prob.part6.13_14 - 1.1. SIMULATION OF DISCRETE...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
1.1. SIMULATION OF DISCRETE PROBABILITIES 5 5 10 15 20 25 30 35 40 -10 -8 -6 -4 -2 2 4 6 8 10 Figure 1.1: Peter’s winnings in 40 plays of heads or tails. One can understand this calculation as follows: The probability that no 6 turns up on the ±rst toss is (5 / 6). The probability that no 6 turns up on either of the ±rst two tosses is (5 / 6) 2 . Reasoning in the same way, the probability that no 6 turns up on any of the ±rst four tosses is (5 / 6) 4 . Thus, the probability of at least one 6 in the ±rst four tosses is 1 - (5 / 6) 4 . Similarly, for the second bet, with 24 rolls, the probability that de M´ er´ e wins is 1 - (35 / 36) 24 = . 491, and for 25 rolls it is 1 - (35 / 36) 25 = . 506. Using the rule of thumb mentioned above, it would require 27,000 rolls to have a reasonable chance to determine these probabilities with su²cient accuracy to assert that they lie on opposite sides of .5. It is interesting to ponder whether a gambler can detect such probabilities with the required accuracy from gambling experience. Some writers on the history of probability suggest that de M´ er´ e was, in fact, just interested in these problems as intriguing probability problems. Example 1.4 (Heads or Tails) For our next example, we consider a problem where the exact answer is di²cult to obtain but for which simulation easily gives the qualitative results. Peter and Paul play a game called
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 2

prob.part6.13_14 - 1.1. SIMULATION OF DISCRETE...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online