Friday Puzzler -- Three Rolls

Re: Friday Puzzler -- Three Rolls

JohnV

100 with my luck.

Re: Friday Puzzler -- Three Rolls

Alex Y

Kept thinking you could beat that 5, did you? ;-)

Re: Friday Puzzler -- Three Rolls

Larry Barrett

On any one roll of the dice your expected score is 3.5 (sum of the digits 1-6 divided by 6) and your payoff would be $350.

To improve your payoff you might try a strategy like roll the dice and quit if your score is a 5 or 6; if not, roll again and quit if your score is a 4,5,or 6; if not, roll a third time and take whatever you get.

This strategy would yield 5.5(1/3) + 5(2/3x1/2) + 3.5(2/3x1/2) =

1.83+1.67+1.17=4.67 (x100) = $467.

This is a better strategy than the simple one that has an expected payoff of $350, but there are many variations. I don't know how to find the best strategy without trying them all.

Right Answer

Alex Y

Work backwards to show yours is the optimal strategy.

And a similar-sounding problem *LINK*

Alex Y

I say similar sounding, since I do not believe it can be solved in the same way. I certainly couldn't. I'll give a link to the answer, so feel free to go there, or try it on your own if you'd like.

In this case, we have a 20-sided die. You are paid the amount you roll. Or you can pay $1 and re-roll. There is no limit on the number of times you can pay $1 for an additional roll.

What is your strategy and expected gain?


d20 Stopping Puzzle

Re: And a similar-sounding problem

Larry Barrett

I thought about this before looking at the link and my simple strategies yielded expected returns barely better than just rolling the d20 die one time (10.5).

So thanks for the link. It was very interesting to read Nick's solution. Looking around a little, I think I saw some familiar 'what is the angle in this triangle' problems, and the link to the TED talk about passwords was also interesting.

Proof

Alex Y

Larry gave the correct strategy and expected value. To show that it is indeed the best strategy, work backwards:

If you reject the first two throws, there is no strategy to apply, you just have to take what you roll on the third try, and the expected value of that roll is 3.5 ($350 in the game as described, but let's look at the dots on the die and multiply at the end.)

If you reject the first throw, you do have a strategy for accepting or rejecting the second throw. Pretty clearly, if you roll a 4,5,or 6, you would want to keep that versus going for a third roll where your expected value is only 3.5. But if you roll a 1,2, or 3 on that second roll, taking a chance on a third roll is the best way to go. So the expected value on the second roll is

1/6 * 6 + 1/6 * 5 + 1/6 * 4 + 1/2 * 3.5 = 4.25.

Similar logic tells you that on the first roll, you should cash in a roll of 5 or 6, or go for the expected value of 4.25 if you roll a 1, 2, 3, or 4.

1/6 * 6 + 1/6 * 5 + 2/3 * 4.25 = 4 2/3