A fair die is rolled up to three times. At the end of each roll, you have the option to roll again or take the number showing in dollars. What is the optimal strategy that maximizes your expected earnings? What are the expected earnings?
save this religious garble for a different site this site is for advanced math and science !! the expected earnings on each roll range from 1-6 dollars. the probability of getting any given dollar amount is equivalent to that specific number on the die which is 1/6. since rolled 3 times fc ( fundamental counting ) gives (1/6)^3 . but thgis is probability answer when the expected earnings can me one of 6^3 possabilites, say example player rolls a 1, then a 5 and quits. his expected earnings are 1-6 then 1-6 again, so the earnings depend on how many rolls he decides which changes each play since he expects he will 'hit the six'.the probability of getting a six in either roll is 1/6 just as each other number is equally likely at each roll hence there is no optimal strategy. This may not be coorect but as a data scientist I will return here if my solution changes $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
ok after re reading this problem I see it can be done with taking the average . on the first roll the expected payoff is thus (1+2+3+4..+6)6= 3.5 Now the trick is if player gets a 6 he would of course stop :) same goes for 4 and 5 . thus there is a more complex way to determine the expected earnings and an optimal strategy, why since jan orf 2014 no one has commented on a solution to this oldie but goldie makes me not want to reveal more , as math like this is self taught with determination and abstract thought, not just feeding the kilobytes of the solution to a reader :).
save this religious garble for a different site this site is for advanced math and science !!
ReplyDeletethe expected earnings on each roll range from 1-6 dollars.
the probability of getting any given dollar amount is equivalent to that specific number on the die which is 1/6. since rolled 3 times fc ( fundamental counting ) gives (1/6)^3 . but thgis is probability answer when the expected earnings can me one of 6^3 possabilites, say example player rolls a 1, then a 5 and quits.
his expected earnings are 1-6 then 1-6 again, so the earnings depend on how many rolls he decides which changes each play since he expects he will 'hit the six'.the probability of getting a six in either roll is 1/6 just as each other number is equally likely at each roll hence there is no optimal strategy. This may not be coorect but as a data scientist I will return here if my solution changes
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$
ok after re reading this problem I see it can be done with taking the average . on the first roll the expected payoff is thus (1+2+3+4..+6)6= 3.5 Now the trick is if player gets a 6 he would of course stop :) same goes for 4 and 5 . thus there is a more complex way to determine the expected earnings and an optimal strategy, why since jan orf 2014 no one has commented on a solution to this oldie but goldie makes me not want to reveal more , as math like this is self taught with determination and abstract thought, not just feeding the kilobytes of the solution to a reader :).
DeleteSteve that is in-advanced , semitic jibberish . Please keep such text off a math and science blog .
ReplyDelete