# Drawing a random number with chance of redrawing a second time. Best strategy that will never lose long term?

Q: We're going to play a game. You draw a random number uniformly between 0 and 1. If you like it, you can keep it. If you don't, you can have a do-over and re-draw, but then you have to keep that final result.

I do the same. You do not know whether I've re-drawn and I do not know whether you've re-drawn. Decisions are made independently. We compare our numbers and the highest one wins $1.

What strategy do you use?

I know the answer is 0.618..., tested this multiple times (0.5 will give the highest average value, but it won't guarantee you don't lose)

How do I come to this answer?

I was thinking I call the chosen threshold for A : "p", threshold for B: "q".

There is a (1-p)*(1-q) that they both take their first # : chance A wins = p-q+0.5

There is a (1-p)*q chance A takes his first #, B redraws a random # : chance A wins = 0.5p + 0.5

There is a p*(1-q) chance B takes his first #, A redraws a random # : chance A wins = 0.5-0.5q

There is a p*q chance they both redraw a #: chance A wins: 0.5

Total chance A wins:

(1-p)(1-q)(p-q+0.5) + (1-p)*q*(0.5p+0.5) + p*(1-q)*(0.5-0.5q)+0.5pq

I rewrote this and maximized this and after maximizing this, set p=q (equilibrium) and come out a different result. I ran tested above formula with some python programming and the formula is wrong. **What is wrong with my reasoning here and how would this be done correctly?** I've been trying to figure this out for a day. Thank you!

*individual probabilities for different possibilities seem to be correct based on testing in Python with 10 million runs each

## Answer

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Low bounty!

I'm asking what's wrong with my reasoning. Someone might see what I did wrong and could write it down in 1 sentence.

To see what you did wrong, one needs to first understand what you wrote here. The offered bounty is even low for just reading your long question and details.

I second that. Very low bounty!

"0.618" is not an answer to the question "What strategy do you use?" ... Just saying ...

e.g. in the 1st case, (1-p)(1-q), what do you mean by "chance that A wins: p-q+0.5 ? This is not a "chance", it can be larger than 1 and also negative. You have to take into account the possible values that A has drawn (in that case, randomly distributed in [p, 1]) and the same for B (randomly distributed in [q, 1]). Similarly for the other cases.

FWIW, probabilities that A wins in the 1st case [that happens with prob.(1-p)(1-q)], depending on p, q:

p\q 0.2 0.4 0.6 0.8

0.2 [1/2 3/8 1/4 1/8]

0.4 [5/8 1/2 1/3 1/6]

0.6 [3/4 2/3 1/2 1/4]

0.8 [7/8 5/6 3/4 1/2]

(replace

with line break if this gets scrambled)