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The game is as follows: Roll a fair six sided die until it lands on 1. Place bets beforehand on how many rolls the game will go before stopping.

I know that this is a geometric($\frac{1}{6}$) rv, so the EX = 6. I also know, that the $p(X = i) = (1 - p)^{i-1}p $ for $p = \frac{1}{6}$, so the most probable outcome is X = 1. So which should I bet on? 1 or 6?

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    $\begingroup$ It's impossible to provide adequate advice without knowing the payoffs. Perhaps you could let us know what they are? $\endgroup$
    – whuber ♦
    Commented Dec 12, 2015 at 21:26
  • $\begingroup$ ok, could you explain more how the payoffs alter the answer? A fixed payout is only given if a bet is correct, nothing is given to incorrect bets. and the game is only played once. $\endgroup$
    – fdzsfhaS
    Commented Dec 12, 2015 at 21:33
  • $\begingroup$ That's what we need to know. In many games--Roulette is a classic example--the payouts go up as the outcome gets less likely. They influence the betting enormously. All you need to do now to answer this question is figure out whether you will win more frequently (in the hypothetical long run) by betting on 6 or on 1. Oh, I see you already gave that answer... . $\endgroup$
    – whuber ♦
    Commented Dec 12, 2015 at 21:37

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I guess it falls to someone to summarize the information you provided.

Between your question and your comments you state the following facts:

  1. the bets being compared are to bet on the outcome being $1$ or bet on $6$

  2. the payout is the same in either case

  3. the most probable outcome is $X=1$

... so this apparently boils down to a question equivalent to one like

"Would you prefer a smaller chance of making a dollar or a bigger chance of making a dollar?"

I leave that remaining issue for you as a rhetorical question to ponder.

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