Bet a Skeptic

Your skeptical friend thinks you claim you break even too much playing Craps.

Bet him even money you can break even or win making exactly 5 bets of "any 7". He ought to go for it and you should clean up.

You have to bet big enough to cover your expected loss though, which never breaks even [g].

Haven't done a mathematical proof of this, but anybody want to bet me?

so in other words the bet is to roll at least one seven in 5 rolls which is:

1-(5/6)^5

which is about 60%.

Any seven pays 4 units. Therefore, to break even or win, you have to have at least 1 7 in five rolls.

The odds of not rolling a 7 on the first roll is 5/6.
The odds of not rolling a 7 by the fifth roll is (5/6)^5.

Let b be the bet value.

The expected value of the bet is: -b*(5/6)^5 + b(1 - (5/6)^5) = 1 - 2(5/6)^5 = .196245b

Q.E.D.

To finish the question, how big must b be to cover the any seven bet (a)?

The expected value of any seven where the bet size is "a" = 1/6*4-5/6 = -1/6a
The expected value of the bet (bet size = b) with your friend is .196245b but it resolves itself over five bets.

Therefore, to win, 5(-1/6)a + .196245b > 0.

and .196245b > 5/6a
.235494b > a
b > 4.246396a.

If "a" is $5, then a bet of $22 or more with your friend covers your losses.

$3 sevens pay $13 or $16 and down. So 4 1/3 not 4 units

If your friend says to make it four bets or six bets, you might suspect he's figured it out. You can't break even.

win one bet and lose 3: that's +4 for the win, other bets add up to -3
win one bet and lose 5: that's +4, -5