The chances of a higher power.

Math question, huh?

My head hurts

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.

Now, now fresco. The question has a mathematical answer and you need to show your working for any glorious credit.

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.

If you mean what are the chances of Leadfoot winning or drawing
the answer is : 1/5 x 5/6 +1/5 +1/6.

slight mistype of last...1/5 x 5/6 + 1/5 x 1/6

First part: 5/9
Second part: 2/5

Thus far we've proven that recursive probability questions can be quite tricky.

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.

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

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."

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.

Incorrect, the three probabilities are displayed.



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



Incorrect, the probability of redrawing is 1/12.



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.

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.

Denominator not numerator.

sour grapes ?

According to your own rules, a draw as a final outcome for the first problem posed is impossible.

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.