Two strong-man competitors, Atlas and Brutus, are smashing large rocks with great blows of a hammer. Their goal is to smash as many rocks as they can with four strikes. They each get one point for each crumbled rock.
An unstruck rock has a 20% chance of crumbling, a rock that has been struck once but didn't crumble has a 40% chance of crumbling on its second strike, a rock struck twice has a 80% chance of crumbling on its third strike, and a rock that survives 3 strikes is guaranteed to crumble on the fourth.
What is the probability that the final score is a tie?
Note: This problem is highly adapted from a regional math competition for students up to grade 9 and calculators were not allowed. The solution was requested to two decimal places.
I made a truth table for the 2^4 ways it can happen
0: FFFF 0
1: FFFT (4/5)(3/5)(1/5)1 = 12/125 = 60/625
1: FFTF (4/5)(3/5)(4/5)(4/5) = 192/625
1: FTFF (4/5)(2/5)(4/5)(3/5) = 96/625
1: TFFF (1/5)(4/5)(3/5)(1/5) = 12/625
2: FFTT (4/5)(3/5)(4/5)(1/5) = 48/625
2: FTFT (4/5)(2/5)(4/5)(2/5) = 64/625
2: TFFT (1/5)(4/5)(3/5)(4/5) = 48/625
2: FTTF (4/5)(2/5)(1/5)(4/5) = 32/625
2: TFTF (1/5)(4/5)(2/5)(4/5) = 32/625
2: TTFF (1/5)(1/5)(4/5)(3/5) = 12/625
3: FTTT (4/5)(2/5)(1/5)(1/5) = 8/625
3: TFTT (1/5)(4/5)(2/5)(1/5) = 8/625
3: TTFT (1/5)(1/5)(4/5)(2/5) = 8/625
3: TTTF (1/5)(1/5)(1/5)(4/5) = 4/625
4: TTTT (1/5)(1/5)(1/5)(1/5) = 1/625
Fortunately the sum is 625/625, so hopefully no mistake.
Note that assuming they started each with zero points, the probablity of getting equal points would be:
Both get 1 (360/625)*(360/625) = .331776
Both get 2 (236/625)*(236/625) = 0.14258176
Both get 3 (28/625)*(28/625) = 0.00200704
Both get 4 (1/625)*(1/625) = 0.00000256
The probability of a tie is the sum = 0.47636736
(1 - 0.47636736)/2 = 0.26181632 each player's probability of a win.
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Posted by Larry
on 2023-05-08 12:51:33 |
Solution
