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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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warspyking
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    Hello, fellow copycat! – Rand al'Thor Dec 28 '14 at 17:20
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    @rand al'thor Well it was highly probable to happen. :D – warspyking Dec 28 '14 at 17:21
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    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
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    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
  • ... Do everybody hate my puzzles? I'm soon gonna quit. I see nothing worse in this one than the previous 2... – warspyking Dec 28 '14 at 19:18
  • @rand Why on Earth did you change my title? – warspyking Dec 28 '14 at 19:21
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    @warspyking Sorry, sorry! I misread. Didn't see the coins bit. – Rand al'Thor Dec 28 '14 at 19:22
  • @rand Do you hate the puzzle? Oh and did you get notified of my previous comment? Or come across it? Wondering because I edited '@rand in after. – warspyking Dec 28 '14 at 19:25
  • It looks like Lopsy and xnor are right (and I did DV), but there could be a simple way of doing it (so I reversed my DV). Do you know the answer? And yes, I did get notified of your 'Why on Earth did you change my title?' comment. – Rand al'Thor Dec 28 '14 at 19:28
  • @rand I guess we'll see what happens, why don't you try it? – warspyking Dec 28 '14 at 19:34
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    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
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    lol . How is it possible to "be the last one standing" when all of us are sitting?!? – Spikatrix Dec 29 '14 at 04:25
  • I agree that this is not a very good puzzle, if you don't know that it has a good answer or an entertaining way of finding the answer. I DV as well. However, I agree that it is not very different from the alternative 'puzzle' posted earlier - which I also DV.(for the same reason.) – BmyGuest Dec 29 '14 at 07:56
  • "Will it better your chances if he leaves?" - your chances *of what?* The rules are the same for everyone. So no matter how many people play, they all have the same chance of winning. Better your chances of getting a result at all? With all the different scenarios for "nobody wins", this would be painful to work out. – Disillusioned Jan 02 '15 at 15:35

1 Answers1

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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.

David Boshton
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  • The 3 plays rule was exactly what I was looking for, this looks right though. If someone finds a problem I'll deaccept. – warspyking Jan 16 '15 at 10:13