Coins, Dice, and Probability! [closed]

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!

closed as unclear what you're asking by AzaAug 4 '16 at 18:30

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Hello, fellow copycat! – Rand al'Thor Dec 28 '14 at 17:20
• 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. – Lopsy Dec 28 '14 at 17:39
• 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. – xnor Dec 28 '14 at 18:02
• 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. – xnor Dec 28 '14 at 19:46
• lol . How is it possible to "be the last one standing" when all of us are sitting?!? – Spikatrix Dec 29 '14 at 4:25

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.
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...