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At a gambling joint, you are invited to be a player in a game with identical unfair but consistent coins flipped simultaneously in each round, with no cost to enter. Each player gets a different number of coins to flip.

In the first set of games, there are 7 players, having 2, 3, 4, 5, 6, 7, or 8 coins each. In each round, one player (the highest-scoring one) gets a payoff equal to their score, and others get nothing. The score of a player with n coins who gets k heads is k/(n+1-p), where p = 5/17 is the probability of a head appearing. There are many rounds per set of games.

Example (with Cn referring to player n with n coins): Suppose C3 flips 3 coins and gets 1 head, scoring 1/(3+1-p) ~ 0.2698; that C4 flips 4 coins and gets 3 heads, scoring 3/(4+1-p) = 0.6375; and C8 flips 8 coins, gets 5 heads, and scores 5/(8+1-p) ~ 0.5743. If everyone else scores less, C4 receives $0.6375.

Given a choice in this first set of games, how many coins will you use? If that first choice is gone, what's your second choice?

In the second set of games, instead of 7 players with 2, 3, 4, 5, 6, 7, or 8 coins each, there are only 5 players, having 3, 4, 5, 6, or 7 coins each. Again, if given the choice, how many coins will you use, and if that choice is gone, what's your second choice?

Addendum: Now that I've accepted Florian F's correct answer to the question as I asked it, here's the question I meant to ask (but unfortunately mixed up head and tail probabilities in original post). This has a more interesting result, that can be separately answered as seen below. Question: With p = 12/17 as the probability of a head appearing, and score being k/(n+p) for k heads on n coins, what are your first and second choices in the 7-player and 5-player cases?

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  • $\begingroup$ To clarify: payoff is one and the same same as the highest score of each round? $\endgroup$
    – C8H10N4O2
    Jan 10 at 0:53
  • $\begingroup$ @C8H10N4O2 Just one same, but yes, the payoff in each round equals the highest score in that round. Also, a player's n value can change only at the beginning of a set of games; it does not change during a set. $\endgroup$ Jan 10 at 1:24
  • $\begingroup$ It is possible to compute the answer by just brute forcing all the cases (there are $3*4*5*6*7*8*9=181440$ cases for the first problem as we only need to know how many heads there are), but is a simpler answer intended? $\endgroup$
    – littlecat
    Jan 10 at 3:04
  • $\begingroup$ Yes, a simpler answer is intended, although straight calculation is ok too. Note, each of those 181440 cases has its own probability of occurring; for example, the case of Cn getting n heads (for all n) is far less likely than that of Cn getting n-1 heads (for all n). As another example, C(n+1) is more likely to get k>0 heads than is Cn. $\endgroup$ Jan 10 at 6:38
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    $\begingroup$ @DrXorile I think I'll not ask it separately after all (due to already-answered status) - although that's probably what I should have done, instead of including it here. Hopefully you will get some upvotes anyway! Thanks. $\endgroup$ Jan 11 at 5:09

2 Answers 2

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OK, gave it a try.

Intuitively:

The score is roughly $k\over n$ and averages the number of heads. It should be close to $5/17$ on average. Adding $1 - p$ in the denominator reduces the score. A little for a large n, but significantly for a small n.

Let's take the approximation of $k\over n$ for the score. It should be $5/17$ on average for all players.
To win you need a good chance of having a larger score than all others. For this the trick is to aim for a large variance. A large variance gives relatively high and relatively small scores. When the score is very low, you don't loose more. But when it is very high you improve your chance to outscore all other players. What is more, when you win, your score is higher on average. Players with many coins often score nearer to the average. In that case it is likely that the one or the other player scores better.

So, on first approximation, you should choose as few coins as possible.

Let's now add the (1+p). The score is now $k\over (n+1-p)$. You see that with few coins your score is reduced significantly and might offset the advantage of having a large variance. There must be a tipping point, a number of coins below which you loose more from the reduction in the average score than you win on the number of wins. But it is difficult to compute exactly where that is.

So I ran a simulation.

Simulation

A simple monte carlo simulation shows the following results. It is the payout for the 7-player and 5-player game.

2: 0.088468
3: 0.093272
4: 0.080928
5: 0.067717
6: 0.070351
7: 0.061642
8: 0.058534

3: 0.112919
4: 0.099512
5: 0.092455
6: 0.097203
7: 0.077289

As I suspected, 2 coins is not optimal with 7 players. The 2-coin player scores less than the 3-coin player. 3 seems to be the best choice. 2 is second-best.

But one surprising thing is the score of 6 compared to 5. 6 coins score better than 5 coins, which doesn't match with my intuition that scores should got down as n increases. I thought this might be an effect of the monte carlo simulation. But running the code again a few times shows this is not random.

What could be the reason? First, to be sure, I did an exact calculation over all possible outcomes.

Calculations:

The calculations give the following result.

2: 0.088465
3: 0.093328
4: 0.080926
5: 0.067681
6: 0.070351
7: 0.061637
8: 0.058535

3: 0.112924
4: 0.099505
5: 0.092453
6: 0.097210
7: 0.077288

Pretty much the same result. There is no monte carlo anomaly.

I am not sure what is the reason behind the score with 6 coins. It could have to do with the fact that the score $2/6 = 0.333$ is still over the average $5/17 = 0.294$, and therefore only 7/64 outcomes are below average. With 5 coins it is 6/32 outcomes, which is more often. So with 6 coins you score more often over average than with 5.

Anyway, the answer to the original question is:

In the 7-player game choose 3 coins or 2 as second choice.
In the 5-player game choose 3 coins or 4 as second choice.

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  • $\begingroup$ In the 5-player game, choosing 2 coins is not an option. $\endgroup$ Jan 10 at 21:17
  • $\begingroup$ You are absolutely right, I concentrated on the 7-player game and didn't pay enough attention to the 5-player version. I fixed this. $\endgroup$
    – Florian F
    Jan 10 at 22:04
  • $\begingroup$ Note, please see addendum in question - you might be able to adapt your code in another answer $\endgroup$ Jan 10 at 23:11
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I replicated @Florian's results to check my calc and then changed to the new game rules.

My results for the 2,3,4,5,6,7,8 game are:

2: 0.040476
3: 0.132761
4: 0.133910
5: 0.132450
6: 0.140919
7: 0.131355
8: 0.119017

My results for the 3,4,5,6,7 game are:

3: 0.144246
4: 0.154364
5: 0.170298
6: 0.168679
7: 0.175720

It's definitely surprising:

2-8 game has 6 and then 4 as the winning positions.
3-7 game has 7 and then 5 as the winning positions.
Meaning that the winning position shifts even though the winning position isn't removed from the game. Huh?! I need to think more about this...

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  • $\begingroup$ The shift can be explained by the removed candidate (2 and 8 coins) beating some winning chances of 7 coins. So if 8 > 7 > 6, then 7 doesn't win in 7-player, but wins in 5-player case. By removing 8, 7 gets more probability mass as a result. So from here we can see that "if 7 beats 6, it's also likely that 8 beats 7" $\endgroup$
    – justhalf
    Jan 11 at 6:41
  • $\begingroup$ The point is that 8<7<6 in the 7 player game. Then with 8 removed, 7>6. $\endgroup$
    – Dr Xorile
    Jan 11 at 13:10
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    $\begingroup$ Non-transitive (two-sided) dice. ^_^ $\endgroup$ Jan 11 at 20:21
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    $\begingroup$ @DrXorile I was talking not about the overall count, but individual count. So for example, in 6,7,8 players, the 6 got 3 heads, 7 got 4 heads, 8 got 5 heads. 8 wins in this case, but remove the 8, and 7 wins. I'm saying that there are more combinations where 7 wins over 6, but 8 took that away by winning over 7. Remember that the final probability you have is not individual game. What matters is the result of individual games, then aggregated. So you can have 8 cases of 8>7>6, and 15 cases of 6>7>8, and 5 8>6>7, and 14 7>6>8. 6 wins more here, but if you remove 8, 7 wins more. $\endgroup$
    – justhalf
    Jan 12 at 2:56

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