# Stein's game expected value

Problem statement: https://brilliant.org/practice/steins-game/?p=5

Copying here for ease of reader:

Context:

 1 2 3 To play Stein's game with N cards, a dealer takes N cards numbered 1,2,…,n and shuffles them. Then, the player flips them over one by one, and they will be paid 1 dollar every time a new highest card shows up. (The first card always counts as a highest card, since you haven't seen any cards yet.) 

Cue:

 1 2 If you are given the choice to play Stein's game for free either with 3 cards or with 4 cards, which will give you more money in expectation? 

## My Question

Given the computation below (which I am assuming is correct).

Why didn't I have to compute P(win on 2nd and 3rd flip) or P(win on 1st, 2nd and 3d flip) and so on?

## 3 card game

P(win on 1st flip) = 1 (since this is first card)

P(win on 3rd flip) = 1/3 (card 3 has to be 3rd card)

P(win on 2nd flip) = P(1 on 1st position and either of {2,3} in 2nd) + P(3 on 1st position and 3 on 2nd) = 3/3! = 1/2

E(amount won) = 1 + 1/3 + 1/2 = 1.88

## 4 card game

P(win on 1st flip) = 1 (since this is first card)

P(win on 4th flip) = 1/4 (card 4 has to be 4th card)

P(win on 3rd flip) = P(cards in 1st 2 positions are {1,2} and cards in 3rd positing is either of {3,4})

= 2 (ways to permute {1,2} in 1st 2 positions) * 2 (getting either 3 or 4 in 3rd position) / 4!

= 1/6

P(win on 2nd flip) = P(1 on 1st position and either of {2,3,4} in 2nd) + P(2 on 1st position and either of {3,4} in 2nd) + P(3 on 1st position and either 4 on 2nd)

= 3/4! + 2/4! + 1/4! = 1/4

E(amount won) = 1 + 1/4 + 1/6 + 1/4 = 2.166

Note by shanxS Phone
3 weeks, 2 days ago

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