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need help with umat challenging probability question (2 Viewers)

magic123

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in a game of chance, one coin of each denomination is tossed ($0.05, $0.10,$0.20,$0.50,$1,$2). Participants are asked to guess either heads or tails for each coin tossed, and if guessed correctly they will win that coin. And if they manage to guess correctly for all throws, they will also win back their entrance fee.

What amount in dollars needs to be charged by the organizers to break even?

Options:
A) 0.05 +0.1+0.2+0.5+1+2
B) 0.5 (0.05 +0.1+0.2+0.5+1+2)
C) [0.5 (0.05 +0.1+0.2+0.5+1+2)]/(1-1/2^6)
D) [0.5 (0.05 +0.1+0.2+0.5+1+2)]x(1+1/2^6)

Can you plz include the working out for the solution and explanation. thanx
 

Queenroot

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ceebs working out
A is too simple, C the answer will be zero

so it's either D or B
 

Kevink98

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The answer is C, you can work backwards from the answers given.

If [0.5 (0.05 +0.1+0.2+0.5+1+2)]/(1-1/2^6) is the entry price, how much money does a player get per round?

They get 0.5*(all the coins) and there is also the additional probability of the bonus refund.

0.5*(all the coins) + (0.5^6)([0.5 * (all the coins]/(1-0.5^6))

= (0.5*all the coins*1 - (0.5^6)*0.5*all the coins + (0.5^6)([0.5 * (all the coins])/(1-0.5^6)

= 0.5*all the coins / (1-0.5^6)

= C

So since the player gets back the entry fee per round, the company has broken even.

But I'm guessing for Umat you can just find the answer by elimination.
 
Last edited:

mirachael

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Let x=entry fee
a=0.05 +0.1+0.2+0.5+1+2
There are 26=64 possible outcomes from this game
Thus, in 64 games in which every outcome has occurred, the organizer's income would be 64x and their payout would be 64*0.5a+x


Therefore, the answer is C
 

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