of last...1/5 x 5/6 + 1/5 x 1/6
0 Replies
 
markr
 
  1  
Reply Thu 12 Jul, 2018 09:29 pm
@ascribbler,
First part: 5/9
Second part: 2/5
0 Replies
 
ascribbler
 
  1  
Reply Thu 12 Jul, 2018 10:58 pm
Thus far we've proven that recursive probability questions can be quite tricky.
markr
 
  1  
Reply Thu 12 Jul, 2018 11:32 pm
@ascribbler,
It's not tricky or recursive if one reads the question correctly. I did the math for 6- and 5-card decks instead of 7- and 6-card decks.

6/11
5/12

For the first case, there are only 6 (L wins) + 5 (F wins) = 11 outcomes that we care about. All others don't count and don't affect the final numbers.

For the second case, there are only 6 (L wins) + 5 (F wins) + 1 (draw, so L wins) = 12 outcomes that we care about.
0 Replies
 
ascribbler
 
  1  
Reply Fri 13 Jul, 2018 12:28 am
The chance they both draw a Square = (6/7)x(5/6) = 30/42 = 5/7

The chance Farmerman draws a Triangle and Leadfoot draws a Square = (1/7)x(5/6) = 5/42

The chance Farmerman draws a Square and Leadfoot draws a Triangle = (6/7)x(1/6) = 6/42

The chance they both draw a Triangle is = (1/7)x(1/6) = 1/42

The total recursive probability of Farmerman drawing a Square and Leadfoot drawing a Triangle is :

1/7 first draw + (5/7)x(1/7)second draw + (5/7)^2x(1/7)third drawer + (5/7)^3x(1/7)fourth drawer + ...

summing the geometric series

(1/7)/[1-(5/7)] = 1/2 = 6/12

Similarly, the total probability of Farmerman drawing a Triangle and Leadfoot drawing a Square is 5/12

And the total probability of both drawing Triangles is 1/12

The somewhat more elegant solution is to realise that there are 6 ways for Farmerman to lose by drawing a Square, 5 ways for Leadfoot to lose by drawing a Square and only one way to obtain a draw.

Leadfoot has a 6/12 probability of winning
Farmerman has a 5/12 probability of winning
A draw has a probability of 1/12
markr
 
  1  
Reply Fri 13 Jul, 2018 12:53 am
@ascribbler,
You posed two questions and gave one solution. That solution assigns a non-zero probability to an invalid ending with respect to the first question. Your recursive calculation does not account for redrawing when they both draw triangles.

Your "somewhat more elegant" method is exactly what I did, but you, again, didn't treat the draw correctly for the first question.

"If they both draw a triangle from their decks at the same time they redraw."
fresco
 
  1  
Reply Fri 13 Jul, 2018 01:09 am
@ascribbler,
I concur with makr's general points.
Apologies for not reading your pack totals correctly in my calculation, which says something about me being distracted by the irrelevent context.
ascribbler
 
  1  
Reply Sat 14 Jul, 2018 05:11 am
@markr,
Quote:
Leadfoot has a 6/12 probability of winning
Farmerman has a 5/12 probability of winning
A draw has a probability of 1/12


Quote:
You posed two questions and gave one solution.


Incorrect, the three probabilities are displayed.

Quote:
That solution assigns a non-zero probability to an invalid ending with respect to the first question.


Your answer of 6/11 is incorrect the correct answer is 6/12.

Quote:
Your recursive calculation does not account for redrawing when they both draw triangles.


Incorrect, the probability of redrawing is 1/12.

Quote:
Your "somewhat more elegant" method is exactly what I did, but you, again, didn't treat the draw correctly for the first question.


Incorrect, the probabilities are as displayed, you forgot to account for the draw, if your first answer was correct it would have a numerator of 12.
ascribbler
 
  1  
Reply Sat 14 Jul, 2018 05:19 am
@fresco,
Quote:
I concur with makr's general points.
Apologies for not reading your pack totals correctly in my calculation, which says something about me being distracted by the irrelevent context.


The words in word problems are an imperative like markr instead of "makr" and irrelevant instead of "irrelevent".

Your answer(s) were well awry of the mark but to concur with incorrect assertions without mathematical basis is inapt at best or in need of buying avowal at worst.
ascribbler
 
  1  
Reply Sat 14 Jul, 2018 06:00 am
@ascribbler,
Quote:
Incorrect, the probabilities are as displayed, you forgot to account for the draw, if your first answer was correct it would have a numerator of 12.


Denominator not numerator.
0 Replies
 
fresco
 
  1  
Reply Sat 14 Jul, 2018 10:59 am
@ascribbler,
Laughing sour grapes ?
0 Replies
 
markr
 
  1  
Reply Sun 15 Jul, 2018 06:57 pm
@ascribbler,
According to your own rules, a draw as a final outcome for the first problem posed is impossible.
markr
 
  1  
Reply Sun 15 Jul, 2018 07:35 pm
@markr,
The three probabilities you supplied are appropriate for the second problem only since the probability of a draw (final outcome) is impossible for the first problem.

Do you know how to program? If so, try simulating the first problem according to the rules you provided - simultaneous triangles (as well as simultaneous squares) cause a redraw. A million simulations should convince you.
 

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