# Token Drawing [All At Once (per player) vs Round Robin] -- Mathematician / Statistician needed for a question

In Paris Connection (PC), you go player by player drawing a starting number of trains, with each player drawing their entire complement all at once.

One of our players conjectured this draw method is 'unfair/biased' because the first player to draw has a higher probability of gaining a majority of a single color. [Let's set aside whether or not having a majority of a single color actually gives the player an advantage in PC.] Whereas it was my conjecture that while there's variance in the distribution drawn, in the long run, it'd be a wash overall amongst all the players... and in the process of writing this & formulating my thoughts on the matter further, that it's actually the last player who has the highest likelihood of an unbalanced draw.

If there are any Mathematicians / Statisticians out there who could weigh in (and not just speculate) I'd love to hear a scientific reason why one or the other is correct.

``````Simplified Example:
1 Bag Containing:
10 Tokens each of six (6) different colors (ROYGBV), for a total of 60 Tokens total
6 Players drawing six (6) tokens each, each player draws all six (6) tokens at once before passing the bag to the next player
``````

"In general", a player would be predisposed to drawing one of each color, leaving all the other players predisposed to drawing one of each color. Hence "in general", all players would have an even distribution.

However, given a player randomly draws more of a single color (as is bound to happen), then it would predispose the next player to also draw more of a single color, which in turn would ripple through the drawings of each player with each player having "more" of a color; hence negating a single player having "more" of a color while the others do not.

Note #1: I'm certainly not saying that one player cannot draw all six red tokens, while all other players have a more flat distribution of one of each color; after all in probability almost anything 'can' happen... but it doesn't make it likely or probable. What I'm saying is that drawing first doesn't inherently give you an increased chance of a single color majority.

In fact, I would almost argue the opposite, that the last player has the highest probability of a majority of a single color. Here's my thoughts... Player #1 has an equal chance of drawing all colors, lets say she draws R. Now on her second draw, she has a slightly higher chance of drawing any color but R; so she draws an O, etc. So let's say for sake of argument, she draws a flat distribution (ROYGBV). Now her twin sister, Player #2, when starting, has an even distribution, and manages to draw an even distribution as well (VBGYOR). But let's say their cousin, Player #3 - Miss Entropy, draws an uneven distribution (RRRRVV)... from this point forward, the remaining players to draw, are more and more predisposed to draw a non-flat distribution. So it would seem, and I'm not a Mathematician / Statistician, that the first player has the best chance for an even distribution; and with each player the chances worsen for an even distribution.

Note #2: As you'll see below, in PC, the number of tokens drawn is only an even multiple of the number of colors in the 5 player game... which I feel would even strengthen the effects of 'increasing Entropy' as each player draws, as only the first player would have a truly even distribution to draw from.

Finally, all that being said... let's say we do the other Round Robin method, where each player draws a single token, and then passes the bag each time. Does this necessarily increase the 'flatness' of each player's distribution? "Ideally" (ROYGBV). Or does it 'really' put the beginning players on the same level as the last player? Or does it not matter at all? For instance, in my only statics course I took, I remember something about "dependent events"... but I'm not sure if that applies?.?. and whether it's tokens, trains, or cards; does the method of drawing/dealing matter?

All of that being said, I unfortunately do not have the mathematical / statistical rigor to "prove" it one way or another, or some other thing I had not considered.

**

``````\\   \\   \\   \\
\\   \\   \\   \\
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Actual Paris Connection (PC) Example Details:**
1 Bag Containing:
31 Tokens each of six (6) different colors (ROYGBV), for a total of 186 Tokens total
X Players drawing y (Y) tokens each, each player draws all y (Y) tokens at once before passing the bag to the next player

X == Num Players
Y == Num Tokens Drawn

X     Y
3    10
4     8
5     6
6     5
``````
• Welcome to boardgames stackexchange. I think, the probability of getting an awesome answer increases if you post the question on a more math centered site. Commented May 16, 2017 at 5:05
• Hi! Great question. For future questions, be aware of stats.stackexchange.com and math.stackexchange.com if you want to reach probability experts.
– Stef
Commented May 15, 2023 at 20:16

No player has a greater chance of drawing an even or uneven distribution than any other.

One way of looking at it is to consider permutations of the tokens, where they are laid out in some sequence instead of jumbled in a bag or something. Then, if we shuffle the tokens up, so that they can be in any order, we can distribute the tokens, so that the first player gets the first Y tokens in the sequence, and the next player gets the next Y, and so on.

If you assume that any permutation has the same probability as any other (which is as valid as assuming that the bag is uniformly jumbled), then you can see that the player's position doesn't affect what tokens they are more or less likely to receive.

Your sampling sequence is an example of an exchangeable sampling sequence. Exchangeable sequences are not independent. The outcome of earlier samples does affect the outcome of later samples.

However the marginal distributions are identical. i.e. the probability of whether or not the first player gets a single color is identical to the probability the last player gets a single color.

Mathematicians/probabilists call this an urn problem. That should be sufficient to help you find a more detailed solution if this one is too high level for you.

Imagine that we drew the tokens face down. The first player draws their six tokens, but don't look at these tokens and don't show them to the other players. Then the second player also draws their six tokens without looking, etc, until all players have drawn their tokens, but no one has looked at any token.

Then one of your friends, let's call him Terry, complains that with this drawing method, the first player has an advantage.

So, the first player, Fred, offers Terry a choice: you can keep your tokens, or we can exchange. In case of an exchange, Terry takes all six tokens that were in Fred's hand, and Fred takes all six tokens that were in Terry's hand. Note that at this point, neither Fred nor Terry has looked at the tokens yet.

Do you believe that Terry should make the exchange? Well, he can, but it will have no impact whatsoever on the probabilities. Fred's hand and Terry's hand both follow the same probability distribution.

Now, you can look at the tokens, and it doesn't affect the prior probabilities. So the first player does not have an advantage, and neither does the last player.

However, the first player receive the information about their own hand faster. This might be what gives your friends the impressions that the game is not fair. Indeed, there will be an intermediate state in which the first player has already drawn their hand, and already knows whether it's a good hand or a bad hand; and the other players have not drawn their hand yet, so they don't know if they're about to get a good or a bad hand. The other players do have a bit of residual information; for instance, if the first player's hand contained a lot of red tokens, then the other players now have a decreased probability of getting red tokens.

Compare this with a game of Russian roulette for six players. There are six chambers in a revolver, and one of these chambers contains a bullet. There are six players. The chambers have been rolled uniformly at random; so each player has exactly one chance out of six to get the bullet (about 17%). However, after the first player has taken their turn, if they haven't been blown up, then there is an intermediate state where the first player has probability 0% of hitting the bullet, while the other players each have one chance out of 5 to get the bullet (20%). During this intermediate stage of the game, things can feel very unfair.