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)
- 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|
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).