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You have a method rand7 that generates a random integer from 1 to 7. Use it to write a method rand5 that generates a random integer from 1 to 5.
**

rand7 returns each integer with equal probability. rand5 must also return each integer with equal probability.

Your first thought might be to simply take the result of rand7 and take a modulus:

def rand5
rand7 % 5 + 1
end

**However, this won't give an equal probability for each possible result**. We can write out each possible result from rand7 (each of which is equally probable, per the problem statement) and see that some results for rand5 are more likely because they are caused by more results from rand7:

rand7 | rand5 |
---|---|

1 | 2 |

2 | 3 |

3 | 4 |

4 | 5 |

5 | 1 |

6 | 2 |

7 | 3 |

So we see that there are two ways to get 2 and 3, but only one way to get 1, 4, or 5. This makes 2 and 3 twice as likely as the others.

What about calling rand7 five times, summing up the result, and then taking the modulus?

This is *really close* to uniform, but not quite. Since we're
calling rand7 five times, there are
7^5 = 16,807 possible results. That's not
divisible by five, so some outcomes must be more likely than
others. (If you're curious, 1 is the result 3,357 times; 2 and 5
are the result 3,360 times each; and 3 and 4 are the result 3,365
times.)

In fact, no matter how many times we run rand7, we'll never get a number of outcomes that's divisible by five.

The answer takes worst-case infinite time. However, we can get away with worst-case space. **Does your answer have a non-constant space cost?** If you're using recursion (and your language doesn't have tail-call optimization), you're potentially incurring a worst-case infinite space cost in the call stack. But replacing your recursion with a loop avoids this.

rand5 must return each integer with equal probability, but we don't need to make any guarantees about its runtime...

In fact, the solution has a small possibility of *never* returning...

We simply "re-roll" whenever we get a number greater than 5.

def rand5
result = 7 # arbitrarily large
while result > 5
result = rand7
end
result
end

So each integer 1,2,3,4, or 5 has a probability \frac{1}{7} of appearing at each roll.

Worst-case time (we might keep re-rolling forever) and space.

Note that if we weren't worried about the potential space cost (nor the potential stack overflow) of recursion, we could use an arguably-more-readable recursive approach with space cost:

def rand5
result = rand7
if result <= 5
result
else
rand5
end
end

This kind of math is generally outside the scope of a coding interview, but: if you know a bit of number theory you can *prove* that there exists no solution which is guaranteed to terminate. Hint: it follows from the fundamental theorem of arithmetic.

Making sure each possible result has *the same probability* is a big part of what makes this one tricky.

If you're ever struggling with the math to figure something like that out, don't be afraid to *just count*. As in, write out all the possible results from rand7, and label each one with its final outcome for rand5. Then count up how many ways there are to make each final outcome. If the counts aren't the same, the method isn't uniformly random.

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