A and B, two lifelong friends frequently play the following game of their invention:
1st player rolls
a pair of fair dice and if they jointly display 7 or 11
he wins, if not the 2nd player tries his luck and they take turns untill one of the goals is reached.
Since it clear to both of them that 1st player has higher chances to win than the second - the one that started becomes the second and they always play an even number of rounds.
Oney day A suggested the following modification:
1st player is one of us for the whole day, but while for
each of his wins he will get 10 bucks - for 2nd player's winning the reward will be 16 ...
Your choice today, be 1st or second !
What is your (the solver’s) advice?
There are 6 ways for the dice can total 7 and 2 ways to total 11; that's a probability of 8/36 of getting a win on a given roll.
The probability the win comes on the first player's (say A's) first roll is 8/36. The probability that A fail on the first attempt and B wins on that first opportunity is 28/36 * 8/36. Reduced, those probabilities are 2/9 and 14/81 respectively.
The overall ratio of probabilities the first player wins vs second player wins is just the ratio of these two ratios: (2/9) / (14/81) = 9/7.
A fair ratio of winnings would be 9/7 of 2nd players reward to 1st player's reward. That comes to about 1.286. What the proposal calls for a ratio of 1.6, generously rewarding the second player more than is called for to make the game fair. So choose to be second.
|
|
Posted by Charlie
on 2023-07-29 08:50:51 |
solution
