raph.thefinanceguy
Region: US
Thursday 13 November 2025 17:19:45 GMT
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Arooooon12 :
The first is 3.5. If i can roll twice, i want the sum over k of the probability of earning k for 2\leq k \leq 12 times k, were the probability is the number of ways we can decompose k as a sum of 2 positive integers/36 (where order of addition matters). For example, for 3 we can write 3=2+1=1+2 and for 5 we get 3+2=2+3=1+4=4+1 so P(getting 3 dollars) = 2/36 and P(5) = 4/36 and so on. This gives E = 7.
2025-11-13 23:31:08
4
lucad :
4.25?
2025-11-13 18:50:48
3
SimSim :
How much i am WILLING to pay to roll dices ? I pay 1$ to be sure to at least break even, and in all other cases i get at least +1$. For rolling twice, i’d pay 2$ with the same logic
2025-11-14 10:49:56
1
Tristan Gaspar :
3.5 et 4.25
2025-11-13 23:18:37
2
Terminator :
3.5, 0
2025-11-14 06:48:29
0
47 :
Hey brother! Random question but what type of microphone are you using?
2025-11-23 04:36:23
0
sosoxedl :
3.5 et 5.25 ?
2025-11-13 19:59:12
1
Yi Chen Chong6 :
ppl should not be willing to pay 3.5 dollars to play this game... it should be strictly less than 3.5 (likewise for second game) unless u just enjoy risk
2025-11-15 04:04:33
0
Fastdam :
If you have a second chance => you reroll it only when you get lower then 3.5 because this would be the expected value of the second roll. Now it means that you roll it a second time only when the first roll is 1, 2 or 3 (50% of the time). Hence the expected value is 1/2*3.5 + 4/6 + 5/6+6/6=4,25
2025-12-06 02:19:55
0
Isha :
3,5 and 4,25
2025-11-14 08:19:04
0
chrispy :
Estimated value should be 3.50, therefore i would be willing to pay < 3.50 to make a profit. If i can roll twice then it should be < 7 if I understood correctly
2025-11-13 17:45:55
2
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