Question

17-2: Game Show Uncertainty: In the final round of a TV game show, contestants have a chance to increase their current winnings of $1 million to $2 million. If they are wrong, their prize is decreased to $500,000. A contestant thinks his guess will be right 50% of the time. Should he play? What is the lowest probability of a correct guess that would make playing profitable?

Please double check these calculations and help me to develop the rational for why it is so.

What is known:

Current winnings: 1M

Probability of winning: 0.5; if he wins, he gets: 2M (get 1M).

Probability of losing 0.5; if he loses, he gets: $500, 000 (loses 500,000).

Therefore: 0.50 x $1,000,000 = $500,000 0.50 x $500,000 = $250,000

Since $500,000, (the probability of him winning is greater than) $250,000, he should play.

I need help with the answer below: I'm not sure which equation is bet for this answer. How do I develop the rationale for the right calculation?

Equation #1

Lowest Probability:

If lowest probability =p,

then: p(2)+(1-p)(0.5)>1

=(2p+0.5-0.5p)>1 1.5p-0.5>1

=p>0.5/1.5

p=0.33

Then the lowest probability that would make it profitable to play is 33%.

OR Equation #2

1,000,000(x/100)-500,000x(1-(/100)=0

10,000x-500,000+5000x=0

15,000x=500,000

15x=500

X=0.33

Answer 1

0 1.95 milion HU enpefti Assuming LApa^d pay Tha ncr Sen