It's just on the edge of calculable (I think I know the answer), but -- while there's no doubt WHICH die to take, it's a little subtle as to which die you should capture it with. Test your intuition, then see if you can do the math. As long as you capture the obvious die, you'll have at least 90% chance of winning no matter which die you do it with-- but if you take it with the correct die, it shoots up to almost 96%.
Button Men Game #700184
Skills in this game: Focus, Reserve, V Swing
2 comments:
OK, I think you're implying the choice of the d4 or the d12 to take the d4 showing 3.
If I were in this situation, my first instinct would be to use my d4. Why? This means that if I lose my d4 (75%), I am left with a decent chance that my d12 will be able to take the two remaining d4s without loss.
I can see the other route, using the d12 initially, so that there is only a 25% chance of losing a die on the first go, leaving the d10 to take the other two dice without loss. A problem here is that the d12 might roll a 3, meaning that the two d4s can reroll to 4 and 4, resulting in certain loss. Though the probability of this occurring is very small, it increases the cognitive complexity (for me) of considering this option, and thus it ends up in the "too hard" basket.
OK, let's try the maths.
Scenario 1: use the d4 to take the d4 showing 3.
25% chance: my d4 rerolls to 1. opponent uses the low d4, leaving 2 and 1-4 showing. The d12 takes the higher die: 75% chance of a 2 remaining and 25% of a 1 remaining. Loss chance: (1/4) * ((3/4 * 2/12) + (1/4 * 1/12))
25% chance: my d4 rerolls to 2. opponent uses the high d4, leaving 1 and 1-4 showing. The d12 takes the higher die, leaving a 1 remaining. Loss chance:
(1/4) * (1/12)
25% chance: my d4 rerolls to 3. opponent uses both d4s, leaving 1-4 and 1-4 showing. The d12 takes the higher die. Loss chance:
(1/4) * ((7/16 * 1/12) + (5/16 * 2/12) + (3/16 * 3/12) + (1/16 * 4/12))
25% chance: my d4 rerolls to 4, ensuring a win. Loss chance = 0.
Adding up these probabilities, the total loss chance ends up at 9.64%.
Scenario 2: use the d12 to take the d4 showing 3.
1/12 chance: my d12 rerolls to 1. opponent uses the low d4, leaving 2 and 1-4 showing. The d10 takes the higher die: 75% chance of a 2 remaining and 25% of a 1 remaining. Loss chance: (1/12) * ((3/4 * 2/10) + (1/4 * 1/10))
1/12 chance: my d4 rerolls to 2. opponent uses the high d4, leaving 1 and 1-4 showing. The d10 takes the higher die, leaving a 1 remaining. Loss chance:
(1/12) * (1/10)
1/12 chance: my d12 rerolls to 3. opponent uses both d4s, leaving 1-4 and 1-4 showing. The d10 takes the higher die. Loss chance:
(1/12) * ((7/16 * 1/10) + (5/16 * 2/10) + (3/16 * 3/10) + (1/16 * 4/10))
9/12 chance: my d12 rerolls to 4 or more, ensuring a win. Loss chance = 0.
Adding up these probabilities, the total loss chance ends up at 3.85%.
Oh, I just wanted to add that it's obvious to me now why to use the d12 instead of the d4, but it only became obvious after I had done the maths. :)
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