This happens with probability for integers. If Jason picks up dice and leaves a die showing, he will need the other two to sum to. If Jason decides to pick up all the dice to re-roll, by the stars and bars rule ways to distribute,, there will be bars and stars ( of them need to be guaranteed because a roll is at least ) for a probability of. We need to assume he does not have a to begin with. Jason will pick up no dice if he already has a as a sum.
Note that any of the three dice can be the remaining die, so we need a factor of for all counts in the third column: We will analyze these cases by considering the initial outcomes of the two dice rerolled. The first three cases in the table above satisfy this requirement. If Jason needs to reroll at least two dice to win, then he rerolls exactly two dice if and only if the probability of winning is greater than the probability of winning for rerolling all three dice.If Jason needs to reroll at least one die to win, then he rerolls exactly one die.If Jason needs to reroll at least zero dice to win, then he rerolls exactly zero dice.Note that the dividers' positions and the ordered triples have one-to-one correspondence, and holds for all such ordered triples. From left to right, the numbers of balls in the piles correspond to and respectively. There are gaps between the balls, and placing dividers in of the gaps separates the balls into piles. If Jason (re)rolls all three dice, then the probability of winning is alwaysįor the denominator, rolling three dice gives a total of possible outcomes.įor the numerator, this is the same as counting the ordered triples of positive integers for which Suppose that balls are lined up in a row.If Jason rerolls exactly two dice, then the outcome of the remaining die must be or Applying casework to the remaining die produces the following table:.If Jason rerolls exactly one die, then the sum of the two other dice must be or The probability of winning is always as exactly of the possible outcomes of the die rerolled results in a win.Jason rerolls exactly zero dice if and only if the sum of the three dice is in which the probability of winning is always.We conclude all of the following after the initial roll: There are ways in which rerolling two dice is optimal, out of possibilities, Therefore, the probability that Jason will reroll two dice is.
The possible triplets that satisfy these conditions, and the number of ways they can be permuted, are Thus, rerolling two dice is optimal if and only if and. In order for rolling two dice to be more favorable than rolling three dice. By stars and bars, there are ways to do this, making the probability of winning. To find the probability of winning if we reroll all three dice, we can let each dice have dot and find the number of ways to distribute the remaining dots. If we reroll two dice, the probability of winning is, once again. Rerolling one die will not help us win since the sum of the three dice will always be greater than. However,, so rolling one die is always better than rolling two dice if. If we reroll two dice, we will roll the two maximum dice, and the probability of winning is, as stated above. If we reroll one die, we can reroll any one dice in the hope that we get the value that makes the sum of the three dice. If, then we do not need to reroll any dice. Thus, we can let be the values of the three dice. This means that if Jason rerolls two dice, he must choose the two dice with the maximum values. In order to maximize the probability of winning, must be minimized. Therefore, if one die has a value of and Jason rerolls the other two dice, then the probability of winning is. There are pairs that satisfy this, namely, out of possible pairs. What is the probability that he chooses to reroll exactly two of the dice?Ĭonsider the probability that rolling two dice gives a sum of, where. After rerolling, he wins if and only if the sum of the numbers face up on the three dice is exactly Jason always plays to optimize his chances of winning.
Then he looks at the rolls and chooses a subset of the dice (possibly empty, possibly all three dice) to reroll. Jason rolls three fair standard six-sided dice. 7.3 Video Solution 3 (pls subscribe to my youtube channel thanq).
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