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# The chances of a higher power.

ascribbler

1
Wed 18 Jul, 2018 07:37 am
@markr,
Quote:
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.

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
markr

1
Wed 18 Jul, 2018 09:20 am
@ascribbler,
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.
ascribbler

1
Wed 18 Jul, 2018 11:53 pm
@markr,

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.
markr

2
Thu 19 Jul, 2018 01:36 am
@ascribbler,
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
ascribbler

1
Sun 22 Jul, 2018 12:00 am
@markr,
Got it.
0 Replies

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