Question

I roll a die 4 times. I win if a six appears. To make this game more interesting, I decide to add a second die and target the appearance of a double six. I reason as follows: a double
six is one-sixth as likely as a six — 1/36 compared to 1/6. I should be able to increase the number of rolls by a factor of 6 (now 24 rolls) and still maintain the same probability of
winning. Is this true? (Probabilistic justification is required here!)

Answer 1

Let's calculate the probabilities of both the scenarios explicitly and then see if they are same or not.

The first case is where we roll a single die 4 times. We win if we get a 6 in at least one of the 4 rolls. Rolling a die is an independent event and probability of success in each role is 1/6. Thus in order to win we need a 6 in 1st roll OR 2nd roll OR 3rd roll OR 4th roll. If the experiments are independent and there is an 'OR' condition then we add the probabilities to get the total probability of success. Therefore the probability of winning is (1/6 + 1/6 + 1/6 + 1/6) = 4/6 = 2/3

In the second case we use two dice and consider getting a double 6 as the win. Now we have to find out the probability of winning when we have 24 rolls to get at least one double 6. Again, each rolling will be an independent event. Using the same explanation given above if the probability of getting a double 6 in a roll is 1/36 then for 24 rolls it will be 1/36 + 1/36 + ..... + 1/36 (24 times) = 24/36. Thus the probability of winning in this scenario will be 24/36 = 4/6 = 2/3

Thus we have shown that both the scenarios mentioned above will have the same probability of winning. Hence the given statement in the question is true.

Hope this helped! All the best!