- 12 Apr '13 04:18A game.

The player tosses a coin.

If its a head he adds 1 point to his total score and may "stick" or continue.

If its a tail his total score is 0 and he may not continue.

When a player "sticks" or flips a tail it is the turn of the next player.

Going first is obviously a disadvantage - but if you have to, what is the best strategy?

In a 2 player game its simple. Flip the coin once and you have at least a 50% chance of winning if its a head.

What about a 3 player game? 4 player? N? - 12 Apr '13 15:51I would say you should continue if your chances of improving your score are greater than the chance of everybody behind you failing.

Let's assume in all cases that your first flip is heads

1 other player:

You have a 50% chance of improving your score.

The Other player has a 3/4 chance of failing to beat you. You should stop.

2 other players:

50% chance to improve.

The other two players have 3*4 * 3/4 = 9/16 chance of failing to win. You should stop.

3 other players:

50% chance to improve

The other 3 players have a (3/4)^3 = 27/64 chance of failing to win. You should flip again. - 20 Apr '13 17:06 / 2 editsYou win the game if

* you get score n. Everyone else has to try to get to n+1, they try and fail.

* what if you fail? Do you automatically lose so in a two-player game player 2 just looks at your failed coin toss and says "I win" without tossing the coin? Or do you still win if everyone else fails too

(as you got zero pts first)? Will the others go for a win (possibly multiple rolls if there are 3+ players) or just try to defeat you (if you fail, everyone else rolls only once) ?

A Russian roulette variant of that.. you accept n chances of one in six of dying, all your opponents have to face the odds n+1 times; of an opponent survives n+1 clicks he'll use the gun on you and you lose. I find three clicks to be optimal vs one opponent (29% chance of win), five against two opponents 17% chance), seven vs three (12% chance), 8 vs 4, 9 vs 5, 10 vs six or seven, 11 vs six, 12 vs seven or eight, and 13 vs nine or ten, at which point survival chances are a humble 3%. - 05 May '13 19:42 / 1 editAll right. So at least one of the x opponents has so score n+1 when you scored n, otherwise you win. If you drop to zero pts, at least one of them needs to score 1 point, or you win. They know what you scored before they do their turn, as you go first.

A player going for n+1 pts has a chance of 0.5^(n+1) of succeeding in that and 1 - 0.5^(n+1) of failing. If you have x opponents, the chance that they all fail is (1 - 0.5^(n+1))^x. The decision you face in the game is that when you have some amount of n points, do you take one more 50-50 chance of going got n+1 points or zero pts.

Chance that you win if you stay at n points:

(1 - 0.5^(n+1))^x

Change that you win if you go to n+1 points:

0.5 * (1 - 0.5^(n+2))^x + 0.5*0.5^x

It's sensible to take another turn if the latter probability is higher.

0.5 * (1 - 0.5^(n+2))^x+0.5 * 0.5^x - (1 - 0.5^(n+1))^x > 0.

An exact solution of the form n(x) for that would be rather hellish. A numerical one shows that you should always go for at least one point.. you have nothing to lose.. and go for 3 points with 4-10 opponents, 4 points with points with 12-22 opponents, and 5 points with 23-44 opponents. At that point the chances of winning are so dismal that your strategy doesn't really matter much.