What is the statistical odds of winning the 1 BTC which is a 100x payout?

Assuming that each card has 6

BTC's and 9 X's, and that each card is completely random and fair:

The odds of getting 6

BTC's out of 6 is 6/15 x 5/14 x 4/13 x 3/12 x 2/11 x 1/10 = .01998%

The odds of getting 6

BTC's out of 7 (ie 6

BTC's and 1 X) = .1398%

The odds of getting 6

BTC's out of 8 = .5594%

The odds of getting 6

BTC's out of 9 = 1.678%

Therefore, the overall odds of choosing all 6 with 9 picks is 2.39748%. If you are getting paid 100x (and it's fair), that is massively +ev.

EDIT: Sorry, I misread the rules. 3 X's = you lose. The odds of you choosing all 6 with 8 picks is .71918%. A massive house edge. Stay away.

28% house edge.

The house edge on all games other than Megacoin is 1.9%. If you click the "odds and payouts" link on the site, you can see the odds for all games (except Megacoin) and the payouts. The payouts are 0.981*(1/$odds_of_winning), which is the 1.9% house edge.

Regarding Megacoin, the odds of winning are 0.559440559441%, or about 1:178. This leaves us with a relativly high house edge of about 43.8%. Before everyone freaks out about the high house edge on this one game, let me describe why we chose it. We have a relatively high exposure to statistical deviations on this one. In theory we could be down 7 or more BTC before the laws of statistics bring us back up. We were asked by our users to provide a game where you could win a lot on a small bet. In order for us to be able offer this kind of game, we needed (at least for now) sufficient buffer to deal with the statistical deviations we are faced with. We most likely will lower the price of this scratchticket (and therefore reduce the house edge) in the future as we see how things go. But for now, if you are playing for efficiency, the other games on the site have a very economical house edge. If you are playing for excitement and a chance to win big, and you are feeling lucky, play Megacoin.

Now, an apology. The Megacoin odds should be listed along with the other scratchtickets on the site. It isn't there right now because of a bug. The Megacoin ticket operates differently than the others (it has a set purchase price, while the others allow for variable betting). Because of this, it is tagged differently in the database and didn't show up in the list which is auto generated from the database. Our fault. We will be fixing this today. Again, my apologies.

Finally for those who are interested, here is some info on calculating the odds.

The calculation of the odds is not a simple N!/M! problem (except in the case of having to choose 6 "BTC" without revealing any "X"s). It is actually the number of paths you could choose to achieve a win times the probability of a single path being traversed. The # of paths you could choose to win is a statistical combination. And the probability of a single path being traversed is a factorial ratio. This PHP code gives more details:

// Factorial function, not all PHP implementations

// have GMP math extensions installed.

//

function factorial($in) {

// 0! = 1! = 1

$out = 1;

// Only if $in is >= 2

for ($i = 2; $i <= $in; $i++) {

$out *= $i;

}

return $out;

}

// Set your own values here

//

$b =

<# of bitcoin symbols hidden on the scratchticket (that must be revealed to win)>;

$x =

<# of "X" symbols hidden on the scratchticket>;

$end_game_x_symbols =

<# of "X" symbols that, if revealed, will end the game in a loss>;

$allowable_x_selections = $end_game_x_symbols - 1;

// Calculate $c which is $k combinations in a set of $n where:

//

// $k = # of allowable X selections (1 less than the # that end game)

// $n = bitcoins required to reveal to win + allowable X selections

//

// $c tells us how many possible sequences of choices can be made

// by the player to achieve a win.

//

// Remember from your statistics studies:

//

// N!

// C(N,K) = -------------

// K!*(N - K)!

//

$n = $b + $allowable_x_selections;

$k = $allowable_x_selections;

$c = factorial($n)/(factorial($k)*factorial($n - $k));

// Now, we calculate the probability of successfully traversing a

// single path.

//

$total_scratchticket_locations = $b + $x;

$successful_path_prob = factorial($b)*factorial($x)/factorial($total_scratchticket_locations);

// The probability of winning the game equals the number of possible

// sequences a player could choose times the probability of successfully

// traversing a single path;

//

$probability_of_winning = $c * $successful_path_prob;

--- vineyard