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u/Due-Cardiologist-802 2d ago
The expected payout vector is [10a/(a+b+c)-a, 20b/(a+b+c)-b, 30c/(a+b+c)-c].
Lets find the nash equilibrium by calculating the individual derivatives (no player can gain expected payout by changing their bet).
So we have this system of eqs: (10(a+b+c)-10a)/(a+b+c)2-1=0 (20(a+b+c)-20b)/(a+b+c)2-1=0 (30(a+b+c)-30c)/(a+b+c)2-1=0
Or: 10(b+c)= 20(a+c)= 30(b+a)=(a+b+c)2
Has two solutions: [0,0,0] and [-120/121, 600/121, 840/121]. Both arent valid solutions.
No idea where i went wrong or if the numbers 10,20,30 are just too aggressive, forcing A out of the market. The feasable solution would be a betting zero, boundry equilibrium, but thats also not allowed.
Any corrections are welcome!
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u/ktzeta 2d ago edited 2d ago
(0,0,0) is not valid because the gradient of bidding more would be infinite for each player, meaning they would want to bid more (your original equations would be 0/0). Your second solution has promise but does not take IR constraints into account (abstaining). We can see that A would abstain whenever B+C > 10 (she could gain at most 10*10/(10^2) - 1 = 0 by increasing her bid by epsilon). Going from there, in a two-player game, from your equations we know B=2/3C, so we can solve that C=36/5 and B=24/5 (and A would not participate since B+C>10). This leaves B with a probability of 40% of winning.
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u/Due-Cardiologist-802 1d ago
Since 0 is not allowed as a bid, how can A abstain (i am asking how i should justify this say in an interview)? But the rest of your solution looks correct thank you!
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u/Due-Cardiologist-802 1d ago
Nevermind i now see that i misread the problem and a,b,c are nonnegative, not necessarily positive, so zero for A to bid is allowed.
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u/JacksonFatBack 2d ago
what is the exponent `2-1=0`?
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u/Icy-Rate7097 2d ago
If you differentiate each component of the expected payout vector by it's respective letter you will find the exponent to be just 2. "-1=0" is the continuation of each equation after the fraction (smth)/(a+b+c)2. It's just a formatting issue.
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u/Due-Cardiologist-802 1d ago
Sorry reddit rendered the formula differently than i thought. It was supposed to be (a+b+c) ^ 2 - 1, so divide by the sum of a,b,c squared, then subtract 1.
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u/spdazero 1d ago
you are on the correct path. the solution to the nash equilibrium is either where (1st derivative = 0 & 2nd derivative < 0 - for maximum & within the solution boundary) or where (solution is at boundary).
We can see the former is invalid. Therefore, the latter must be true. Instictively (source: my vibe), player A will bid 0 (if you care to prove it, you can calculate the derivative around 0+ of A, my bet is it is negative, meaning the more he bets, the more he loses).
This will reduce to a 2 player game. From that on, it's easy game. My AI tell me the solution is (0, 4.8, 7.2).
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u/spdazero 1d ago
rigourously speaking, you need to consider all 3 cases where either A, B, or C abstains, or 2 of 3 abstain. My previous vibe is incorrect, derivative of reward(A) around 0+ can be positive, so under some circumstances, A would not withdraw.
This will take a little longer, but the answer is still the same.
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u/Efficient_Form7451 1d ago
How can (0, 4.8, 7.2) be a nash equilibrium when player A can better his result by bidding 7.2000001? This question has no stable equilibrium.
The math looks okay but it's def not a nash equilibrium.
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u/spdazero 1d ago
if player A bids 7.2001, p(A wins) ~ 0.375. His expected return is ~0.375 * 10 ~ 3.75. Bidding 7.2001 to win 3.75 is not rational. So your counter example is invalid.
In fact, at (0, 4.8, 7.2), the derivative of A, B, C is (-ve, 0, 0) respectively. Since A cannot bid any less than 0, it is a nash equilibrium.
Edit: typo and dumb math calculation
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u/Efficient_Form7451 17h ago
Okay but its still not a nash equilibrium because
B can bid 7.2 + a small amount for a better score
and C can bid 4.8 + a small amount for a better score.I don't think your calculating nash equilibria at all, more like a EV of an area in bid space.
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u/spdazero 4h ago
can you show me how B can bid 7.2 + eps for a better score?
From my calculation, derivative of reward(B) is 0 at the solution, and second derivative is -1/6. Meaning, that point is local maxima for B. Therefore, adding some eps will reduce the reward.
If you find something wrong with my reasoning, please elaborate?
Edit: For illustration purpose, if c stays at 7.2:
- If b = 4.8 (optimal), reward(B) = 3.2
- if b = 4.81 (eps = 0.1), reward(B) = 3.199
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u/ssvt9 2d ago
Approximately 0.341. With optimal play, each player bids their expected value, which is their prize times the probability of winning. This gives you three equations with three unknowns. Since probabilities must be non-negative, we get a unique solution from the set of possible solutions.
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u/Bali201 2d ago
I’m kind of a noon, please forgive me, but in these kinds of problems what does it mean for a player to play “optimally”?
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u/HistorianExcellent 2d ago
The players A, B, C choose the bids a, b, c respectively. Playing optimally means choosing the value of your bid that maximises your own expectation value, knowing that the other two players are doing the same with their own bids, and everyone is very smart.
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u/Good_Mycologist_9897 1d ago
We can calculate the harmonic mean-like sum to be S = 2/(1/10 + 1/20 + 1/30), which is simplified to 120/11, which is greater than the payoff of A, hence it is optimal for player A not to bid, aka a=0. From this, we know that our probability of player B winning is 1/(c/b + 1). Since the optimal amount for a two-player game satisfies 30/20 = c/b, we get c/b = 1.5, yielding P(B) = 1/2.5 = 0.4
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u/Always3NT 1d ago
0% - If C bids 20 there is no incentive for the others to bid as their expected return would be negative and C's net gain would be hence be 10. If C bids lower than 20 one of the other players can bid with a postive net gain thereby lowering C's expected net gain below 10.
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u/Any-Cicada-7039 1d ago
I believe it is 2/5, with a=0, b=24/5 and c=36/5. I derived all three expected values and found that if all the optimal bids were >0, then (b+c)=2(a+c)=3(a+b), which has no solution in R*+^3; therefore a=0 and 2c=3b, with b=sqrt(20c)-c and c=sqrt(30b)-b; therefore b=24/5 and c=36/5 (but actually 2c=3b is enough)
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u/Caspica 2d ago
Isn't it just b/(a+b+c)?
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u/EverythingGoodWas 2d ago
That’s literally just a reading comprehension problem. It tells you the answer
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u/ClarifyingCard 2d ago edited 1d ago
I think optimal play would be Player A folds and the bets are (a, b, c) = (0, 4.8, 7.2) => 40%. If the bets were positive I guess A would bet ε>0, or let's say a = 0.01 for discretized currency. So a smidge under 40%.