A Game of Dice
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
Your friend offers to play a game of dice with you. He explains the game to you.
"There are two dice and the goal is to have the highest total of two dice. You roll first. If you don't like your roll, then you can roll both dice again. If you roll a second time, you must keep the second roll. I'll roll second. I have to keep my first roll, but if we both roll the same number, I win."
On your first roll you roll a 7. Should you roll again or keep the 7?
Bonus question: if you play the optimal strategy, what is the probability that you will win?
Answer
You should roll again.
The probability of winning with a particular roll is the cumulative probability of your friend rolling one of the lower numbers.
There are 36 possible outcomes of rolling the dice (6 for the first die x 6 for the second die). The number of ways and probability of rolling each number is:
2,12: 1 = 1/36
3,11: 2 = 2/36
4,10: 3 = 3/36
5,9: 4 = 4/36
6,8: 5 = 5/36
7: 6 = 6/36
Since your friend wins ties, he wins with any number 7 or greater. There are 1+2+3+4+5 = 15 ways to roll a 2, 3, 4, 5, or 6. This gives you a
15/36 = .41666...
probability of winning if you keep the 7.
If you roll a second time, then you have to take the sum of the probabilities of each roll multiplied by the probability of winning with that roll. This gives:
2: 1/36*0 = 0
3: 2/36*1/36 = 2/1296
4: 3/36 * 3/36 = 9/1296
5: 4/36 * 6/36 = 24/1296
6: 5/36 * 10/36 = 50/1296
7: 6/36 * 15/36 = 90/1296
8: 5/36 * 21/36 = 105/1296
9: 4/36 * 26/36 = 104/1296
10: 3/36 * 30/36 = 90/1296
11: 2/36 * 33/36 = 66/1296
12: 1/36 * 35/36 = 35/1296
Adding these probabilities together gives a
575/1296 ~ .44367
probability of winning if you roll again.
Bonus question:
Since you now know that the optimal strategy is to reroll when you get a 7, and to keep anything 8 or higher, you can now calculate your chances of winning. Take the probability for each roll for the first throw of the dice multiplied by the probability of winning after that first roll. Any number 8 or higher has the same probabilities calculated above:
8: 5/36 * 21/36 = 105/1296
9: 4/36 * 26/36 = 104/1296
10: 3/36 * 30/36 = 90/1296
11: 2/36 * 33/36 = 66/1296
12: 1/36 * 35/36 = 35/1296
Adding these probabilities gives a 400/1296 ~ .31 probability of winning with the first roll.
Any number 7 or lower will be rerolled, and from above you know that the probability of winning on the second roll is 575/1296. The probability of rolling 7 or lower is 21/36 (from above: 1  15/36 = 21/36). This gives a probability of
27: 21/36 * 575/1296 = 12159/46656 ~ .26
And a
.31 + .26 = .57
probability of winning on the first or second roll.
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Comments
melzerh
Jan 13, 2009
 Its ok but you won't always beat the odds. 
Malika123
Jan 15, 2009
 It's nice. 
here2
Jan 22, 2009
 very nice and thanks for the full explanation. 
mumbojumbo1234
Nov 15, 2009
 First part was simple enough but couldn't follow the chance of winning playing optimally. I get .58.. 
BOBdaMAN
May 16, 2010
 This is way to confusing. this is me, reading the answer :p 
opqpop
Sep 28, 2010
 There is an error.
The probability you win when you roll second time should be 575/1296. There are two arithmetic errors, when you do 4*6 and 4*26. You can also get 575/1296 from (12[(1/36)^2(2/36)^2  ...  (5/36)^2]  6/36) / 2 via symmetry. The probability you win is [1  P(both of you roll same sum)] / 2.
Subsequently, the error carries over to answer for the bonus question. 398/1926 should be 400/1926 and 12159 should be 12075.
Luckily, the final answer still comes out to .57 if you round.
This is an interesting question because at first glance, EV(sum of 2 rolls) = 7, and since you want the highest sum, you'd think one would be indifferent to rolling again, but this turns out to be false. I'm having trouble figuring out why this is so. Anybody want to try and give a good explanation? 
opqpop
Sep 28, 2010
 I suppose if one is familiar w/ the game where you roll a dice and win the number of dollars equal to your roll, or can choose to roll again and win the number of dollars equal to your new roll, one's intuition would be more likely to trick him here.
In that problem, we know EV(roll) = 3.5, so you roll again if you get 3 or lower because 3 < 3.5. Here, we should not be concerned with EV, but instead the probability of winning, which is why we can't use the same approach. 
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