Question

In a casino in Blackpool there are two slot machines (machine 1 and machine 2): one that pays out 10 % of the time, and one that pays out 20 % of the time. Obviously, you would like to play on the machine that pays out 20% of the time but you do not know which of the two machines is the more generous. You thus adopt the following strategy: you assume initially that the two machines are equally likely to be the generous machine. You then select one of the two machines at random (this turns out to be machine 1) and put a coin into it. Given that you loose that first bet estimate the probability that machine 1 is the more generous of the two machines
What do I know?
What do I want to find out?
What do I expect the answer to be?
How do I go from what I know to what I want to find?
Is the answer consistent with what I expected?

Answer 1

Answer:

Given that;

a)

2 machines, 1 and 2

Machine 1

pays 10% of times when the machine is generous

So the probability that the machine 1 pays given that its generous is

P (Pays/machine 1 is generous) = 10% = 0.10

Machine 2  

pays 20% of times when its generous

i.e the probability that the machine pays given that its generous is;

P (Pays / machine 2 is generous) = 20% = 0.20

Also we assume there is equal chance of being generous

i.e

P(machine 1 is generous) = P(machine 2 is generous ) = 0.5  

b)

this is to help obtain the probability that machine 1 is generous given that the player loose in the first bet,

i.e P(machine 1 is generous / lost)  

c)

Since, the probability that the machine is generous is 0.5,

it can be said that there is 50% chance that machine 1 is generous when the player loses the first bet

d)

therefore The required probability is calculated as;

P(machine 1 is generous/lost) = p(machine 1 is generous ∩ lost) / p (lost)

= [p(lost/machine 1 is generous) × p(machine 1 is generous)] / [{(1-p(pays/machine 1 is generous)) × p(machine 1 is generous)} + {((1-p(pays/machine 2 is generous)) × p(machine 2 is generous)}]

= [(1 - p(pays/machine 1 is hereous)) × p(machine 1 is generous)] × [(( 1- 0.10) × 0.5) + ((1 - 0.20) × 0.5)]

= ( 1- 0.10) × 0.5) / 0.85

= 0.5294

so  the probability that the player loses the first bet given that the machine is generous is 0.5294

e)

Since the gotten probability that the player loses the first bet given that machine 1 is generous is close to 0.50 then it can be said that the probability is consistent with the expectations.