# Some probobability questions for mathniks

My family has decided to take up Bunco as a “party game” and we played our first game this weekend. Great fun. But it got me to thinking of probability.

One of the suggested prizes on the prize sheet was \$8 for whoever got an equal number of wins and losses. My dad and I thought “that will be a tough one to get” so we came up with a different way to distribute the cash if no one had an equal number of wins and losses. To our surprise, FOUR people got 12 wins and 12 losses.

To make is simple: the game has 24 total rounds. 12 people are playing. In each round you either get a W (win) or an L (loss). In each round, 6 people will get W’s and 6 will get L’s. What is the probability that you (a single player) will end up with 12 W’s and 12 L’s?

In our game, 4/12 people got 12 wins ans 12 losses. Does that mean their chances were 1/3?

Here’s an even trickier one, from the same game. I’m not sure if I can even describe the problem correctly:

You are randomly placed at 1 of 4 seats, at 1 of 3 tables (so essentially, 1 of 12 seats) at the beginning. This gives you a random partner. After each round, you and your partner go together to another table but you are no longer partners (you switch seats). So for example…

Round 1
A1&A2 - Wins (Table 1)
B1&B2 - Loses (Table 1)
C1&C2 - Wins (Table 2)
D1&D2 - Loses (Table 2)
E1&E2 - Wins (Table 3)
F1&F2 - Loses (Table 3)

Round 2
A1&C1 (Table 1)
A2&C2 (Table 1)
D1&E1 (Table 2)
D2&E2 (Table 2)
B1&F1 (Table 3)
B2&F2 (Table 3)

You keep mixing it up from there. Wether you win or lose determines which table you go to next, and you never keep the same partner 2 times in a row. Also in the above example, in Round 2 A1 could have ended up with C1 or C2, A2 with C1 or C2, D1 with E1 or E2, etc. It’s sort of random which person you end up with in the next round.

Is there some sort of probability formula that could apply to this situation to see if there is a probability of NEVER ending up with a specific person as a partner? I recall that in our game Saturday I only ended up with my best friend once, in the very last round. I don’t think I ended up with my mom at all.

I really suck at probability so I am not even sure if this is a real type of situation that can be written out in a forumula. But, it seemed interesting to me.

It’s simpler than you think. Forget the number of players. You’re asking what the probability of getting exactly 12 wins and twelve losses in 24 tries is, when the probabability of winning and losing are equal. It’s the same as asking what’s the probability of exactly 12 heads and 12 tails are in 24 flips, regardless of their order. So it’s the probability of twelve heads in 24 flips. That’s the number of [possibilities of having 12 heads arranged in a set of 24, divided by the possible number of results. The former is (12 24) (Imagine that one two lines – the binomial term for 12 items arranged in a set of 24), divided by 2 raised to the 24th power (since there are two possibilituies for each flip). I get a bit under 1/6 for the overall

Of course, you won’t get that rsult all the time. Your results will have a distribution. A binomial distribution, in fact, with its peak value right at – 12 heads and twelve losses. Turns out that it’s the most likely result. Pick any other distribution and it’d be less likely o show up.

What reason do we have to think the odds of winning and loosing are equal?

Assuming that you’re equally likely to win or lose in a given round (which seems necessary), and that the current round doesn’t depend on any previous rounds, the probability of getting exactly 12 wins is [sub]24[/sub]C[sub]12[/sub]/2[sup]24[/sup], or 2704156/16777216 (which is just slightly less than 1/6).

That’s your best estimate for the probability in lieu of further information. Now, it’s possible that one of my assumptions is wrong–after all, I don’t know anything about the game. Or it’s possible that you had an event of probability .09 or so occur.

They have to be – it’s stated that in each game you have 6 winners and six losers out of 12 people. That’s a 500/50 chance either way. If there were a possibility of unequal winners and losers the probabilities might get tweaked, but this is clearly and truly a zero sum game, with just as many winners as losers. The OP says nothing about possibility of ties games, but implies that they don’t exist.

Since I am so slow at math I am going to have to slowly read your replies so far a few times to grasp them (plus the odd coding in ultrafilter’s post).

But yes…there are NO TIES and every round (24 in all) has exactly 6 winners and 6 losers.

Once the custom tags work again, my post will make a whole lot more sense.

Of course, I’m assuming that all the players are unaware of your award system, and about equal in ability. Things get weird if they vary greatly in their ability and know in advance that you’re paying off on equal wins and losses. Your system rewards mediocrity – neither really good nor really bad players profit. Knowing this, your best players will start throwing games so that they end up with equal losses and wins. Your worst players, on the other hand, can’t play better and therefore win more games. Of course, your best players have to lose to somebody, but unless they work it out so that there’s an equal number of wins and losses for everyone, your poorest players are gonna get screwed. So if your players know the score and don’t act like automatons, the whole probability argument is off.

It’s also possible that strategy and ability don’t make that much difference in the outcome. If the game consists of repeated plays of Chutes and Ladders or Candyland (probably not), there’s nothing but luck involved.

It’s a dice game with absolutely no skill involved. The only human interference could come from playing too slowly, and that only works if you’re at table 2 or 3 and way ahead of the other team at your table. And you and your teammate would have to both roll slowly.

There’s no way to be good or bad at this game. Just lucky or unlucky.

And there are 5 other prizes, not just the equal win/loss prize.

As for the second question, it may depend on how the table advancement works. If partner/table selection is sufficiently close to random over the whole night, then you could estimate it pretty simply.

Take one specific person (you haven’t yet partnered with). After each round, you have a 1/10 chance for them to be your partner, and a 9/10 chance that they won’t be (10/11 at the start of the game). Then the probability that you never play them is (10/11)*(9/10)^23, or about 8%. So the probability that you do play any one particular person is around 92%.

I also did a quick calculation (so it might have been off) for the number of partners you end up with. For this I ignored some of the constraints, instead assuming the partnering to be random each time.[sup]*[/sup]

(Rounding errors have occurred, so this doesn’t quite equal 1)
If P(x) is the probability that you have x partners, then after 24 rounds :
P(11) = 0.26
P(10) = 0.44
P(09) = 0.24
P(08) = 0.06
P(07) = 0.01
P(6 or less) is negligibly close to 0.

As you can see, the most likely outcome is that you play all but one person.
[sup]*[/sup]I suspect there’s an easier method, but I modeled this as a Markov process. With i= number of partners you’ve had, the matrix has diagonal a[sub]ii[/sub] = i/11, and the superdiagonal has 1-a[sub]ii[/sub] in the ith row. This was then raised to the 23rd power (This ignores the 0->1 transition since that always occurs on game 1; I suppose I could have ignored 1->2 as well).

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