Best Answer bonanova, 20 May 2014 - 07:39 PM

**49 cards**that you might hold.

On average my three cards will divide those remaining 49 cards into equal intervals.

The interval below my lowest card will be on average 12.25 cards

The interval above my highest card will be on average 12.25 cards.

The interval between my lowest and highest card will be on average **24.5 cards**.

That interval size is the break-even point.

The "between" interval, being in each case an integer, will never be 24.5 cards.

Rather, it will be an integer * n* between 0 and 49, and

*will be symmetrically distributed.*

**n**The number of hands with interval = * n* will equal the number of hands with interval = 49-

*.*

**n**

This means that if I mindlessly bet **between **on every hand, my winnings and losses will cancel.

That is, I will break even. That strategy alone keeps you from winning.

But I can do better, of course.

One eighth of the time, the interval **less **than my lowest card will exceed 24.5 cards,

One eighth of the time, the interval **greater **than my highest card will exceed 24.5 cards.

**Three-fourths** of the time, my hand will contain an interval that **exceeds 24.5 cards**.

Three-fourths of the time, I will be favored to win my bet.

By simulation, the average number of cards in the largest interval in my hand is **30.36 cards**.

My average gross winnings for each hand is thus $20 x (30.36/49) = $12.39+

My average net winnings (after paying you $10) will thus be **$2.39+ per hand**.

When do we play?