## Imp 1 Pow What’s on Back Essay

Sage Wentzell-Brehme Problem Statement ?In this problem I am looking at strategies for a game called What’s on Back? This game has 3 cards. One card has an X on both sides, one card has an O on both sides, and one has an O on one side and a X on the other. On your turn you choose one of the cards at random from a bag. You hold it up with one side facing you.

You are not allowed to look at the other side or the other cards. You have to predict whether it is an O or an X on the other side of the card. My task is to look at a bunch of different strategies for playing and find the probability of success for each one.

For each strategy I look at I have to find the probability using both an experimental and theoretical model. My real goal is to find the strategy with the highest probability of success or the one most likely to help me win. Strategy # 1 a.

Always choose the same thing the card says. So if it is an O choose O, if it is an X choose X. b. 30 trials 1. yes 6. no 11.

no 16. yes 21. yes 26. yes 2. yes 7. yes 12.

yes 17. es 22. no 27. yes 3. yes 8. no 13.

yes 18. no 23. yes 28. yes 4. no 9. no 14. yes 19. yes 24.

no 29. yes 5. no 10. yes 15. yes 20. no 25. no 30.

Yes P (right) – 18/30 or 6/10 or . 6 Success rate = . 6 c. (XX)-works-1/6 (XX)-works-1/6 (OO)-works-1/6 (OO)-works-1/6 (OX)-doesn’t work (XO)-doesn’t work total successful – 4/6 or 2/3 or . 66 Success rate = . 66 Strategy #2 a.Always choose O no matter what.

b. 30 trials 1. yes 6. no 11. yes 16.

yes 21. no 26. yes 2.

es 7. no 12. yes 17. yes 22. no 27.

no 3. no 8. yes 13. no 18. yes 23. no 28.

yes 4. no 9. no 14. yes 19. yes 24. yes 29. yes 5. no 10.

yes 15. yes 20. yes 25. no 30.

no P (right) – 17/30 or . 56 Success rate = . 56 c.

(XX)-doesn’t work (XX)-doesn’t work (OX)-doesn’t work (XO)-works-1/6 (OO)-works-1/6 (OO)-works-1/6 total successful – 3/6 or . 5 Success rate = . 5 Strategy # 3 a. Always choose the opposite of the card you chose. So if it is an O choose X, if it is a X choose O.

. 30 trials 1. no 6. no 11. no 16.

no 21. no 26. yes 2. yes 7. no 12. no 17. no 22.

no 27. no 3. no 8. no 13. no 18. no 23. no 28.

yes 4. no 9. no 14. yes 19. no 24. yes 29. yes 5.

no 10. yes 15. yes 20. no 25. no 30.

no P (right) – 7/30 or . 23 Success rate = . 23 c. (XX) – doesn’t work (XX) – doesn’t work ? (XO) – works – 1/6 (OX) – works – 1/6 (OO) – doesn’t work OO) – doesn’t work total successful 2/6 or . 33 Success rate = . 33 Strategy # 4 a.

Always choose X no matter what. b. 30 trials 1. no 6. yes 11. yes 16.

no 21. no 26. yes 2.

yes 7. no 12. yes 17. yes 22. yes 27. no 3.

no 8. yes 13. no 18. yes 23. no 28. no 4. yes 9.

no 14. yes 19. no 24. yes 29. yes 5. no 10. yes 15. no 20.

no 25. no 30. no P(right) – 14/30 or . 46 Success rate = . 46 c. (XX)- works – 1/6 (XX) – works – 1/6 (OX) – works – 1/6 XO) – doesn’t work (OO) – doesn’t work (OO) – doesn’t work total successful – 3/6 or . 5 Success rate = .

5Summarization a. The best strategy is strategy number one. To choose the same option as the one you see. So if it’s O choose O, if it’s X choose X. This is the most successful because the theoretical and experimental probabilities are higher than the others. b. I trust theoretical more because it is more accurate. Extensions a.

If you had 12 cards, 4 with all O’s, 4 with all X’s, and 4 with one X and one O you would get the same probabilities. . It produces the same probability because 12 is a multiple of 3. So it is just all the numbers quadrupled but that of course doesn’t change the average.

So my variation has the same probabilities of success. Evaluations a. This was a very easy problem because experimental probability is very simple and the theoretical probability was simple once you understood how to do it. b. I thought it was kind of boring.

c. The best part was the experimental probability because it was easy. The worst part was finding a variation that fit all of the requirements.

x

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