Interesting Puzzle

I am glad that I rushed down to take part in the Brewerkz pub quiz after ringing Jake up mainly because we managed to win, with a team of just four people. That worked out to a cool 80 over dollars per person on the night, which was quite wonderful. I also won a free beer to boot after betting with Jake that Yokohama was the second largest city in terms of population in Japan after Tokyo, though Jake rightly pointed out that much of that depended on things like how you defined city and metropolitan limits.

I ended up having a long chat with the quiz masters afterwards, and it turned out that one of them was a banker, dealing with derivitives and thus analytical by nature. His background in gambling actually helped a great deal, at least in terms of securing him his job, because some of the interview questions were actually based on logic and probability. He challenged us with two specific puzzles which I only fully worked out afterwards, and it makes enormous intuitive sense, I thought I would share them.

The first puzzle involved a game of chance with a die. Let us say that you will pay in cash the amount equivalent to the die roll (e.g. if it lands on a six, the person wins six dollars). What price should you make a person pay in order to compete in the game? The answer is derived by using simple math. First, there is an equal probability of each outcome (the die landing on any number from one to six). So the average payout can be calculated by taking the total payout in each individual outcome (i.e. $1 in the case of die roll 1, $2, in the case of die roll 2) and dividing it by 6. So what you get is $1+$2+$3+$4+$5+$6= $21 in total, dividing by 6 gives you $3.50. So you should charge at a very minimum $3.50 for a roll.

He then added a more challenging twist. Say you give the gambler a chance at a second roll of the die. The number that then comes up on the second roll is the payout will be given (i.e. if you roll 3 on the first go and 2 on the second, the payout is $2). What price would you set for a person to compete in this 2 roll game? Would it be the same as in the first case, more expensive, or less expensive?

The answer of course is that you have to set it more expensive. Calculating the exact amount is a matter of logically predicting the behaviour of the gambler. To begin with, the gambler will not re-roll unless he has a even or better chance of improving his payout. Because there is always a risk that he will throw lower the second time around. So the gambler will probably hold if he rolls 4,5 or 6 the first time, and re-roll if he rolls 1,2,3. If he re-rolls the scenario exactly mirrors the first one above. However, given that the gambler has the option of holding on a high number and improving his payout on a low one, it logically means that the price has to be set higher.

How to calculate this? Very simple, first deal with the first roll which has a payout of 4+5+6 divided by 3 which is 5 - the gambler will re-roll otherwise. As calculated above, the average payout on a second roll (should the gambler roll 1,2 or 3) is 3.5. So the price set for a gambler to take part in this second game is the average of these two which is 8.5 divided by 2 which is 4.25. So you should make someone pay $4.25 in order to take part in the double roll game.