## Poll

16 votes (51.61%) | |||

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10 votes (32.25%) | |||

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**31 members have voted**

Quote:chevy

Amy wins 3 rock against scissors

Amy loses 6 paper against scissors

Amy wins 5 scissors against paper

Amy loses 6 scissors against rock

Total

Amy wins 8, Bob wins 12.

(Because no ties, Bob's 9 scissors must be against Amy's rock and paper (3+6=9). Leaving Amy's 11 scissors against Bob's rock and paper (6+5=11))

I agree!

Quote:Ace2Starting from a come out roll, what is the expected number of rolls for all six points to be won? The points can be won by various shooters

This one was asked in Ask the Wizard #331.

The answer is 8706865474775503638338329687/39730260732259873692189000 = Approximation: 219.1494672902

Suppose you play a game with a die where you roll and sum your rolls. You may stop any time, and the sum is your score. However, if your sum is ever a multiple of 10, your score is zero and the game is over.

What strategy will yield the greatest expected score?

Quote:GialmereSuppose you play a game with a die where you roll and sum your rolls. You may stop any time, and the sum is your score. However, if your sum is ever a multiple of 10, your score is zero and the game is over.

What strategy will yield the greatest expected score?

My first thought was, "Is there one? At any time, either the probability of losing is 1/6 (if the current score mod 10 is between 4 and 9 inclusive) or zero."

However, I did find one through simulation if the strategy is assumed to be, "Stop if your score is (some particular number) or higher."

I'd like to see a mathematical explanation for this answer - especially if the actual strategy is different (say, "Stop after X rolls if your score is Y or higher"):

The best strategy appears to be, stop if your score is 24 or higher. This barely beats out stopping at 34 or higher.

The method of working it our is to use recursion and work out the EVs, whether it's best to stand or hit (think Blackjack!).

For instance on 34 your EV is (35+36+37+38+39+0)/6 = 30 5/6 - so you should stop on 34.

However 39 is different because if you get, say, 41 you reroll, so it's value is 45.76 (this is because of the additional chances of going via 42 and 43). So the EV is (0+45.76+46.08+46.5+44+45)/6 = 37.89. It's a closer decision, but you still still stop.

In the 20's this effect does change your decision.

So you hit 21, 22, 23. Working back from 29 you hit 26-29, but stand on 24 and 25.

The net EV seems to be 13.217 (I'm sure someone can work out the really long fraction!)

Summary

Hit 23 or less.

Stand on 24 or 25.

Hit 26 to 33.

Stand on 34 or more.

Quote:charliepatrickI'm not sure how to solve this without resorting to a spreadsheet so here goes.

Initially consider the last digit of your current total. If you have 1, 2, 3 then you cannot lose. If you have 4,5,6,7,8,9 then the average score by rolling is the next six numbers but then ignoring the multiple of 10. There comes a point where the risk of losing what you've got means you should stop rolling.

The method of working it our is to use recursion and work out the EVs, whether it's best to stand or hit (think Blackjack!).

For instance on 34 your EV is (35+36+37+38+39+0)/6 = 30 5/6 - so you should stop on 34.

However 39 is different because if you get, say, 41 you reroll, so it's value is 45.76 (this is because of the additional chances of going via 42 and 43). So the EV is (0+45.76+46.08+46.5+44+45)/6 = 37.89. It's a closer decision, but you still still stop.

In the 20's this effect does change your decision.

So you hit 21, 22, 23. Working back from 29 you hit 26-29, but stand on 24 and 25.

The net EV seems to be 13.217 (I'm sure someone can work out the really long fraction!)

Summary

Hit 23 or less.

Stand on 24 or 25.

Hit 26 to 33.

Stand on 34 or more.

Correct!

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I lost my kids' college fund playing dice in Vegas.

I admit it was a crappy thing to do.

You’re a contestant on the hit new game show, “You Bet Your Fife.” On the show, a random real number (i.e., decimals are allowed) is chosen between 0 and 100. Your job is to guess a value that is less than this randomly chosen number. Your reward for winning is a novelty fife that is valued precisely at your guess. For example, if the number is 75 and you guess 5, you’d win a $5 fife, but if you’d guessed 60, you’d win a $60 fife. Meanwhile, a guess of 80 would win you nothing.

What number should you guess to maximize the average value of your fifing winnings?

My gut instinct tells me the answer should be 49.999999 (9 repeating), but I'm guessing that that is too easy an answer. Given that a real number is chosen, I'm assuming that some calculus is involved.

So if the number selected is "n" and our guess is "x", then I thought about I could solve this as the integral from 0 to n of (x) times (x/100) (value of the win times the probability of the win). Then taking the derivative of the function with respect to n to determine the maximum. That's not working. What am I missing?

Quote:Gialmere

Suppose you play a game with a die where you roll and sum your rolls. You may stop any time, and the sum is your score. However, if your sum is ever a multiple of 10, your score is zero and the game is over.

What strategy will yield the greatest expected score?

That die appears to have two 2's and two 3's and no five or six.

Now think that 49*51 is 50*50-1 (50-1)*(50+1). So the best value to pick is $50, as you have a 50% chance of winning (EV=$25).

Using calculus

EV=100x-x

^{2}

d(EV)/dx = 100 - 2x. This is zero when x=50.

Quote:charliepatrickIf you pick $75 then there's a 25% chance of winning. Similarly if you pick $25, there's a 75% chance of winning. Generally speaking if you pick $X then there's a (100-X)% chance of winning, and the EV is the same as picking $(100-X) with X% of winning.

Now think that 49*51 is 50*50-1 (50-1)*(50+1). So the best value to pick is $50, as you have a 50% chance of winning (EV=$25).

Using calculus

EV=100x-x^{2}

d(EV)/dx = 100 - 2x. This is zero when x=50.

Thanks... I set up my integrand incorrectly.