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

To estimate the probability that machine 1 is the more generous of the two machines, we can use Bayes' theorem. Given that the first bet was lost on machine 1 and initially assuming both machines are equally likely to be generous, the probability that machine 1 is more generous is found to be 0.9 or 90%.

Step-by-step explanation:

To estimate the probability that machine 1 is the more generous of the two machines, we can use Bayes' theorem. Let's define the following:

• A: Machine 1 is the more generous machine
• B: Losing the first bet on machine 1

From the given information, we know that both machines are equally likely to be the more generous machine initially. So, P(A) = P(B) = 0.5. We want to find P(A|B), which is the probability that machine 1 is the more generous machine given that the first bet was lost on machine 1.

Using Bayes' theorem, we have: P(A|B) = (P(B|A) * P(A)) / P(B).

P(B|A) is the probability of losing the first bet on machine 1 given that machine 1 is the more generous machine. Since machine 1 pays out 10% of the time, P(B|A) = 0.9.

P(B) is the probability of losing the first bet on either machine. Since both machines are equally likely to be the more generous machine, the probability of losing the first bet on either machine is the same. Therefore, P(B) = 0.5.

Substituting the values, we have: P(A|B) = (0.9 * 0.5) / 0.5 = 0.9.

Therefore, the probability that machine 1 is the more generous of the two machines, given that the first bet was lost on machine 1, is 0.9 or 90%.

Answer 2

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.