need help with umat challenging probability question (1 Viewer)

[magic123]

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 please include working out of the solution and explanation. Thanx

 

[FrankXie]

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 please include working out of the solution and explanation. Thanx

Correct answer is C.

First of all, there are six coins so there are 2^6 different results of guess, because each coin has two possible outcomes: correct guess or wrong guess. And for each of these 2^6 results, the probability is all same, equal to 1/2^6.

Let x be the entrance fee. For the organizers to break even, we should have



Let me explain a little about the LHS (the RHS just means "break even", no profit, no loss): the first term meaning the 6 coins are all guess wrongly; the terms in the first square brackets meaning only one coin is guessed correctly; similarly the terms in other square brackets; the last term meaning all 6 coins are guessed correctly.

Because each single term has coefficient 1/2^6, we can divide it or equivalent ingore it.

Here comes the key step: we need to find a way to sum up the LHS. For each value of the denomination, say 0.05, it appears in the sum 2^5 times(this is because when this coin is guessed correctly, all other 5 coins still have 2^5 different results). And x appears (2^6-1) times(because in the last term x cancel out). Therefore the sum of LHS is





Dividing both numerator and denominator by 2^6, we have the result C.