Math, asked by ritikdhiman206602, 1 month ago

12 dice are thrown 3086 times and a throw of 2, 3, 4 is reckoned as a success. Suppose that 19142 throws of 2, 3, 4 have been made out. Do you think that this observed value deviates from the expected value ? If so, can the deviation from the expected value be due to fluctuations of simple sampling?​

Answers

Answered by frenzy87
0

Answer:

Let's suppose we have only 1 roll. What is the expected payoff? Each roll is equally likely, so it will show 1,2,3,4,5,61,2,3,4,5,6 with equal probability. Thus their average of 3.53.5 is the expected payoff.

Now let's suppose we have 2 rolls. If on the first roll, I roll a 66, I would not continue. The next throw would only maintain my winnings of 66 (with 1/61/6 chance) or make me lose. Similarly, if I threw a 55 or a 44 on the first roll, I would not continue, because my expected payoff on the last throw would be a 3.53.5. However, if I threw a 1,21,2 of 33, I would take that second round. This is again because I expect to win 3.53.5.

So in the 2 roll game, if I roll a 4,5,64,5,6, I keep those rolls, but if I throw a 1,2,31,2,3, I reroll. Thus I have a 1/21/2 chance of keeping a 4,5,64,5,6, or a 1/21/2 chance of rerolling. Rerolling has an expected return of 3.53.5. As the 4,5,64,5,6 are equally likely, rolling a 4,54,5 or 66 has expected return 55. Thus my expected payout on 2 rolls is .5(5)+.5(3.5)=4.25.5(5)+.5(3.5)=4.25.

Now we go to the 3 roll game. If I roll a 

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