Problem of the week #71 - August 5th, 2013

[Jameson]

This problem comes from a game I saw in a restaurant while visiting Montana recently.

A game costs 50 cents to play. You are given 6 dice and roll them all at once. If you get at least 5 of the same value, you win the jackpot. If you don't, your 50 cents is added to the jackpot. How big should the jackpot be in order for this to be a good bet?
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Remember to read the POTW submission guidelines to find out how to submit your answers!

 

[Jameson]

Congratulations to the following members for their correct solutions:

1) MarkFL

Solution (from MarkFL):

Spoiler

We want the jackpot amount $J$ to be equal to or greater than that value for which the expected gain $E$ is zero. Thus, to find this amount, we may use:

\(\displaystyle E=-0.5\cdot P(\text{lose})+J\cdot P(\text{win})\ge0\)

(1) \(\displaystyle J\ge\frac{P(\text{lose})}{2\cdot P(\text{win})}\)

To find the probability of winning, we may compute the number of ways $N$ to win as follows

There are six ways to get a six of a kind, one for each rank.

For a five of a kind, there are six ranks that can match, five choices for the odd die and six places for the odd die to be in the array, hence:

\(\displaystyle N=6+6\cdot6\cdot5=6\cdot31\)

The total number of outcomes $T$ is:

\(\displaystyle T=6^6\)

Hence, we obtain:

\(\displaystyle P(\text{win})=\frac{N}{T}=\frac{6\cdot31}{6^6}= \frac{31}{6^5}\)

Using the fact that it is certain that we will either win or lose, we may write:

\(\displaystyle P(\text{lose})+P(\text{win})=1\)

\(\displaystyle P(\text{lose})=1-P(\text{win})=1-\frac{31}{6^5}=\frac{7745}{6^5}\)

Now, plugging the win/lose probabilities into (1), there results:

\(\displaystyle J\ge\frac{\frac{7745}{6^5}}{2\cdot\frac{31}{6^5}}=\frac{7745}{62}\approx124.92\)

Since the jackpot amount will only be a multiple of 50 cents, we then find the jackpot should be at least \$125.00 in order for the bet to statistically be a good one.

Extra comments:

Spoiler

What MarkFL did can also be represented by treating this as a binomial random variable. The probability of 5 1's is: $$\binom{6}{5}\left[\frac{1}{6}\right]^5 \left[\frac{5}{6} \right]^1$$

This should be multiplied by 6 to account for the values 1-6.

To calculate the probability of all 1's we can write it as: $$\binom{6}{6}\left[\frac{1}{6}\right]^6 \left[\frac{5}{6} \right]^0=\left[ \frac{1}{6} \right]^6$$

This should also be multiplied by 6 to account for the values 1-6.

This will lead to the same answer as MarkFL submitted and really is the same method, although presented slightly differently.