House edge in this online game?

Hi all, first post here, though I've been a fan of the Wizard's site for a long time. I've been enjoying some of the Bitcoin-based gambling sites and found this game, which is proving to be rather addictive. Right now I'm up slightly, but worry about a steep house edge in this game. Can anyone tell me how to calculate the eV of the various bets it offers?

You can see the game for yourself at www.bitcoin-kamikaze.com. Read the rules, then click on "free play" to play for free.

The game is pretty simple. It offers you a grid 5 squares wide by 8 squares long.

In each row of 5 squares one of the squares has a "bomb." You click on a square and if there is no "bomb" in the square you chose (4/5 squares don't have a bomb) then you win. You can keep clicking on squares until you make it all the way across the board (maximum of 8 rounds). The longer you go, the higher the payout. It's sort of a very simple parlay system. At any time after round 1 you can end the game and take your winnings.

Payouts times the original bet are: 1.2, 1.4, 1.6, 2, 2.5, 3, 4, 5

In other words, if you bet 1 bitcoin and only play through round 1, you can end the game and it will return 1.2 bitcoins to you. If you make it through 3 rounds, it will return 1.6 bitcoins. Make it through all 8 rows and it will return 5 bitcoins.

I've tried calculating this a few ways, but am not a statistician and suspect I'm doing it wrong. I also suspect the house edge changes depending on how many rounds you play through.

BTW, the first way I tried to calculate the house edge was to take the probability and divide by 1 for a "fair" payout (no house advantage).

So the probability of winning just one round is 80%. 1/0.8 = 1.25 so a "fair" payout would be 1.25X your original bet. Actual payout is 1.2X.

The probability of winning 2 rounds is 64% (0.8 x 0.8 = 0.64). 1/0.64 = 1.56 for a "fair" payout. Actual payout is 1.4X.

Like I said, I'm not a statistician so am just stabbing in the dark here. Something doesn't seem right when I calculate this way, though.

Multiply the payout times the odds os 1 win (0.8) raised to the power of columns (1 to 8).
You are on the right track. 1.2*0.8 = 0.96. The game has a 4% edge.

HOWEVER... 1.4* (0.8)^2 = 1.4*0.64 = 0.896. Selecting 2 columns has a 10.4% edge. Slot machines pay better!

Selecting 3 or more columns is awful... 1.6*(0.8)^3 = 1.6*0.512 = 0.8192 for a 18.08% edge. Even betting on Big Red 7 at a Craps Table is better!

nasty game if playing 2 or more columns. JMH2c

EDIT: This payout is "more fair" 1.2 , 1.5 , 1.8 , 2.3 , 3.0 , 3.7 , 4.6 , 5.7 but there's a good place to stop, though...

Since you can quit at any point I think it's more fair to look at it one step at a time. An interesting thing is that 1.6 (third win) to 2 (fourth win) is perfectly fair as is 2 (fourth) to 2.5 (fifth). In fact three (sixth) to four (seventh) is actually a positive HE bet. All in all a bad game though.

Quote: ahiromu

Since you can quit at any point I think it's more fair to look at it one step at a time. An interesting thing is that 1.6 (third win) to 2 (fourth win) is perfectly fair

How is it fair? (4/5)^3*1.6 = 0.8192
It seems that the best you can do is always cash out after the first win for 1.2*4/5 = 0.96 return (4% house edge).

Quote: ahiromu

In fact three (sixth) to four (seventh) is actually a positive HE bet. All in all a bad game though.

I don't understand what you mean by this either. First, again, I don't see any positive bets (e.g. (4/5)^6*3 = 0.79), and second, if there really was a positive HE, then why would you call it a "bad game"?

Ahiromu is right.

The game is positive at some points, but it is conditional upon previous wins.

Because you can always cash out, you need to treat each step as a new wager. Let's assume that you already won 6 times, and you have 3 credits. The return on the next step is as though you are wagering 3 credits to win 4 and the return is 4*.8/3=1.0666. Although this is a good bet, you had to suffer a negative return to get there, which makes the game bad overall.

If you make it to the third win, it is beneficial to keep going until the 7th or 8th win. But, overall, it is best to stop after the first win.

There are two ways to look at this: either "each row is a separate bet", or "use a strategy of always stopping after N rows."

If you consider one bet at a time:
First bet: 4/5 chance of winning 0.2 (+0.2), and 1/5 chance of losing 1 (-1), so the EV = 4/25 - 1/5 = -1/25, or a house edge of 4%.
Second bet: technically, you are now betting 1.2 (since you could have stopped after your first bet), so it's 4/5 x (+0.2) + 1/5 x (-1.2) = 4/25 - 6/25 = -2/25, or 8%.
Third bet: 4/5 x (+0.2) + 1/5 x (-1.4) = 4/25 - 7/25 = -3/25, or 12%
Fourth bet: 4/5 x (+0.4) + 1/5 x (-1.6) = 8/25 - 8/25 = 0.
Fifth bet: 4/5 x (+0.5) + 1/5 x (-2) = 2/5 - 2/5 = 0
Sixth bet: 4/5 x (+1) + 1/5 x (-2.5) = 4/5 - 1/2 = +3/10, or a player edge of 30%
Seventh bet: 4/5 x (+1) + 1/5 x (-3) = 4/5 - 3/5 = +1/5, or a player edge of 20%
Eighth bet: 4/5 x (+1) + 1/5 x (-4) = 4/5 - 4/5 = 0.

However, if you stick to a strategy of, "I will play N rows and then stop", your chance of winning after N rows is (4/5)^N, and each loss is considered a loss of just your original bet of 1:
One row: 4/5 x (+0.2) + 1/5 x (-1) = 4/25 - 1/5 = -0.04
Two rows: (4/5)² x (+0.4) + (1 - (4/5)²) x (-1) = -0.104
Three rows: (4/5)³ x (+0.6) + (1 - (4/5)³) x (-1) = -0.1808
Four rows: (4/5)⁴ x (+1) + (1 - (4/5)⁴) x (-1) = -0.1808
Five rows: (4/5)⁵ x (+1.5) + (1 - (4/5)⁵) x (-1) = -0.1808
Six rows: (4/5)⁶ x (+2) + (1 - (4/5)⁶) x (-1) = -0.213568
Seven rows: (4/5)⁷ x (+3) + (1 - (4/5)⁷) x (-1) = -0.1611392
Eight rows: (4/5)⁸ x (+4) + (1 - (4/5)⁸) x (-1) = -0.1611392

The strategy here is to stop after one row, and even then, the house edge is 4%.