![]() ![]() The odds that you get two straight heads on your 4rd roll is (1/4)(1/2)*(1/2)ĭowntown, you're overlooking the fact that there are two ways to create the event X=4 ![]() The odds that you get two straight heads on your 3rd roll is (1/4) (1/2). Then, as the last poster mentioned, the EV would be 3.5+3.5 = 7.ĭowntown:the odds that you get two straight heads on your 2nd roll is 1/4 Note: If you had to keep the first roll and add it to your second score, it would always make sense to re-roll. Since you have two options, which are equally likely, the EV is:ġ/2(the chance you don't reroll) + 1/2(the chance you re-roll and ignore your first score) Again, on your second roll, your EV is 3.5. The other possibility (which is equally likely) is that you get 1, 2, or 3. If you roll over 3.5, then the score you got must have been either 4, 5, or 6. Since the expected value (EV) of one dice roll is 3.5, it makes sense to keep your score if it is higher than 3.5, and re-roll if you get a score less than 3.5. The first time you roll the dice, you have two options: This means that you if you decide to re-roll, you should disregard your first roll and just keep the second one. The key here is that the problem says "you can roll the die again and get paid what the second roll shows INSTEAD of the first" (caps added). I think the previous poster forgot one fact that changes things a bit. Are you the kind of person who thinks the "7 8 9" joke is funny? In the long run, the average value of two dice rolls using regular dice is 7. The random variable X assumes a value equal to the sum of two dice rolls. The random variable X has the following probability distribution: x pX(x) Multiplying the values with their respective probability gives:ĮX. For a discrete random variable, E(X) is calculated as The expected value of a random variable X is denoted by E(X). I don't know why this called "Hardest Probability Questions". If you treat each event as a pair of rolls the expectation would be 36 rolls, but you have to count rolls 2 and 3, 4 and 5, etc. The easiest/quickest idea to pull out of this is that the expectation of a geometric random variable is just 1/p, where p is the probability that your event will happen in a given roll.įinding the E of 2 sixes in a row would take longer, but it seems that it would be less than 36. 167+2.139.etc) up to an infinite number of rolls and find that the expectation converges to 1/P, where P is 1/6 in our case, so the expected number of rolls to get a six (or any other number on the die) is 6. If you wanted to, you could multiply the #rolls times the respective probability (1. The probability that you will get at least one 6 out of 6 rolls is about 0.668. You can answer what the probability is that out of X rolls, you will get at least one 6 using a geometric random variable. Since each 1st roll outcome has a probability of 1/6, we can get the total expected value by add up the weighted expected values for each conditional outcome. For each possible outcome, there is a corresponding expected value for the "better-of-two" rolls.Ĭonditional expected value: (5/6) 5 + (1/6)6 = 5.1667 (this is expected, i.e. There are 6 possible outcomes for the 1st roll (1, 2, 3, 4, 5, 6), each with a probability of 1/6. This would be zeroĢ - represent number of reds blues and yellows with 'a', 'b' & 'c' where a>b>c. I think I got the answers right?ġ- rephrased, it asks "What is the probability of getting 3 consecutive heads before 2 consecutive heads". Now if we know at a certain time, there are 13 blue, 15 red, and 17 yellow snakes, question: could the snakes meet so that eventually all snakes become eg, if a red snake meet a yellow snake, they both change to blue. Every time 2 snakes of different colors met, they both change their color to the 3rd color. On an island there are snakes of 3 different colors.let's say A keep tossing a fair coin, until he get 2 consecutive heads, define X to be the number of tosses for this process B keep tossing another fair coin, until he get 3 consecutive heads, define Y to be the number of the tosses for this process. ![]()
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