Friday Puzzler -- Dice game

6/23/12, 2:19 AM #1

Alex Y
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Friday Puzzler -- Dice game

Alex Y

Actually, it is a "die game", since you only use one.

A fair die with six sides, numbered 1-6.

At the first turn, you pay a penny to play, and you get a payoff of $1 times the number you roll.

The next turn, you pay $0.02 to play, and again get a payoff of $1 times the number you roll

Each turn, the cost of playing goes up by $0.01, while the payoff is $1 times what you roll.

How many times should you play to maximize your expected winnings, and how much do you expect to win following the optimal strategy?

Re: Friday Puzzler -- Dice game

6/23/12, 4:59 PM #2

Larry Barrett
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Re: Friday Puzzler -- Dice game

Larry Barrett

If you follow the optimum strategy, you should expect to win a little over $600.

Re: Friday Puzzler -- Dice game

6/24/12, 11:00 PM #3

Larry Barrett
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Re: Friday Puzzler -- Dice game

Larry Barrett

Since Alex has not responded, my answer may be wrong.

Here is my thinking:

The expected result of any single roll of the die is $1x3.5 = $3.50, since 3.5 = the average of the 1 thru 6 possible roll outcomes.

So the Expected return (expected result - cost of playing) of the nth turn is

$3.50 - .01n. Thus, the expected return is positive for the first 349 turns, is 0 for the 350 turn, and is negative after that.

The optimum strategy, therefore, is to play for 349 times and then quit.

The average return for each of the 349 turns is $1.75, so the total expected return is 349 x $1.75 = $611.

Re: Friday Puzzler -- Dice game

6/25/12, 1:26 AM #4

Alex Y
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Alex Y

Sorry for the non-response--just had not been online.

Your reasoning and answer are spot-on.

This is one of those problems that intimidate me, although as you pointed out this one is pretty straight-forward.

Friday Puzzler -- Dice game

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