This math trick is called Hummer’s Poker Chip Trick.

The trick uses six poker chips with a number on both sides – the table below shows the values on each for the poker chips. Note 5 is on two different chips.

The sides can be denoted by two different colors such as red/green or bold/regular font or some other method as long as you can determine the side.

In his head the magician calculates:

S = sum of exposed chips; 22 in this case (0 + 9 + 8 + 5)

Uses this formula to calculate the sum of the 3 hidden chips:

(10*R + 15) – S = (10*2 + 15) – 22 = 35 – 22 = 13 which is the sum of the hidden chips (0 + 7 + 6)

**Analysis of mathematics behind the trick**

**Examples:**(underlined number indicates it is hidden).

- Initial – starting arrangement of the chips.
- Flip 3 – Spectator flips over any 3 chips.
- Flip initial Red – Magician asks for any chip which remembered were red in the initial arrangement to be turned over.
- Flip & Hide 3 – spectator flips over and hides any 3 chips.

Initial | Flip 3 | Flip Initial Red | Flip & Hide 3 | ||||||||
---|---|---|---|---|---|---|---|---|---|---|---|

0 | 1 | 2 | 5 | 1 | 2 | 5 | 1 | 2 | 0 | 1 | 2 |

3 | 4 | 5 | 8 | 9 | 5 | 8 | 9 | 5 | 3 | 4 | 5 |

5 | 4 | 8 | 10 | 4 | 3 | 10 | 4 | 8 | 10 | 9 | 3 |

0 | 6 | 2 | 0 | 1 | 2 | 0 | 6 | 2 | 0 | 6 | 7 |

10 | 5 | 7 | 10 | 0 | 2 | 5 | 5 | 7 | 5 | 0 | 2 |

4 | 3 | 6 | 4 | 8 | 6 | 4 | 8 | 1 | 4 | 3 | 1 |

10 | 5 | 7 | 10 | 0 | 2 | 5 | 5 | 7 | 5 | 5 | 7 |

4 | 3 | 6 | 4 | 8 | 6 | 4 | 8 | 1 | 9 | 3 | 6 |

10 | 5 | 7 | 10 | 5 | 7 | 5 | 0 | 2 | 10 | 5 | 7 |

4 | 3 | 6 | 9 | 8 | 1 | 9 | 8 | 6 | 9 | 8 | 6 |

In step 2 the spectator can only flip 3 chips. To understand the mathematics behind the formula we need to look at the relationship between Step 2 and Step 3 (there is a relationship between them).

Rules we can state:

In step 3 we can have the following scenarios and state of chip at the end of step 3:

- A Red chip in the step 1 and NOT flipped in step 2 by the spectator will be Black at end of step 3.
- A Red chip in the step 1 and flipped in step 2 by the spectator will be Red (max of 3 since only 3 chips can be flipped) at end of step 3.
- A Black chip in step 1 NOT flipped in step 2 will be Black at end of step 3.
- A Black in step 1 flipped in step 2 will be Red at end of step 3.

Therefore: The table below shows the number of possible Red and Black (color in initial arrangement of step 1) chips flipped in Step 2 and the resulting Red in step 3 (using the 4 rules we just listed):

Actions in Step 2 | 1 | 2 | 3 | 4 |
---|---|---|---|---|

# RED flipped in step 2 | 0 | 1 | 2 | 3 |

# Black flipped in step 2 | 3 | 2 | 1 | 0 |

Resulting # of Red in Step 3 | 3 | 3 | 3 | 3 |

In the table note in column labeled 2: If we flip 1 in step 2 which was Red in step 1 that means we flipped 2 Black (must flip 3 chips) and thus will have 2 Black at the end of step 3. See rules 2 and 4.

The other columns show all other possible scenarios.

At step 3, with 3 Red, the total for the six numbers is 30.

The logic for this is the following:

If we have only Black side chips showing the total of all of them is 15 (0 + 1 + 2 + 3 + 4 + 5). Since we know we have 3 Red chips showing and a Red chip is 5 more than its corresponding Black side we need to add 3*5 = 15 to the total for a grand total of 30 when we have 3 Red chips showing.

Last step will turn over 3 and hide them.

Look at number of Red showing, name it R.

Then 3 – R are number of Black(low) showing – we know this since there had to have been 3 Red at the end of step 3.

Since we started with 3 Red from step 3 and R are showing then 3 – R were turned over and hidden in the last step then we will have:

R Black are hidden (3 – R are number of Red turned over and thus become Black).

Thus we have R Red hidden (3 – (3-R) = 3 -3 + R = R.

So total Red showing in last step is R + R = 2R and sum of values of all chips turned up for last step is (hidden and showing):

15 + 2R*5 = 15 + 10R

And total of sum of values of the hidden chips is = 15 + 10R – (total value of chips showing).