The chances of a higher power.

Incorrect.

The probability of a draw is the same viz. 1 /12 in both instances.

1/42 + (1/42)x(5/7) + (1/42)x(5/7)^2 + ...

(1/42)/(1-5/7) = 1/12

Reread your own problem. They redraw whenever they both draw the same card (triangle/triangle or square/square). Therefore, THE GAME CAN NEVER END IN A DRAW (TIE).

Explain why you're treating triangle/triangle differently than square/square when they redraw in both cases.

You read it again.

A no point did I state that the game can end in a tie.

The game is won when one player draws a triangle at the same time that the other player draws a square. Two triangles drawn simultaneously by each player is specifically defined as not being a win and requiring another round.

You implied that the game can end in a tie when your "solution" gave the probability of a draw as 1/12. The probabilities for the two players must sum to one - one of them has to win. 6/12 + 5/12 != 1.

Here's what you should have written:

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 chance that they have to redraw is 5/7 + 1/42 = 31/42

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

1/7 first draw + (31/42)x(1/7)second draw + (31/42)^2x(1/7)third drawer + (31/42)^3x(1/7)fourth drawer + ...

summing the geometric series

(1/7)/[1-(31/42)] = 6/11

Got it.