How Much Free Money Can You Win?

Fiddler

From Bowen Kerins comes a chance at free money:

A casino offers you $\$55$ worth of “free play vouchers.” You specifically receive three $\$10$ vouchers and one $\$25$ voucher.

You can play any or all vouchers on either side of an even-money game (think red vs. black in roulette, without those pesky green pockets) as many times as you want (or can). You keep the vouchers wagered on any winning bet and get a corresponding cash amount equal to the vouchers for the win. But you lose the vouchers wagered on any losing bet, with no cash award. Vouchers cannot be split into smaller amounts, and you can only wager vouchers (not cash).

What is the guaranteed minimum amount of money you can surely win, no matter how bad your luck? And what betting strategy always gets you at least that amount?

Hint: You can play vouchers on both sides of the even money game at the same time!

Solution

I first looked at a simple strategy, where you are always guaranteed at least $25.

• Bet $25 and $30.
  • Half the time you win $25.
  • Half the time you win $30.

I modified and extended this to guarantee at least $30.

• Bet $20 and $25.
  • Half the time you win $20.
    Bet $10 and $20.
    • Half the time you win $10.
      Total winnings = $30.
    • Half the time you win $20.
      Total winnings = $40.
  • Half the time you win $25.
    Bet $10 and $25.
    • Half the time you win $10.
      Total winnings = $35.
    • Half the time you win $25.
      Total winnings = $50.

This led me to the following decision tree, also guaranteeing at least $30.

• Bet $10 and $10.
  You always win $10.
  
  Bet $10 and $10 again.
  You always win $10.
  Total winnings = $20.
  
  Bet $10 and $25.
  • Half the time you win $10.
    Total winnings = $30.
  • Half the time you win $25.
    Total winnings = $45.

This led me to the following two decision trees, both guaranteeing at least $35.

• Bet $10 and $10.
  You always win $10.
  
  Bet $20 and $25.
  • Half the time you win $20.
    Total winnings = $30.
    Bet $10 and $10 again.
    You always win $10.
    Total winnings = $40.
  • Half the time you win $25.
    Total winnings = $35.

• Bet $20 and $25.
  • Half the time you win $20.
    Bet $10 and $10.
    You always win $10.
    Total winnings = $30.
    Bet $10 and $10 again.
    You always win $10.
    Total winnings = $40.
  • Half the time you win $25.
    Bet $10 and $25.
    • Half the time you win $10.
      Total winnings = $35.
    • Half the time you win $25.
      Total winnings = $50.

Answer

Your guaranteed minimum winnings is $\boxed{\$35}$.

Note, given $v$ voucher(s) totaling $D$ dollars, the lower and upper limits are:

• If $v = 1$, that is one voucher of $D$ dollars, the guaranteed minimum amount of money won is $0$.
• If $v = D$, that is $D$ vouchers of $1$ dollar each, the guaranteed minimum amount of money won could be as high as $D-1$, if $D$ is a power of $2$.

Extra Credit

From Bowen Kerins also comes some Extra Credit:

You have the same $\$55$ worth of vouchers from the casino in the same denominations. But this time, you’re not interested in guaranteed winnings. Instead, you set your betting strategy so that you will have at least a 50 percent chance of winning W dollars or more. As before, you cannot split vouchers and cannot wager cash.

What is the maximum possible value of W? In other words, what is the greatest amount of money you can have at least a 50 percent chance of winning from the outset, with an appropriate strategy? And what is that betting strategy?

Solution

I first looked at a simple strategy, where you will have a 50 percent chance of winning $55.

• Bet $0 and $55.
  • Half the time you win $0.
  • Half the time you win $55.

Extending led me to the following two decision trees,

• Bet $0 and $55.
  • Half the time you win $0.
  • Half the time you win $55.
    Bet $10 and $10.
    You always win $10.
    Total winnings = $65.
    Bet $20 and $25.
    • Half the time you win $20.
      Total winnings = $85.
      Bet $10 and $10 again.
      You always win $10.
      Total winnings = $95.
    • Half the time you win $25.
      Total winnings = $90.

• Bet $0 and $55.
  • Half the time you win $0.
  • Half the time you win $55.
    Bet $20 and $25.
    • Half the time you win $20.
      Total winnings = $75.
      Bet $10 and $10.
      You always win $10.
      Total winnings = $85.
      Bet $10 and $10 again.
      You always win $10.
      Total winnings = $95.
    • Half the time you win $25.
      Total winnings = $80.
      Bet $10 and $25.
      • Half the time you win $10.
        Total winnings = $90.
      • Half the time you win $25.
        Total winnings = $105.

You always win at least $90!

Answer

The greatest amount of money you can have at least a 50 percent chance of winning is $\boxed{\$90}$.

Again, given $v$ voucher(s) totaling $D$ dollars, the lower and upper limits are:

• If $v = 1$, that is one voucher of $D$ dollars, the greatest amount of money you can have at least a 50 percent chance of winning is $D$.
• If $v = D$, that is $D$ vouchers of $1$ dollar each, the greatest amount of money you can have at least a 50 percent chance of winning could be as high as $2D-1$, if $D$ is a power of $2$.

It is best to bet all vouchers as the first bet. The sum of $D$ and the answer from Fiddler yields the answer to the Extra Credit, for any $(v, D)$.

Rohan Lewis

2025.08.18