Let's say you are dealing from a normal 52 card deck and you place two cards up: 3 and a Jack.
As a player, you are placing a bet as to whether the next card up will be in-between these two cards. If you are correct, you win even money. If an A, 2, Q or K comes up you lose your bet. If a 3 or Jack comes up you must pay double your bet.
So, you have 7 cards that are winners and 6 that are losers. But, two of the 6 make you pay double. So, my question is… is this a good bet? The math behind it would also be appreciated.
For just the J-3 scenario:
7 possible winning numbers, but 4 suits of each, so 28 total.
4 possible losing numbers, but 4 suits of each, so 16 total.
2 possible double losing numbers, but only 3 remaining suits of each, so 6 total.
This makes 50 possible outcomes.
The average outcome for a bet of $x$ is:
($x \cdot$$\frac{28}{50}$) - ($x \cdot$$\frac{16}{50}$) - ($2x \cdot$$\frac{6}{50}$)= $0x$
The law of large numbers suggests that over time the payout will approach the average of breaking even. Obviously, changing the J-3 start will influence the probabilities and payouts.