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# Spin the bottle (no kissing involved)

## Question

Eight friends at a party, including the host, sit in a circle on the floor. There's a nice bottle wine that has not been opened that one of the seven guests should take home. The host will spin the bottle and pass it to the person on his left or right, depending on the direction the bottle points when it stops. The guest who receives the bottle will do the same - spin the bottle and in the same way pass it to his left or right. The game continues as long as there is at least one guest who has not received the bottle. Whoever receives the bottle when that is no longer the case (whoever is the last guest to receive the bottle for the first time) wins the bottle of wine. Numbering the positions clockwise (with the host being #1) which position will you sit in to win the bottle with greatest probability?

Edited by bonanova

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I believe I posted a very similar problem years ago involving a roulette wheel and a ball with paint on it. It can be solved with just logic. No math is needed:

The seating position of the guests does not matter. All guests are equally likely to win.

For a guest to win, exactly two things must happen:

1. At some point the bottle must move to the location 1 to their right or 1 to their left.

2. Then the bottle must travel completely around the circle, in the opposite direction, until it has arrived at that guest's location. There is a chance the bottle could do 1, move all the way around to the other side of the guest and then reverse directions again, but by then the game is decided, as all other guests have been visited and we are just waiting for that guest to eventually get the bottle.

Here is an example of how 3 might win. 1, 2, 1, Host, 8, Host, 8, 7, 6, 7, 6, 5, 6, 5, 4, 3.  All wins will look something like this, or its opposite.

It's clear that for any guest, if 1 happens, there is a fixed percentage chance that 2 will happen, and that percentage is the same no matter where that guest sits. And if you think about how the game works, the probability that 1 will happen for any given guest is 100%. At some point, all guests will have the bottle one to their right or left before they get the bottle. Therefore, all guest are equally likely to win the game.

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Based on this exacting 1000 time simulation of the best night of your high school life, you got very lucky!! BTW, this code works fine.

from random import randint

def he_gets_the_bottle():
if randint(0,9) == 1:
return True

def it_points_to_uma():
if randint(0, 9) == 1:
return True

def bonanova_gets_a_kiss():
if he_gets_the_bottle() and it_points_to_uma():
return True

kisses = 0

for i in range(1000):
if bonanova_gets_a_kiss():
kisses += 1

print kisses

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My guess

Position #5. My rationale-the closest position (either 2 or 8) cannot win. The next closest cannot win. Th

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(sigh--hard to post from phone...)

rationale was twofold--the positions are symmetric about the axis 1-5 AND the numbers closer to 1 (ie, 2 and 8) are the most likely the first to see the bottle, so LEAST likely to be the last to see it. So, impressionistically, it feels like the further from 1, the better. 5 is the furthest from 1, in either direction)

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Does the host who spends it originally count as "receiving the bottle"?

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My first thought was like Capt Ed.

BUT thinking about it some more.

I suspect that the person sitting at 4 or 6 will probably have a better chance.

Rationale: after a few moves, the passing is going to tend to go one direction [say CCW] and will tend to go back and forth on that side of the host. By that logic, #5 may often be the next to last guest to get the bottle.

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Dgreening, I see, sort of like if you have 8 trumps out in the other two hands, they're more likely to split 5-3 than 4-4.

This may account for my simulation results in 38 trials, for positions 1(0), 2(4), 3(4), 4(9), 5(7), 6(5), 7(4), 8(5). Then again, this might merely indicate sloppy coding on my part or just small sample size. Still, your point is very plausible: each individual run is less likely to be perfectly symmetrical, hence more likely to end off-center. Perhaps an analytical solution will quantify that result.

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Does the host who spends it originally count as "receiving the bottle"?

Yes. The host cannot win.

No satisfactory analysis yet.

Try the case with three guests (two guests is trivial) and compute the probability that after n spins you have not received the bottle.

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Another good insight! It's funny to introspect--I thought the guests in positions 2 and 8 were particularly disadvantaged, but they start the game with condition ( 1 ) already true, so in some sense it's EASIER for them to win.
Nice analysis, Bubbled!
Nice puzzle, Bonanova

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Just for giggles I ran a simulation for 1,000,000 games. Here's my output and Python code:

As expected, the results are very close to 1/8 for each spot.

{1: 125123, 2: 124812, 3: 125169, 4: 125239, 5: 125046, 6: 125019, 7: 125028, 8: 124564}

from random import randint

wins = {1 : 0, 2 : 0, 3 : 0, 4 : 0, 5 : 0, 6 : 0, 7 : 0, 8 : 0}

def play_game():
locations = [8, 7, 6, 5, 4, 3, 2, 1, 0, 8, 7, 6, 5, 4, 3, 2, 1]
where = 8
visited = [0]
direction = [-1, 1]
while len(visited) < 8:
move = direction[randint(0, 1)]
where = where + move
guest = locations[where]
if guest not in visited:
visited.append(guest)
for i in range(9):
if i not in visited:
return i
for i in range(1000000):
winner = play_game()
wins[winner] += 1
print wins

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Credit for the puzzle doesn't go to me. I didn't recall specifically the roulette wheel version, bubbled; rather, I saw an even different embodiment recently in Quora. I knew already of the uniform distribution of results and thought it would be fun to discuss it here. so I created a bottle spinning flavor text for it. I recall my first and only game of spin the bottle, back in high school. It was memorable because my single kiss of the evening was with the homecoming queen, a beautiful Uma-Thurman-like blonde named Hope. Calculate the odds of that.

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Very cool.

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