<p>This should happen to you about 1 in every 620 games that have 3 players, 14 tiles that an Abbot can be played on, 1 start tile, and 57 other tiles.</p>
<p>Given that there are 72 tiles, in a 3-player game, you'd draw either 23 or 24 tiles, depending on turn order. Although the 72 tiles divide evenly, one of them is the start tile and is removed. I'll assume that in 1/3 of games you go last, and only get 23 tiles.</p>
<p>Without changing the statistics, we can pretend you draw all your tiles first, without any other players taking their turns. You can think of their turns as simply showing you what was never in the tiles you were going to get without affecting it, besides that you're stopping your draws early and not drawing the whole stack.</p>
<p>Drawing 23 tiles out of 71 without replacement, 57 of which are not Abbot-eligible, your chance of only drawing from those 57 is 57/71 * 56/70 * ... * 35/49. This comes out to about 0.200% (or about one in 499) of games where you go last.</p>
<p>Drawing a 24th tile multiplies this by 34/48, for about 0.142% (or about one in 705) in games where you don't go last.</p>
<p>Weighting the probabilities together based on your chance of actually going last it's 0.161% (or about one in 620).</p>
<p>Obviously changing the number of players, introducing river tiles, or caring about whether anyone at the table can't play an Abbot rather than just you would change the chances.</p>

CC BY-SA 4.0
2021-01-05T14:50:00.297Z
You could write it more compactly as `C(57,23) / C(71,23)` and `C(57,24) / C(71,24)`
L. Scott Johnson
CC BY-SA 4.0
2021-01-05T02:48:45.873Z
2021-01-05T02:48:45.873Z
Joseph Kendall
<p>I just played a frustrating 3 player game of Carcassonne with the Abbot expansion, where I didn't get a single opportunity to use my Abbot Meeple.</p>
<p>What are the odds a player doesn't draw a single tile usable by the Abbot in a 3 player game? Note there are 6 monastery tiles and 8 garden tiles for a total of 14 tiles the abbot can use (out of 72 total tiles).</p>

<carcassonne>
What are the odds a player doesn't draw a monastery or garden tile?
JCKE
# What are the odds a player doesn't draw a monastery or garden tile?

(Asked by: username_0 on 2021-01-04)

I just played a frustrating 3 player game of Carcassonne with the Abbot expansion, where I didn't get a single opportunity to use my Abbot Meeple.

What are the odds a player doesn't draw a single tile usable by the Abbot in a 3 player game? Note there are 6 monastery tiles and 8 garden tiles for a total of 14 tiles the abbot can use (out of 72 total tiles).

**Tags:** carcassonne


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## 1 answers

### Answer by username_1 on 2021-01-05. Score: 5

This should happen to you about 1 in every 620 games that have 3 players, 14 tiles that an Abbot can be played on, 1 start tile, and 57 other tiles.

Given that there are 72 tiles, in a 3-player game, you'd draw either 23 or 24 tiles, depending on turn order. Although the 72 tiles divide evenly, one of them is the start tile and is removed. I'll assume that in 1/3 of games you go last, and only get 23 tiles.

Without changing the statistics, we can pretend you draw all your tiles first, without any other players taking their turns. You can think of their turns as simply showing you what was never in the tiles you were going to get without affecting it, besides that you're stopping your draws early and not drawing the whole stack.

Drawing 23 tiles out of 71 without replacement, 57 of which are not Abbot-eligible, your chance of only drawing from those 57 is 57/71 \* 56/70 \* ... \* 35/49. This comes out to about 0.200% (or about one in 499) of games where you go last.

Drawing a 24th tile multiplies this by 34/48, for about 0.142% (or about one in 705) in games where you don't go last.

Weighting the probabilities together based on your chance of actually going last it's 0.161% (or about one in 620).

Obviously changing the number of players, introducing river tiles, or caring about whether anyone at the table can't play an Abbot rather than just you would change the chances.

> **Comments:**
> **username_2**: You could write it more compactly as `C(57,23) / C(71,23)` and `C(57,24) / C(71,24)`  

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