*We wonder if our conjecture is secure, if the escrow isn’t enough, we could double it. Our honest player gets happy to be cheated on. Our malicious players go broke!*

Here is a fundamental explanation some of the cryptographic protocols for Cypher-Poker. Beyond that there is also an explanation of the problem posed in regard to securing the release of each players relevant private key.

This article is an attempt to bridge Cipher-Poker, or any mental poker implementation, with ethereum, or any comparable infrastrcture that might support p2p decentralized secure poker. This conjecture levates the “function(claim or refund)” mechanism described in this paper: How to Use Bitcoin to Play Decentralized Poker

In specific regard to poker we can pose the problem like this, which the most malicious case a single player might face:

## We might assume that the table is full of malicious players except the hero. The worst scenario would be: all players are all-in, hero wins the hand, but 1 malicious player refuses to hand in a key in order to verify the game.

## Hero could have potentially won his initial chipstack times the number of other players. The malicious group must not have anything to gain from colluding in this fashion: 1) they must not have any monetary gain 2) The honest player therefore must get back at least what he was supposed to win.

This suggests the escrow simply needs to be (n-1)(full_chipstack), where the chipstack is the most our honest player can lose in a hand and n is the number of players at the table.

The actual pot (not the escrow) then can simply be divided among the players that DID hand in their keys.

We can pose the problem succinctly like this:

## The escrow needs to cover the worst case scenario: the maximum # of malicious players (n-1 for a table of n player) times the maximum amount the honest player can lose (total equity or chipstack).

And so the succinct solution would be, at a table of n players, that escrow for each player is simply:

## (n-1)(chipstack)

If this conjecture holds there is a natural objection many will have that might lead to some interesting implications.

Edit: It’s actually going to be:

## (n-1)^2(chipstack)