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.
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
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.
(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.
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 . So the probability of getting at least one 6 is
.
That's quite cool!
The probability of at least one 6 is
or
which is
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
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.