# Math Help - Fair Bet

1. ## Fair Bet

If someone offers you even odds on a 10 dollar bet that in six rolls of a die at least one 6 would be rolled. What would the payouts have to be in order to make the expected value of the equal to 0? Is this a good bet to take?

Thanks!

2. Originally Posted by MathNoob23
If someone offers you even odds on a 10 dollar bet that in six rolls of a die at least one 6 would be rolled. What would the payouts have to be in order to make the expected value of the equal to 0? Is this a good bet to take?

Thanks!
The probability of rolling at least one 6 among six consecutive die rolls is the probability of rolling a 6 first, plus the probability of rolling not a 6 and then a 6, plus the probability of rolling two not-6s and then a 6, etc.

$P = \frac{1}{6} + \left(\frac{5}{6}\right)\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^3\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^5\left(\frac{1}{6}\right)$

So basically the $10 is irrelevant because even money just means, if you lose then you lose your bet, and if you win then you gain twice your bet, which can be represented as -1 and 2. So we have $E(X) = (2)(P) + (-1)(1-P)$ where X is winnings (for a bet of$1, or 1 of any unit of currency you choose), a discrete random variable.

If the number you obtain for E(X) is greater than 0, then this is a good bet.

In order to find what the payout would have to be to get E(X) = 0, you must set E(X) = 0 and solve the following equation for m.

$E(X) = (m)(P) + (-1)(1-P)$

(I just chose the letter m arbitrarily.)

I usually work with probabilities rather than odds. I'll leave it to you to express m as payout. Even money is 2 for 1, or 1 to 1. So you'll have m for 1, but it seems you should convert this to something to 1, not for 1. If I were a gambler maybe I'd see the point of using odds; probabilities seem so much easier to work with.

Edit: @awkward: as soon as I read "Here is a short-cut" I realized the much easier way. Thanks.

3. Here is a short-cut to computing the probability of getting at least one 6 in 6 rolls, which should give you the same answer as the previously posted method without as much computation.

Consider the complementary event, i.e. rolling 6 times without getting any 6's. This happens with probability $(5/6)^6$. So the probability of getting at least one 6 is

$1 - (5/6)^6$.

4. Thanks guys! You two saved my life!

5. Originally Posted by undefined
The probability of rolling at least one 6 among six consecutive die rolls is the probability of rolling a 6 first, plus the probability or rolling not a 6 and then a 6, plus the probability or rolling two not-6s and then a 6, etc.

$P = \frac{1}{6} + \left(\frac{5}{6}\right)\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^3\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^5\left(\frac{1}{6}\right)$
That's quite cool!

The probability of at least one 6 is $1-P(no\ 6)$

or $P(one\ 6)+P(two\ 6's)+P(three\ 6's)+P(four\ 6's)+P(five\ 6's)+P(six\ 6's)$

which is $\binom{6}{1}\frac{1}{6}\left(\frac{5}{6}\right)^5+ \binom{6}{2}\left(\frac{1}{6}\right)^2\left(\frac{ 5}{6}\right)^4+.........$

Undefined's method calculates the probability of the first being a 6,
followed by no 6's, one 6, two 6's, three 6's, four 6's, five 6's.

That's the initial $\frac{1}{6}$

His second term in the sum calculates the probability of the first not being a 6, the 2nd being a 6 and the remainder can be any of the six values.
Hence it calculates probabilities of 1 to 5 sixes that do not have a 6 in the first position.

The remaining terms calculate the other sixes not already counted.

Anyway, the bet could be worth taking, since the probability of 6 coming up
is > 0.5.