Monday, April 09, 2012

Liar's Dice

I played a game of Liar's Dice at Lounge Day which lasted a mere 2 turns for me. I lost 6 dice on my second action! I thought the bid was very reasonable to make at the time so the question I need to answer is if I merely got very unlucky or if I played badly in combination with getting unlucky.

I bid 11 2's with 4 2's in my own cup. There were 22 outstanding dice that I couldn't see and a 1 in 3 chance that each of those dice could be a 2 or a wild. The expected value from my point of view was a little more than 7 more 2's which when added to my own 4 made my bid of 11 2's conservative! It turned out Pounder on my left had none at all and challenged my bid. From his point of view there were also 22 outstanding dice so I'd likely overbid by 4. Even if you assign 4 of my dice to be 2's there are only 17 outstanding dice after those. This means there aren't even going to be 10 of them and challenging my bid is still pretty safe for him. Even if I have all 5 there's a good chance I'm going down. The chance that I'm bidding on air needs to be considered as well! So my first flaw is looking at it from my point of view and not from Pounder's likely point of view.

Of course, if he has a couple 2s he can't possibly challenge me. Even a single 2 means his challenge has to be based on putting me on a stone cold bluff instead of on a reasonable EV bid. Maybe he'd do that. Maybe he wouldn't. Heck, if he would do that I really like my bid since we were playing with an odd exacta rule which makes me really want to have a reasonable bid get challenged.

The actual result was that there was a single 2 between the 22 outstanding dice. So I lost 6 dice and was out of the game. That result felt astronomically unlikely and I wanted to take a look at the odds of that happening and the odds of getting reasonably called by Pounder at each of his possible number of 2s.

My first thought was to just build a spreadsheet like I did for Can't Stop but then I stopped and considered just how big a number 6^22 is. Yeah, not going to happen. But since each die roll is independent I should be able to make use of Markov Chains!

With 22 rolls the odds of getting 1 or fewer of a specific number is .16%. Pretty unlikely for sure, but not nearly as astronomical as I was thinking. The odds of there being 6 or fewer (and therefore my bid being a loser) is 36.2%. Those odds aren't too bad, especially with almost 15.6% being a 1 die loss and another almost 11% being a mere 2 die loss. Of course, this is still treating Pounder as a vacuum who doesn't take his own dice into account when he challenges...

First thing's first... What are the odds of each of Pounder's possible die rolls?

0 - 13%
1 - 33%
2 - 33%
3 - 16%
4 - 4%
5 - .4%

And with each of his possible results, what are the odds I have a loser if he challenges?

0 - 67%
1 - 48%
2 - 28%
3 - 13%
4 - 4%
5 - 1%

I'm a little surprised at the 1 number. Even if I have 4 matching dice under my cup if Pounder only has a singleton of my bid he'll be correct to challenge almost half the time. Depending on the psychology of the player in Pounder's spot this may or may not be ok. If they're willing to show the one die and re-roll the rest the bid is a big winner. If they're going to challenge then it isn't. And if I ever make big bluffs they'll be even more correct to challenge borderline bids like this one! Assuming Pounder will make the challenge at 0 and 1 and let the rest slide what is my total expected outcome?

53.4% - game keeps going as Pounder makes a bid
24.6% - I lose some dice
 6.3% - I win an exacta
15.7% - Pounder loses some dice

Considering this is a 6 player game and I have 18% of the dice it seems less than optimal to lose dice 24.6% of the time. Especially since I could very well lose dice in the 'game on' category as well. Playing it a little safer is probably a wiser thing to do. I like putting the screws to the person on my left (the idea being it likely won't get back around to me) but if the cost of putting the screws is to lose more than my fair share of the time it seems suboptimal. Consider what would happen if I'd bet only 10 2's... At this point Pounder has to bid on with anything except 0 2's...

86.4% - game keeps going as Pounder makes a bid
 6.2% - I lose some dice
 2.5% - I win an exacta
 4.9% - Pounder loses some dice

This is now a very safe bet. The odds of losing anything at all aren't very high. Even with 0 2's Pounder might even bid on anyway since there are lots of reasonably safe options for him if he has any set of numbers to bid himself. The only problem is I may well be put to a real test on my next turn around the table. Hopefully by then some dice will be revealed and the variance will be reduced!


David Nicholson said...

I think you are being a bit paranoid that if you bid 10 2s that it will be tight when it gets back to you. With 4 more bids you are likely in the outrageous range when it comes back to you.

pounder said...

My approach is to bid more conservatively when there are more people playing on the theory that bids won't get back to me, and then as we move to fewer players to start to stick the screws to lefty.

Sthenno said...

Bidding conservatively with more people also makes sense because with more people at the table risking your dice to try to make the person to your left lose dice is a bad risk. At the beginning of a six player game it only makes sense to risk one of your dice to take out one die of another player if you have a (rounded) 85% chance or better of winning.

Robb said...

Bidding more conservatively also makes sense in that you can hope a future player tries to bid to screw lefty :)

Also, you want to get challenged as rarely as possible - may as well give early lefty somewhere to bid to, rather then force them to challenge.