### The Puzzle

Here’s an interesting puzzle: The hostess of a baby shower devises a game in which sixteen tokens are placed in an opaque bag. The tokens are all either pink or blue, and they’re otherwise identical in shape, weight, texture, and so on. Two guests are chosen. The first guest chooses a single token from the bag and holds onto it. The second guest then chooses a single token from the bag. If the tokens are the same color, then the first guest wins; otherwise, the second guest wins. The hostess wants to make sure that each guest has an equal chance of winning. How many of each color token should the hostess put in the bag?

### Interesting Part #1: The “immediately obvious” answer is wrong

The most obvious answer is, 8 of each color. This appears to give each player a 50/50 chance of drawing each color, which would lead to a 50/50 overall chance of winning.

Except… it doesn’t. The key part of this that makes it “problematic” in a pedagogical sense is that the first guest (let’s call her Amy, and yes, I’ll make the sexist assumption that the guests are all female) doesn’t put the token back in the bag. Once she takes a token, there are fifteen left in the bag, and the second guest (let’s call her Beth) has eight chances in fifteen to pull the other color. The game has a slight advantage to the house.

This is somewhat the opposite of the Gambler’s Fallacy. The Gambler’s Fallacy is the belief that previous random events change the likelihood of current random events. For instance, if I flip a coin five times and get heads each time, I’ll most likely be inclined to expect the next flip to be tails. But while the odds of flipping six heads in a row with a properly weighted coin are indeed quite low (1/64), the odds of flipping a head after having flipped a head five times in a row is the same as that of flipping a tail: 1/2.

Previous random events don’t generally affect the likelihood of current random events when the events are independent. In the Baby Shower game, though, the events are ** not** independent. When Amy pulls out a token, she changes the poll from which Beth will be choosing.

Let’s look at it another way. Instead of pulling tokens, both players silently think of a number from 1 to 4, with Amy winning if the total is even. Then there’s a 50/50 chance that the sum will be even. The table below shows all of the possibilities, with the even sums in bold and the odd ones in italics:

1 | 2 | 3 | 4 | |

1 | 2 |
3 |
4 |
5 |

2 | 3 |
4 |
5 |
6 |

3 | 4 |
5 |
6 |
7 |

4 | 5 |
6 |
7 |
8 |

There are eight odds and eight evens, so it’s fair. But if we restrict the rules so that the two players must guess different numbers, this removes certain possibilities.

1 | 2 | 3 | 4 | |

1 | — | 3 |
4 |
5 |

2 | 3 |
— | 5 |
6 |

3 | 4 |
5 |
— | 7 |

4 | 5 |
6 |
7 |
— |

All of the ones we removed were even, leaving us with eight odds and four evens. It’s no longer fair; Beth has a 2/3 chance of winning.

### Interesting Part #2: There is an answer; actually, two symmetrical answers

The most obvious assumption of fairness is that the game has to have 50% odds for each player before either player draws. If we use this assumption, then there might be a condition in which Amy is favored after her draw, and another where Beth is favored after Amy’s draw. If these conditions balance each other out completely, it’s a fair game.

Less sophisticated solvers might want to use a trial-and-error method; more sophisticated solvers may want to try something more algebraic. Either way, though, it should be obvious that if we have more pinks than blues, Amy has a higher chance of pulling a pink, but in so doing decreases the odds that Beth will do so.

For instance, imagine there are nine pinks and seven blues. Let’s shorten this in a mathematically friendly way: There are 9P & 7B. Amy could pull either a P or a B. Her odds of pulling a P are 9/16; her odds of pulling a B are 7/16.

In the first scenario, Beth pulls from a pool of 8P & 7B. Her odds of pulling a P are 8/15; her odds of pulling a B are 7/15. In the second scenario, Beth pulls from a pool of 9P & 6B, with odds of P = 9/15 and B = 6/15.

This leads to four possibilities, each with the following odds of happening:

Amy: P, Beth: P. \( \frac{9}{16} \cdot \frac{8}{15} = 0.3000 \)

Amy: P, Beth: B. \( \frac{9}{16} \cdot \frac{7}{15} = 0.2625 \)

Amy: B, Beth: P. \( \frac{7}{16} \cdot \frac{9}{15} = 0.2625 \)

Amy: B, Beth: B. \( \frac{7}{16} \cdot \frac{6}{15} = 0.1750 \)

Amy wins in the first and last scenarios, giving her a 47.5% chance of winning. slightly higher than the 8/15 or 43.3% chance she had when there were eight of each in the bag.

Imagine there are 10P & 6B. This leads to these four possibilities:

Amy: P, Beth: P. \( \frac{10}{16} \cdot \frac{9}{15} = 0.3750 \)

Amy: P, Beth: B. \( \frac{10}{16} \cdot \frac{6}{15} = 0.2500 \)

Amy: B, Beth: P. \( \frac{6}{16} \cdot \frac{10}{15} = 0.2500 \)

Amy: B, Beth: B. \( \frac{6}{16} \cdot \frac{5}{15} = 0.1250 \)

Amy wins in the first and last scenarios, giving her a 50% chance of winning. This is the point of balance. We could of course reverse the colors, so that there are ten blues and six pinks, and get the same answer.

### Generalizing the solution

Trial-and-error gives us the answer in this case, but what if there are 100 tokens in the bag? We wouldn’t want to use trial and error to find if there’s a solution, and if so, what. It would be nice to have a formula, both for the specific case of 16 tokens and for the more general case of any number of tokens.

Let’s start with the general case. Let \(p\) be the number of pink tokens and \(b\) be the number of blue tokens. How would we express the first scenario in terms of \(p\) and \(b\)? That would be \(\frac{p}{p+b} \cdot \frac{p-1}{p+b-1}\). The last scenario would be \(\frac{b}{p+b} \cdot \frac{b-1}{p+b-1}\). Amy’s overall odds of winning, then, are \[\frac{p}{p+b} \cdot \frac{p-1}{p+b-1} + \frac{b}{p+b} \cdot \frac{b-1}{p+b-1} \\ = \frac{p\cdot(p-1) + b\cdot(b-1)}{(p+b)(p+b-1)} \\ = \frac{p^2 – p + b^2 – b}{p^2 – p + b^2 – b + 2pb}\]

This is more complicated than the trial-and-error method, and we’re not done yet. It has the advantage, though, of producing a generalized solution. Note, importantly, that the numerator and denominator of the last line are nearly identical. This expression represents a fair game when it equals 1/2 (that is, 50%), so let’s solve and see what emerges*: \[\frac{p^2 – p + b^2 – b}{p^2 – p + b^2 – b + 2pb} = \frac{1}{2} \\ \implies 2(p^2 – p + b^2 – b) = (p^2 – p + b^2 – b) + 2pb \\ \implies p^2 – p + b^2 – b = 2pb \\ \implies p^2 – p + b^2 – b + 2pb = 4pb \]

This last line is the expanded form of our denominator, and so we can quickly factor back to the original: \( (p+b)(p+b-1) = 4pb \).

So, let’s go back to the original problem. There are 16 tokens, so we are looking for a case where \( 16 \cdot 15 = 4pb \). In that case, \( 4pb = 240 \implies pb = 60 \). Finding two factors of 60 that add to 16 is fairly trivial for most Algebra II students: 6 and 10.

If there are 100 tokens, we are looking for a case where \( 4pb = 100 \cdot 99 \), that is, \( pb = 2475 \); \( p + b = 100 \). This is a bit more challenging, but factoring out we find 45 and 55.

Note that not all token counts have solutions. Indeed, astute readers may notice that 6, 10, 45, and 55 belong to a particular series: {1, 3, 6, 10, 15, 21, 28, 36, 45, 55, …}. Each element *i _{n}* in the series is the sum of the preceding

*i*and its index; said another way, each element is \(\sum_{x=1}^n x \). Add any two consecutive elements from this series and you’ll have a fair game: 4, 9, 16, 25, 36, 49, 64, 81, 100… hey, wait a second, that looks familiar!

_{n-1}To summarize: If you start with a number of tokens that’s a square number greater than 1, it’s possible to create a fair game. Otherwise, it isn’t.

### Interesting part #3: Even though the game’s odds are even at the very beginning, one player has a clear advantage after the first draw.

To take the smallest case, if there are four tokens in the bag, the mix is 1P and 3B (or vice versa). If Amy pulls the pink token, the game’s over and she’s lost (since all the remaining tokens are blue). If she pulls a blue token, she has a 2/3 chance of winning. Either way, the odds are strongly skewed. (To check that it’s fair: Amy has a 1/4 · 0/3 = 0/2 chance of winning via a pink token and a 3/4 · 2/3 = 2/4 = 1/2 chance of winning via a blue token, for an overall odds of 1/2. Perfect!)

People who play craps might be familiar with a similar situation. While craps rolls are independent (as opposed to token draws in this game), there are certain rolls that win off the bat and other rolls that require multiple rounds. This is because the goal was to make the cumulative odds of winning craps close to 50% (with, naturally, a slight advantage to the house). Because certain dice rolls have higher likelihoods than others, the rules were devised in such a way as to balance that over the course of multiple rounds.

* Thanks to Sakari Maaranen for inspiring this portion of the analysis.