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Background:

It's time for a little bit of fun, your friends are playing a game and you decide to join them!

Gamplay:

You each sit down and play on a flat surface. You sit in a circle and each throw your die. If the person to your left gets a number within 2 number range (e.g; a 5 when you get 3) You are out. Then you all flip a coin, if you get the same as the person to your right you are out.

How to Win:

Be the last person standing.

How ties are handled:

If there are 2 people left nobody wins. If there are nobody left nobody wins. If the same result happens 3 times in a row nobody wins.

Notes:

You are using six sided dice.

Puzzle:

You have 11 friends, but one of them has to go, he says he can stay one more round if you want him to; will it better you chances if he leaves? Why or why not?

Motivation:

What? You're not motivated? Fine! You can not have infinite rep on Puzzling if you solve!

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    $\begingroup$ Hello, fellow copycat! $\endgroup$ Dec 28, 2014 at 17:20
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    $\begingroup$ Please clarify the rules, there are some ambiguities. For example, what does "same result" mean? Also, just eyeballing it, there is no reason for this puzzle to be mathematically interesting, you obviously just put in a bunch of random rules and hoped it would work out because the last puzzle that looked like this did. $\endgroup$
    – Lopsy
    Dec 28, 2014 at 17:39
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    $\begingroup$ I agree with @Lopsy. I think what makes a math problem a puzzle is that is has a clever and elegant solution, an unexpected answer, or both. I think for this, there's nothing better to do than computer simulation or a laborious calculation, neither of which is very interesting or puzzley. $\endgroup$
    – xnor
    Dec 28, 2014 at 18:02
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    $\begingroup$ I'll remove my downvote if a nice solution is found, but still it's up to the puzzle maker to create a puzzle with a nice solution, not to hope that one happens to turns up. $\endgroup$
    – xnor
    Dec 28, 2014 at 19:46
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    $\begingroup$ lol . How is it possible to "be the last one standing" when all of us are sitting?!? $\endgroup$
    – Spikatrix
    Dec 29, 2014 at 4:25

1 Answer 1

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So you have to survive the dice-roll, which turns out by counting, to be a $11/36$ chance of being kicked off, and then a $50\%$ chance of being kicked off by a coin toss.

The other players have the same probabilities, and each person being kicked off is dependant only on those next to them. So it actually benefits everyone if there are fewer players purely since the method of removing oneself from the game is independent of the number of players.

However: The only meaningful (and likely) way to achieve "The same result three times" is if no-one is removed from the game. This is clearly more likely with fewer players, however impossible with an odd number of players; someone must leave because the ranges allowable on the dice roles alternate between $\{1,2,3\}$ and $\{4,5,6\}$, so there has to be an even number of people to achieve this; assuming the table is round etc. Therefore...

Since this affects every player, he should be asked to stay and play as this will improve your chances of winning due to the 3 consecutive plays rule -- 11 players as opposed to 10.

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  • $\begingroup$ The 3 plays rule was exactly what I was looking for, this looks right though. If someone finds a problem I'll deaccept. $\endgroup$
    – warspyking
    Jan 16, 2015 at 10:13

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