ROR question

[DSchles]

I thought the formula should be, bet size = kelly ratio * bankroll * edge / variance ? where kelly ratio = 1.0

No. I just made a similar post to Nyne's on another thread. The Kelly ratio isn't the operative factor. The important point is that the payoff for a winning bet is a hefty multiple of that for a losing one. For example, it's very easy to conceive of several different scenarios, all of which have the same e.v., but where that e.v. is attained in various ways: win less than you lose, but win VERY frequently; even money payouts where you win more often than you lose; or longshot payouts, where you win much less frequently than you lose, but your win is a large multiple of the loss amount.

As you must understand, the Kelly-optimal wagers in these three circumstances are all quite different, with the first being the highest, and the last being the lowest. And it's easy to understand why: the latter scenario is much more risky than the former, with very frequent losses; so you need to be cautious with your bet size. In those situations, the PRECISELY correct Kelly wager is (bank x edge)/ratio of winning bet to losing one.

Don

[James989]

No. I just made a similar post to Nyne's on another thread. The Kelly ratio isn't the operative factor. The important point is that the payoff for a winning bet is a hefty multiple of that for a losing one. For example, it's very easy to conceive of several different scenarios, all of which have the same e.v., but where that e.v. is attained in various ways: win less than you lose, but win VERY frequently; even money payouts where you win more often than you lose; or longshot payouts, where you win much less frequently than you lose, but your win is a large multiple of the loss amount.

As you must understand, the Kelly-optimal wagers in these three circumstances are all quite different, with the first being the highest, and the last being the lowest. And it's easy to understand why: the latter scenario is much more risky than the former, with very frequent losses; so you need to be cautious with your bet size. In those situations, the PRECISELY correct Kelly wager is (bank x edge)/ratio of winning bet to losing one.

Don

Don, I learn something new from you again! Thanks.

For my specific case above(I think it is third circumstance), nos of winning bet = 3(1 pay 12.5 units), nos of losing bet = 34(loss 1 unit), ev = (3*12.5 - 34)/37 = 0.0946.

Nyne proposed bet size = bankroll * edge / payoff_odds, 1 = bankroll * 0.0946/12.5, bankroll = 132.14

Your Kelly wager = (bank x edge)/ratio of winning bet to losing one, 1 = bankroll * 0.0946/(34/3), bankroll = 119.80

Which one is correct ? I hope I am not too troublesome, I just want to learn the correct way to calculate the required bankroll.

Thanks in advance.

[DSchles]

Don, I learn something new from you again! Thanks.

For my specific case above(I think it is third circumstance), nos of winning bet = 3(1 pay 12.5 units), nos of losing bet = 34(loss 1 unit), ev = (3*12.5 - 34)/37 = 0.0946.

Nyne proposed bet size = bankroll * edge / payoff_odds, 1 = bankroll * 0.0946/12.5, bankroll = 132.14

Your Kelly wager = (bank x edge)/ratio of winning bet to losing one, 1 = bankroll * 0.0946/(34/3), bankroll = 119.80

Which one is correct ? I hope I am not too troublesome, I just want to learn the correct way to calculate the required bankroll.

Thanks in advance.

No, you misinterpreted what I wrote. The denominator is the ratio of the AMOUNT of a winning payoff (12.5) to the amount of a losing bet (one unit). The 34 and the 3 have nothing to do with the denominator. They're used to calculate the e.v.

Don

[James989]

Thanks Don.