Rollie
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Posts posted by Rollie
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3 hours ago, CaptainEd said:
Rollie, your friend is "logical" and "competitive". But is your friend "greedy" as well?
Does your friend prefer making more money than you? or more money, period?
You suggested cooperating with your friend
what if you...
both bet a penny each time
my holdings friend holdings my bet friend bet $0.50 $0.50 $0.01 $0.01 $0.99 $0.99 $0.01 $0.01 $1.47 $1.47 $0.01 $0.01 $1.96 $1.96 $0.01 $0.01 $2.44 $2.44 $0.01 $0.01 $2.93 $2.93 I think BMAD wants to introduce us to the fact that there can be an outcome that is better for both players than the "logical", "competitive", "rational" outcome.
Thanks, BMAD!That goes back to my previous path of reasoning, which briefly described the $0.01 scenario (i.e., if my friend believes I will bet 0.01, he can bet 0.02 and win everything from then on without my cooperation). It depends if my friend is trustworthy, and also whether competitive means making as much more than me as possible, maximizing their own profit, or simply guaranteeing they have funds >= to mine. It also depends whether my friends thinks that I am trustworthy, because even if he's willing to do the 0.01/0.01 trade over and over, if he thinks I'm going to betray him, he will follow the same reasoning as I had above, leading to the previously mentioned 0.50/0/50 battle.
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But given that the prize is only cut in half if there is a tie, if during round one, I bet 0.50, and he bets 0.01, I get a prize of 0.50, leaving me with 0.50. He however forfeits his 0.01 bet, leaving him with 0.49. As such, I'm guaranteed to win the next bet as well, which will bring him down to 0.48, and leaving me with 0.50. At this point, I can start lowering my bid 0.01 each round, increasing my earnings each time.
So the results of the 0.50/0.01 betting scheme
My Money Friend's Money My Bet Friend's bet
0.5 0.5 0.5 0.01
0.5 0.49 0.5 0.01
0.5 0.48 0.49 0.01
0.52 0.47 0.48 0.01
0.56 0.46 0.47 0.01
0.62 0.45 0.46 0.01
(my win)
Results of 0.5/0.5 betting scheme (recalculated after looking yet again at the problem description more carefully)
My Money Friend's Money My Bet Friend's bet
0.5 <same> 0.5 <same>
0.25 <same> 0.25 <same>
0.37 <same> 0.37 <same>
0.31 <same> 0.31 <same>
0.34 <same> 0.34 <same>
0.33 <same> 0.33 <same>
0.33 <same> 0.33 <same> (repeats forever)
(tie)
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Ohh, misunderstood
In that case, I don't think there is a real answer for this. A game theory approach would say we both continuously bet $0.01 every round, ensuring we rack up $0.24 each per round. However, it's been stated my friend is competitive, so knowing this, he will want to win, not just draw for mutual benefit. So he will bet $0.02, to get all the money, and then just outbid me for every subsequent round. Knowing this, I could say I will then bet $0.03 instead, to which he would deduce he should bet $0.04. This continues until we both realize we should bet $0.50, a price at which we both realize we will lose money, but realize we will lose the competition if we ever bet less than our total savings; so .25 after one round, and then it would stay at .25 because we would continue to all-in for a payout of 0.25.
If my friend was even more cooperative and trustworthy, we would alternate games winning, with one person winning each round, maximizing mutual gain.
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Spoiler
I will bet $0.50 the first round. If my friend bets anything but $0.50, he will lose and I will win the rest by simply bettering $0.51, so he is guaranteed to also bet $0.50. Each round, we will both bet this, pay it, and receive $0.50 back, keeping our totals the same.
Number of kids
in New Logic/Math Puzzles
Posted
Answer already given, but some comments on how to find it.
Let the starting number be N. As Sushma noted,
N % 7 = 0
N-2 % 6 = 0
N-5 % 5 = 0
N-10 % 4 = 0
N-17 % 3 = 0
From N % 7 = 0, we know N is divisible by 7. From N-5 % 5 = 0, we also know it is divisible by 5, which means it's also divisible by 35 (the product of 7*5). This gives us only 21 numbers (35, 70, 105,...) below 750, but we can do better.
From N-2 % 6 = 0, we know N is even, giving us 10 numbers (70, 140, 210,...)
From N-10 % 4 = 0, and knowing all our numbers end in 0, we know the 10's place must be odd, giving us 5 numbers (70, 210, 350, 490, 630)
Since the number we are incrementing by at this point is not divisible by 3 (we are adding 140 each iteration), we can break them up into groups of three, test which one minus 17 is divisible by 3, and take that one for all other groups of 3. In our example, {70, 210, 350} is the first group, meaning only the 3rd number in each set of 3, after subtracting 17, will be divisible by 3 (350), changing our increment to 140 * 3 (420). This gives the equation from tojo where any number that fits in 350 + 420*k will match our criteria.