Since I like probability questions, and tttpp brought it up, I figured it might be interesting to calculate how likely you are to lose on the first turn in Forbidden Island, because Fool's Landing sinks into the abyss. This question is only interested in figuring the odds of losing the game during the first players turn. It would be slightly more difficult to figure the odds of losing during the first round, where each player has taken one turn at most. If this question gets enough attention, I might decide to ask that question and figure out the odds. To figure out the odds of losing, you will need to calculate the following:

  • What are the odds that the first player cannot shore up Fool's Landing? (depends upon the first player's character, their starting location, and Fool's Landing location)
  • What are the odds that no player has received a Sandbag/Helicopter (depends upon number of players)
  • What are the odds that a Waters Rise card is drawn?
  • What are the odds that Fool's Landing is redrawn after Waters Rise? (difficultly level effects this)

What are the odds range of losing on the first turn for a 2-4 Player game? Calculate each individual component separately, and then provide a range of maximum/minimum odds of losing on the first turn based upon the individual components above. Assume the following:

  • The all players will attempt to shore up Fool's Landing. (use Sandbags/Helicopter)
  • The default island map is used.
  • The first player and their character are randomly determined, per the rules.
  • Ignore the odds of losing because a player sinks into the abyss. (optional:)
  • Ignore the odds of losing because both tiles for a particular treasure sink into the abyss. (optional)

Note: Calculate this if you want, but I am not uninterested in the minimal effect on the result of losing first turn.

  • 2
    +1 What a good question :-) I hadn't thought of the possibility of losing first turn due to someone sinking into the abyss. Definitely possible, but you've got to be really unlucky!
    – tttppp
    Aug 9, 2012 at 11:30

2 Answers 2


A friendly warning: This is a looong answer.

I've calculated the chances of Fools' Landing being lost immediately after the first turn with all players playing to prevent it and a random tile setup.

A few variables:

  • n = number of players
  • w = water level after rising once (2 for Novice, 3 for everything else)

Special abilities/actions

First, of all, there are a couple of special abilities and cards that make it impossible to fail:

  • If the first character is the Pilot (1/6), they start at FL and can always shore it up, or
  • If anyone starts with a Helicopter Lift (there are 3) or a Sandbag (2) in their hand (1-(20 C (2n))/(25 C (2n))), they can shore up or move the first player to shore up (it's OK to ignore WR cards in the initial deal)

1/6 + 1-(20 C (2n))/(25 C (2n))

i.e. (1 - (1/6 + 1-(20 C (2n))/(25 C (2n)))) chance that the following permutations can end the game.

Floods, Waters Rise and More Floods

What the characters do on their turns is tricky, so we'll deal with all the card probabilities together first:

  • 6 areas are flooded to start. To immediately fail, FL must be one of them (6/24)
  • At the end of the first player's turn, regardless of what they did, a fail requires a WR card be drawn from the remaining deck. They draw two cards. The first has a 3/(28-2n) chance of being WR. If it's not ((25-2n)/(28-2n)), the second card has a 3/(27-2n) chance of being WR. After some balancing, that becomes (156-12n)/((28-2n)(27-2n)) chance that a WR card is drawn on the first turn
  • If it is, the (6-card) discard pile is shuffled into the deck and there is a w/24 chance that FL comes out again.

So, the odds that the cards are against you and only a perfect first turn can prevent a fail is:

6/24 * w/24 * (156-12n)/((28-2n)(27-2n)) = (w(156-12n))/(96(28-2n)(27-2n))

The First Turn: Explorer Edition

This is where it starts to get very hairy, so bear with me. If the first player is the Explorer, they can move diagonally, so that makes everything a bit more reachable. There is a 1/5 chance that the first player is the Explorer (because we've already taken into account that they're not the Pilot). To fail as the Explorer, you have to be more than three spaces away from FL including diagonally.

  • From the outside 8 spaces (hereafter, "outside"), there are six you cannot reach. The odds of placing FL in one of the 6 unreachable spaces is 6/23. Therefore, there is a 8/24 * 6/23 = 48/552 chance you fail for this starting position
  • If you start in a corner of the inner square ("corner"), there are four spaces you cannot reach as an Explorer (the opposite outsides). That's a 4/24 * 4/23 = 16/552 chance of failing
  • If you start in a space that is not in the centre four, "outside" or in a "corner" (which I'll refer to as "inside" for clarity), there are two unreachable spaces. That provides 8/24 * 2/23 = 16/552
  • Finally, your first option could be one of the central spaces (the centre four). If it is, congratulations! You can reach any space!

So, the odds of being the Explorer and being unable to shore FL on your first turn, based on all tiles being placed randomly is:

(48+16+16)/552 * 1/5 = 16/552

The First Turn II: Everyone Else Edition

But what if you're not the Explorer?! Let's look at all the potential starting spaces again (groan...) given you're one of the other non-Pilot characters (4/5):

  • There's a 80/552 chance that FL is unreachable from outside
  • There's a 40/552 chance that FL is unreachable from a corner
  • There's a 72/552 chance that FL is unreachable from inside
  • There's a 12/552 chance that FL is unreachable from the centre.

So, the odds of being a non-Explorer, non-Pilot and being unable to shore FL on your first turn is:

(80+40+72+12)/552 * 4/5 = 816/2760

Therefore, the odds of being unable to shore up FL (given that you're not the Pilot) is:

16/552 + 816/2760 = 896/2760


The Grand Conclusion, Or TL;DR

So, we now take all three sections' results and multiply them together:

  • The odds that you're not the Pilot and no one has cards that can help you = (1 - (1/6 + 1-(20 C (2n))/(25 C (2n))))
  • The odds that you all the flooding will go against you, and you'll draw at least one WR = (w(156-12n))/(96(28-2n)(27-2n))
  • The odds that you won't be able to reach and shore up FL under normal circumstances = 896/2760

Given that w=3 (which it is for most setups), the odds of failing on the first turn by number of players is:

  • 2 players: approximately 0.0768%
  • 3 players: approximately 0.0483%
  • 4 players: approximately 0.0288%
  • 1
    Maybe I've misread, but I don't think the chance of getting a helicopter/sandbags card to start with is 5/2n. This is >1 for two players :-) Instead I think it's 1-(20 C (2n))/(25 C (2n)) (where C is the "combinations" function), which is ~0.62 for 2 players, ~0.78 for 3 players, and ~0.88 for 4 players.
    – tttppp
    Aug 9, 2012 at 11:44
  • 1
    "If the first character is the Pilot (1/6), they can always move to FL and shore it up" Minor quibble, but doesn't the Pilot always start at FL to begin with?
    – bwarner
    Aug 9, 2012 at 13:09
  • 3
    @Johno, great work so far. I haven't checked all your math, but it appears to agree with most of my initial calculations. You might have a mistake, " (6-card) discard pile is shuffled into the deck = w/24" The 6-7 card discard is shuffled and placed onto the flood deck. Whether you draw WR first or second affects if you have 6 (50% chance of FL redrawn if not novice), or 7 cards in the discard. Two waters rise makes the flood level go up by two, this might cause more than 3 tiles to flood (I think) at higher difficulty levels, although the odds are miniscule
    – user1873
    Aug 9, 2012 at 14:15
  • 1
    @Johno, another small technicality is that the diver can move more than a single space through flooded tiles. I was going to answer the question like you (ignoring this ability), because his inability to reach FL is much more complex. I might take a stab at that portion of the question.
    – user1873
    Aug 9, 2012 at 14:22
  • 1
    @Johno, one more small mistake that npmight shave off a tenth of a %. You need to account for one of the two drawn treasure cards being a Sandbag (when the other is WZters Rise), since that player has an opportunity to use it before the tiles flood.
    – user1873
    Aug 9, 2012 at 14:57


The probability of losing in the first turn due to Fool's Landing sinking, assuming all players try their hardest to avoid it, is dependent on the difficultly level and the number of players n:

Difficultly  | n | Probability
Novice       | 2 | 0.00199
Novice       | 3 | 0.00122
Novice       | 4 | 0.00071
Normal/Elite | 2 | 0.00291
Normal/Elite | 3 | 0.00179
Normal/Elite | 4 | 0.00103
Legendary    | 2 | 0.00296
Legendary    | 3 | 0.00182
Legendary    | 4 | 0.00105

The computation of the values in this table was quite long, and an outline of the steps is given below.


I have split the computation into four sections, based on the game stages:

  1. Game setup - Fool's Landing is initially flooded and no one was dealt a special action card.
  2. First Player's Actions - The first player is unable to shore up Fool's Landing.
  3. First Player's Treasure Card Draw - Waters Rise is drawn (and Sandbags is not).
  4. First Player's Flood Card Draw - Waters Rise is drawn a second time and sinks.


  • n The number of players
  • v The number of Waters Rise cards drawn at the end of the first turn
  • w The water level after the first turn

Game Setup

P(Fool's Landing drawn in first six cards) = 6/24
P(No sandbags or helicopters dealt) = (20 C (2n)) / (25 C (2n))

First Player's Actions

Let R be the first player's role:

P(R = Pilot and can't shore up Fool's Landing) = 1/6 * 0 = 0
P(R = Diver and can't shore up Fool's Landing) = 1/6 * 4611612/18574248 = 384301/9287124
P(R = Explorer and can't shore up Fool's Landing) = 1/6 * 10/69 = 5/207
P(R in {Navigator, Messenger, Engineer} and can't shore up Fool's Landing) = 3/6 * 59/138 = 59/276

So summing these:

P(Can't shore up Fool's Landing) = 1945439/6965343

First Player's Treasure Card Draw

For Fool's Landing to sink, one of the two drawn cards must be Waters Rise. The other must not be a sandbags. I'm keeping these separate as the value of v impacts the value of w.

P(v = 2 and no sandbags) = P(v = 2)
                         = 6 / ((28-2n)(27-2n))
P(v = 1 and no sandbags) = P(No sandbags | v = 1) * P(v = 1)
                         = (23-2n)/(25-2n) * 6(25-2n)/((28-2n)(27-2n))
                         = 6*(23-2n)/((28-2n)(27-2n))

First Player's Flood Card Draw

We have now shuffled the six flood cards and put them back on top of the Flood Card Deck. Note that w can be derived from the difficulty and the value of v.

P(Fool's Landing drawn at end of first player's turn) = w/6

The Result

The difficultly level impacts w:

  • Novice: v=1 => w=2, v=2 => w=3
P(Fool's Landing Sinks First Turn) = 6/24 * (20C(2n))/(25C(2n)) * 1945439/6965343 * (6*(23-2n)/((28-2n)(27-2n)) * 2/6 + 6/((28-2n)(27-2n)) * 3/6)
 =  (20C(2n))/(25C(2n)) * 1945439/27861372 * (49-4n)/((28-2n)(27-2n))
  • Normal/Elite: v=1 => w=3, v=2 => w=3
P(Fool's Landing Sinks First Turn) = 6/24 * (20C(2n))/(25C(2n)) * 1945439/6965343 * (6*(23-2n)/((28-2n)(27-2n)) * 3/6 + 6/((28-2n)(27-2n)) * 3/6)
 =  (20C(2n))/(25C(2n)) * 1945439/27861372 * (72-6n)/((28-2n)(27-2n))
  • Legendary: v=1 => w=3, v=2 => w=4
P(Fool's Landing Sinks First Turn) = 6/24 * (20C(2n))/(25C(2n)) * 1945439/6965343 * (6*(23-2n)/((28-2n)(27-2n)) * 3/6 + 6/((28-2n)(27-2n)) * 4/6)
 =  (20C(2n))/(25C(2n)) * 1945439/27861372 * (73-6n)/((28-2n)(27-2n))

Finally, computing these formulae for different difficulty levels and number of players I ended up with:

Difficultly  | n |           Probability
Novice       | 2 | 1355970983/682632941760  ≈ 0.00199
Novice       | 3 | 1223681131/999834661980  ≈ 0.00122
Novice       | 4 |  429942019/608770978200  ≈ 0.00071
Normal/Elite | 2 |   33072463/11377215696   ≈ 0.00291
Normal/Elite | 3 |   99217389/55546370110   ≈ 0.00179
Normal/Elite | 4 |  859884038/837060095025  ≈ 0.00103
Legendary    | 2 | 2017420243/682632941760  ≈ 0.00296
Legendary    | 3 |   33072463/18178812036   ≈ 0.00182
Legendary    | 4 | 3009594133/2869920325800 ≈ 0.00105
  • Breaking it down: P(v=1) = (2 orderings)*(3 ways to pick a WR)*(25-2n ways to not pick a WR)/(total ways to pick two ordered cards) = 6(25-2n)/((28-2n)(27-2n))
    – tttppp
    Aug 15, 2012 at 14:48
  • P(No sandbags | v=1) = P(No sandbags given v=1) = P(The one non-WR card is not a sandbags card) =(23-2n)/(25-2n)
    – tttppp
    Aug 15, 2012 at 14:50
  • you are right. I rechecked my math, have no idea how I got 21.111%. I cleaned up my comments. You might want to keep just your second comment that breaks it down.
    – user1873
    Aug 16, 2012 at 11:27
  • @user1873 Great! Let me know if you want any further bits broken down.
    – tttppp
    Aug 16, 2012 at 16:39

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