This easy puzzle can be solve using a mathematical formula,trial and error method and reverse calculation whichever is suitable for the user.

There are three baccarat tables that we are going to use here to make the puzzle visible in the mind and easy to remember but you can set your own stuffs to refer to.

Here is the deal:

You will enter the first baccarat table and need to pay $10.00,when you bet,your money will be doubled and you need to exit the first table and pay for $10.00.

You will proceed to next table and need to pay $10.00 again,when you bet,your money will doubled and you need to exit the second table and pay for $10.00.

You will proceed to third table and need to pay $10.00 again,when you bet,your money will doubled and you need to exit the third table and pay for $10.00.

This time,after paying $10.00 on the third table during exit,you will have no money left.

Question: How much money should you bring so that after the third table,you have no money left?

Solutions:

Most people use the trial and error method,they think a money more than $10.00 first then start the computations.

Example:Start with $15.00

• 15-10=5
• 5*2=10
• 10-10=0

So they cannot enter on the second table,that means need to increase the $15.00 until they got the correct results.

Some mathematicians use the reverse computations,they started at the last event upfront.

Example:Inside the third table,you need at least $5.00,so when it is doubled ( $10.00 ) and when you pay on exit you have no money.That means after the second table,your money should be $15.00.

The third one is using a formula like this;

[([([(x-10)*2]-10-10)*2]-10-10)*2]-10=0