Willy Plays Craps Roller
Willy goes to Las Vegas and plays the game Craps Roller.
The rules of the game Craps Roller are as follows:
One player rolls two die. If the sum of the numbers of the two die is equal to 7 or 11, the player wins. If the sum of the numbers is 2, 3, or 12, then the player loses. If the sum of the numbers is equal to something else, say , the die are rolled again until either the sum of the numbers is or the sum of the numbers is 7. If is rolled before 7, then the player wins; otherwise, the player loses.
Willy becomes addicted to Craps Roller and plays it 990 times. What is the expected value of the number of times he will win? Round your answer to the nearest integer.
(Dispute: If the player rolls a sum that is, let's say, 4, he keeps rolling until he gets either 4 or 7 - any other valid numbers from the first part will not be considered.)
(Arman Siddique: Programming is not allowed.)
The answer is 488.
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1 solution
Lets begin with the initial state where we roll a pair of dice and take the first value that we see.
There are 36 possible outcomes of this roll; 8 of which grant you a win right away; 4 of which grant you a loss right away; all other rolls we'll count as 'continues' for now.
So (8/36) * 990 = 220, we should expect to win on our first roll exactly 220 times.
Similarly, (4/36) * 990 = 110, we should expect to lose on our first roll exactly 110 times.
It follows that the other 660 times we will Continue on to the Goal Roll as I like to call it.
Once in the Goal Roll state, our chances vary depending on the Goal Number we're trying to reach.
Of the 660 times we get here, we would expect our goal number to be either 4 or 10, 165 times. (6/24) * 660
We'd also expect our goal number to be either 5 or 9, 220 times. (8/24) * 660
Finally, we'd expect our goal number to be either 6 or 8, 275 times. (10/24) * 660
From our initial roll, if we had rolled a 4 or a 10, we would have 3 possible rolls to win (and 6 to lose).
If we had rolled a 5 or a 9, we would have 4 possible rolls to win (and 6 to lose).
And if we had rolled a 6 or an 8, we would have 5 possible ways to win (and 6 to lose).
Putting all of the wins together we'd say ((8/36) * 990) + ((3/9) * 165) + ((4/10) * 220) + ((5/11) * 275) = 488